Solutions of
Electricity & Magnetism
Lesson 20th to 25th
By DC Pandey
20
Current Electricity Introductory Exercise 20.1
2 pr q 1. i = , here q = e, t = t v ev i= \ 2pr 1.6 ´ 1019 ´ 2.2 ´ 106 = 2 ´ 3.14 ´ 5 ´ 1011 = 1.12 ´ 10 = 1.12 mA
3
=
2.0 8.43 ´ 1028 ´ 1.6 ´ 1019 ´ 3.14 ´ (0.5 ´ 103 )2
= 1.88 ´ 106 ms 1 3. Yes. As current always flows in the direction of electric field.
A
2. No. of atoms in 63.45 g of Cu = 6.023 ´ 1023 \No. of atoms in 1 cm 3 (8.89 g) of Cu 6.023 ´ 1023 = ´ 8.89 63.54 = 8.43 ´ 1022 As one conduction electron is present per atoms, n = 8.43 ´ 1022 cm 3 or 8.43 ´ 1028 m 3 As i = neAvd i vd = Þ neA
4. False. In the absence of potential difference, electrons passes random motion. 5. Current due to both positive and negative ions is from left to right, hence, there is a net current from left to right. dq 6. i = 10 + 4 t Þ = 10 + 4 t dt q
10
ò0 dq = ò0
Þ
(10 + 4t ) dt
q = [10t + 2t 2 ]10 0 = 300 C
Þ
Introductory Exercise 20.2 rL 1. R = A
=
= 1.72 ´ 108 ´
35 æ 2.05 ö ´ 103 ÷ 3.14 ´ ç è 2 ø
= 0.57 W 2. (a)
J= Þ
E r
i = JA =
EA r
2
0.49 ´ 3.14 ´ (0.42 ´ 103 )2 2.75 ´ 108
= 9.87 A (b) V = EL = 0.49 ´ 12 = 5.88 V V 5.88 (c) R = = = 0.6 W i 9.87 3. Let us consider the conductor to be made up of a number of elementary discs. The conductor is supposed to be extended to form a complete cone and the vertex O of the cone
3 is taken as origin with the conductor placed along xaxis with its two ends at x = r and x = l + r. Let q be the semivertical angle of the cone. Consider an elementary disc of thickness dx at a distance x from origin. Resistance of this disc, dx dR = r A If y be the radius of this disc, then A = py2 But y = x tan q dx dR = r 2 px tan 2 q \Resistance of conductor l +r rdx R = ò dR = ò 2 r px tan 2 q l +r r é 1ù R= p tan 2 q êë x úû r
=
é1 r 1 ù ê ú 2 p tan q ë r l + r û
R= But,
rl pr ( l + r ) tan 2 q
r tan q = a ( r + l) tan q = b rl R= p ab
4. True.
r=
\
1 s
r´ s=
1 ´ s =1 s
5. RCu = RFe 4.1 (1 + a Cu DT ) = 3.9 (1 + a Fe DT ) 4.1 [1 + 4.0 ´ 103( T  20)] = 3.9 [1 + 5.0 ´ 103( T  20)] 4.1 + 16.4 ´ 103( T  20) = 3.9 + 19.5 ´ 103( T  20) 3.1 ´ 103( T  20) = 0.2 0.2 T  20 = 3.1 ´ 103 = 64.5° C T = 84.5° C
Introductory Exercise 20.3 1. Potential difference across both the resistors is 10 V. i1
VB = VA + 2 = 2 V VC = VA + 5 = 5 V VD = VC + 10 = 15 V V  VB i1 = C =3 A 1 V  VA 15 i2 = D = = 7.5 A 2 2
i2
\ 2W
10 V
4W
3. Current in the given loop is E + 15 i= 8 æ E + 15 ö VAB = E  2i = E  2 × ç ÷ =0 è 8 ø
10 =5 A 2 10 i2 = = 2.5 A 4
Hence,
i1 =
and
1W
i1
E =5 V
Þ
2. As A is grounded, VA = 0 C
and
4. Effective emf, E = 8 ´1  2 ´1 = 6 V Effective resistance of circuit
B
5V 10 V
R1
2V
i1
i2 D
i2
A 2W
10 V
i3 R2
R3
10 V
4 R = Rexternal + 10r = 2 + 10 ´ 1 = 12 W E 6 i= = = 0.5 A R 12
\
5. As R2 = R3 and V1 = V2 Potential difference across R1 is zero. Hence, current through R1 Þ i1 = 0 and current through R2 V Þ i2 = 1 R2 10 = =1 A 10
E 6. i = R+ r Also, V = E  ir i= (a)
I
EV r I
(b)
E r
E r
O
R
O
E
Introductory Exercise 20.4 6 =1  E E = 5V
1. 12 V
r
P E
1W
U
i
Q
1A R
3W T
2A
S
Applying KCL at junction R i=1 + 2 =3 A VST = VRU = VQP Taking VST = VRU
And from VST = VQP 6 =  ir + 12 12  6 6 r= = =2W i 3 2. Power delivered by the 12 V power supply, P1 = Vi = 12 ´ 3 = 36 W and power dissipated in 3 W resister, P3 = i32 R3 = 22 ´ 3 = 12 W
Introductory Exercise 20.5 E1 E2 E3 10 4 6 + + + + r r2 r3 2 2 1. E = 1 = 1 1 1 1 1 1 1 + + + + 1 2 2 r1 r2 r3 10 + 2 + 3 = 2 = 7.5 V 1 1 1 1 and = + + r r1 r2 r3 1 1 1 = + + =2 1 2 2 1 Þ r= 2 = 0.5 W E 2. i = R+ r
r
E
R
Rate of dissipation of energy E2 R P = i2 R = ( R + r )2 For maximum or minimum power dP =0 dR é ( R + r )2  2 R ( R + r ) ù Þ E2 ê ú =0 ( R + r )4 ë û
V
5 Þ
E2
Þ
( R + r )( r  R) ( R + r )4 E 2 ( r  R) ( R + r )3
=0 =0
Þ R=r 3 é ( + ) ( 1 ) 3( r  R)( R + r )2 ù R r d 2P = E2 ê ú 2 ( R + r )6 dR ë û  E2(4r  2 R) = ( R + r )4 d 2P is negative at R = r. dR2 Hence, P is maximum at R = r E2r E2 and Pmax = = 2 4r ( r + r) Clearly
3. When the batteries are connected in series Eeff = 2E = 4V, reff = 2r = 2 W For maximum power R = reff = 2W and Pmax =
2 Eeff (4)2 = =2 W 4reff 4 ´ 2
4. I g = 5 mA, G = 1 W, V = 5 V V 5 R= G = 1 Ig 5 ´ 103 = 999 W A 999 W resistance must be connected in series with the galvanometer.
5. G = 100 W, ig = 50 mA, i = 5 mA igG 50 ´ 106 ´ 100 S= = \ i  ig 5 ´ 103  50 ´ 106 1 1 = = 1  0.01 0.99 100 = W 99 100 By connecting a shunt resistance of W. 99 V 6. ig = G nV and R =  G = ( n  1) G ig 15 E 16 Potential gradient V 15E k = AB = L 16 ´ 600 E V/cm = 640 E E (a) = kL Þ L = = 320 cm 2 2k E 7E (b) V = kl = ´ 560 = 640 8 Also, V = E  ir 7E E  ir = \ 8 E i= 8r
7. VAB =
AIEEE Corner Subjective Questions (Level 1) q ne 1. i = = t t Given, i = 0.7 , t = 1 s, e = 1.6 ´ 1019 C 0.7 ´ 1 it n= = \ e 1.6 ´ 1019 = 4.375 ´ 108 2. q = it = 3.6 ´ 3 ´ 3600 = 38880 C 3. (a) q = it = 7.5 ´ 45 = 337.5 C q (b) q = ne Þ n = e
= 4. T =
337.5 = 2.11 ´ 1021 1.6 ´ 1019
2 pr 1 v Þf = = v T 2 pr =
2.2 ´ 106 2 ´ 3.14 ´ 5.3 ´ 1011
= 6.6 ´ 1019 s 1 q I= = ef T = 1.6 ´ 1019 ´ 6.6 ´ 1019 = 10.56 A
6 8
2
= 1.7 ´ 10 W  m l = 24.0 m 2 2 æ dö æ 2.05 ö A = p ç ÷ = 3.14 ´ ç ´ 103 ÷ è2ø è 2 ø
5. (a) I = 55  0.65 t dq I= dt Þ dq = Idt q = ò I dt Þ
= 3.29 ´ 106 m 2
8
8
0
0
2
R = 1.7 ´ 108 ´
q = ò Idt = ò (55  0.65 t ) dt
\
8
é t2 ù = 55 [ t ]80  0.65 ê ú ë 2 û0 = 440  20.8 = 419.2 C (b) If current is constant q 419.2 I= = = 52.4 A t 8 6. i µ vd \
vd 2
Þ
vd 2 =
vd1
=
= 6.00 ´ 104 ms 1 7.
vd = =
= 0.12 W L 10. R=r A rL A= R If D is density, then D r L2 R 8.9 ´ 103 ´ 1.72 ´ 108 ´ (3.5)2 = 0.125 2 = 1.5 ´ 10 kg = 15 g
m = DV = DA L =
i2 i1 6 . 00 i2 vd1 = ´ 1.20 ´ 104 i1 1 . 20
i neA 1 8.5 ´ 1028 ´ 1.6 ´ 1019 ´ 1 ´ 104
= 0.735 ´ 106 ms 1 = 0.735 mm/s l 103 t= = vd 0.735 ´ 106 = 1.36 ´ 109 s = 43 yr 8. Distance covered by one electron in 1 s = 1 ´ 0.05 = 0.05 cm Number of electrons in 1 cm of wire = 2 ´ 1021 \ Number of electrons crossing a given area per second = Number of electrons in 0.05 cm of wire = 0.05 ´ 2 ´ 1021 = 1020 q ne i= = t t 1020 ´ 1.6 ´ 1019 = = 1.6 ´ 10 = 16 A 1 L 9. R = r A Given, r = 0.017 mW  m
24.0 3.29 ´ 106
11. At 20°C, R1 = 600 W, R2 = 300 W At 50°C, R1¢ = R1(1 + a 1Dt ) = 600 (1 + 0.001 ´ 30) = 600 ´ 1.03 = 618 W R2¢ = R2(1 + a 2Dt ) = 300 (1 + 0.004 ´ 30) = 336 R¢ = R1 ¢ + R2 ¢ = 618 + 336 \ = 954 W R¢  R 954  900 a= = R ´ Dt 900 ´ 30 R = 600 + 300 = 900 W = 0.002 ° C 1 12. As both the wires are connected in parallel, VAl = VCu iAl RAl = iCu RCu L L iAl r Al Al2 = iCur Cu Cu 2 p dAl p dCu Þ
dCu = dAl
iCu r Cu LCu iAl r Al LAl
= 1 ´ 103
2 ´ 0.017 ´ 6 3 ´ 0.028 ´ 7.5
= 0.569 ´ 103 m = 0.569 mm.
7
and
17. The circuit can be redrawn as
Current density is maximum when L is minimum, ie, L = d, potential difference should be applied to faces with dimensions 2d ´ 3 d. V . J min. = rd V VA (b) i = = R rL Current is maximum when L is minimum and A is maximum. Hence, in this case also, V should be applied to faces with dimensions 2d ´ 3 d V (2d ´ 3 d ) 6Vd and imax = . = r ( d) r 15. (a) R = r
0.104 ´ 3.14 ´ (1.25 ´ 103 )2 14.0 = 3.64 ´ 107 W  m V EL 1.28 ´ 14 (b) i = = = = 172.3 A R R 0.104 (c) i = neAvd i vd = neA 172.3 = 8.5 ´ 1028 ´ 1.6 ´ 1019 ´ 3.14 ´ (1.25 ´ 103 )2 = 2.58 ´ 103 ms 1
8 ´ 12 = 4.8 W 12 + 8 V 24 I= = =5A Reff 4.8
18. Here, A and C are at same potential and B and D are at same potential, 8W
A
24V
R1 = 10 R2 Also, R1 + R2 = 20 Þ 10 R2 + R2 = 20
B 6W
4W D
C
12W
Hence, the circuit can be redrawn as A,C
=
16. For zero thermal coefficient of resistance, DR = 0 RC a CDT + RFe a Fe DT = 0 R1  a Fe  5.0 ´ 103 = = = 10 R2 aC  0.5 ´ 103
8W
Reff =
L A
RA r= L d [r = = 1.25 mm = 1.25 ´ 103 m] 2
12W
24 V
24V
4W
12W
r = 2.84 ´ 108 W  m E V 14. (a) J = = r rL
20 W = 1.82 W 11 R1 = 20  R2 = 20  1.82 W = 18.2 W R2 =
Þ
8W
V 0.938 = = 1.25 V/m L 75 ´ 102 E 1.25 (b) J = Þ r = r 4.4 ´ 107
13. (a) E =
6W
B,D
\
1 1 1 1 1 = + + + R 4 8 12 6 6+3 +2+4 = 24 15 5 = = 24 8 8 R= W 5 = 1.6 W V 24 i= = R 1.6 = 15 A
19. Given circuit is similar to that in previous question but 4 W resistor is removed. So the effective circuit is given by
8 V 12 i= = R 36 / 13 13 = A 3
12W
A,C
8W
24 V
6W
21. (a) i =
12 + 6 =3 A 1+2+3 i = 3A
B,D
1 1 1 1 = + + R 8 12 6 1 3+2+4 9 3 = = = R 24 24 8 8 R = W = 2.67 W 3 V 24 i= = =9 A R 2.67 20.
6W
A
B
4W 12W
12V
6W 3W D
C
2W
ßA 6W
4W 4W
12
3W
12W
C
B
2W
A 1W
12 V
B 2W
G
C 3W
6V
D
VG = 0 VA = VG + 12 = 12 V VA  VB = 3 V VB = 12  3 = 9 V VB  VC = 6 V VC = 9  6 = 3 V VG  VD = 6 V, VD =  6 V (b) If 6 V battery is reversed 12  6 i= =1A 1+2+3 i = 1A
D
Wheatstone bridge is balanced, hence 4 W resistance connected between B and C be removed and the effective circuit becomes
12 V B G
A 6W 12V
C
4W
3W
6V
12W
C
B
2W D ß
A
D
VG = 0,
A
VA  vG = 12 V, VA = 12 V 12V
9W
6W
12W
VA  VB = 1 V Þ
Þ 12V
VB = 11 V VB  VC = 2 V
ß
36 W 13
VC = 9 V VD  VG = 6 V
Þ
VD = 6 V
9 200 =5 A 5 + 10 + 25
22. i =
= 42.26 W E i= = 0.102 W Reff
1
Reading of voltmeter V = E  ir = 4.3  0.102 ´ 1 » 4.2 W Reading of Ammeter, V 4.2 i1 = = = 0.08 A R + Ra 42
5W
200 V
2 10W 0 25W 3
(i) V3  V0 = 5 ´ 25 Þ V3 = 125 V (ii) V0  V2 = 5 ´ 10 V2 =  50 V (iii) V2  V1 = 5 ´ 5 V1 =  75 V (iv) V3  2 = 5 ´ 35 = 175 V (v) V1  2 =  5 ´ 5 =  25 V (vi) V1  3 =  200 V
24. Consider the directions of current as shown in figure. 42V
5W
I1
4W
A
I2
6W
10V
8W B
C
1W
I3
23. (a)
6W i
E
i1
R
D
r
4V
Applying KVL in loop 1, 2 and 3, we respectively get, I1 + 6 ( I1  I2 ) + 5I1 = 42 Þ 12I1  6I2 = 42 …(i) Þ 2I1  I2 = 7 4I2 + 6 ( I2  I1 ) + 8 ( I2 + I3 ) = 10 …(ii) Þ 9I2  3 I1 + 4I3 = 5 8 ( I2 + I3 ) + 16I3 = 4 …(iii) 2 I2 + 6 I3 = 1 On solving, we get, I1 = 4.7 A, I2 = 2.4 A, I3 = 0.5 A
A
i2
S V
Reff = R Rv + Ra + r 50 ´ 200 = +2+1 50 + 200 = 43 W E 4.3 i= = = 0.1 A Reff 43 \Reading of ammeter, i = 0.1 A and reading of voltmeter = i ( R Rv ) = 0.1 ´ 40 = 4 V E r (b) i i1
Resistor 5W 1 W 4W 6W Current 4.7 A 4.7 A 2.4 A 2.3 A R = 400W
25.
V
R A
i2
100W 100W i2
Reff = ( Ra + R) Rv + r 52 ´ 200 = +1 52 + 200
200W
400W 200W
Þ A 100W
100W D B 200W 100W
i2
V
8W 16 W 2.9 A 0.5 A
100W i1
i i
10 V
i1 i
10 V
C
10 As Wheatstone bridge is balanced, 100 W resistance between B and D can be removed, ie, 100W
A i2
B
200W
R2¢ =
D 10 V
10 1 = A 300 30 Hence, reading of voltmeter = Potential difference between B and C 20 V = 200 ´ i2 = 3 = 6.67 V i1 = i2 =
R2 RV2 R2 + RV2
E i
i1
i2 V1
i2 V2
V1 =
V2 =
R1 RV1 + RV2 RV2 RV1 + RV2
E=
= 120 V 2000 E= ´ 200 5000
= i2 =
= 80 V
R1
V2 S
R2
i1
1 A 20
R1 2000 1 ´ i= 5000 12 R1 + RV1
1 A 30 \Current flowing through 1 1 S = i1  i2 = 20 30 1 = A 60 =
E
V1
S
E R1 ¢ + R2 ¢ 200 1 i= = A 2400 12 RV1 3000 1 i1 = ´ i= 5000 12 R1 + RV1
3000 ´ 200 5000
(ii) When S is closed,
R1
V2
i=
R2
S
6000 = 1200 W 5
(b) Current distribution is shown in figure
E
R1
=
As R1 ¢ = R2 ¢ Hence, reading of V1 = reading of V2 1200 = ´ 200 = 100 V 1200 + 1200
26. (a) (i) When S is open.
V1
R2'
R1'
C
200W
100W i1
\
E
R2
Now, R1 and V1 are in parallel and their effective resistance R1 RV1 6000 R1 ¢ = = = 1200 W R1 + RV1 5 Similarly, R2 and V2 are in parallel with their effective resistance,
27. Effective emf of 2 V and 6 V batteries connected in parallel E r + E2r1 2 ´ 1  6 ´ 1 E¢ = 1 2 = r1 + r2 1+1 and
= 2V rr 1 r¢ = 1 2 = W r1 + r2 2 = 0.5 W
11 2V
1W
6V 0.5W
ß
4V
0.5W
2V
A
33. In case of charging V = E + i r = 2 + 5 ´ 0.1 = 2.5 V 34. Clearly current through each branch is zero.
Net emf, E = 4  2 = 2 V 28. (a)
2W B 4W 8W 8W Þ 4W
2W – +
E1
– +
E2
As E1 > E2 Current will flow from B to A. (b) E1 is doing positive work (c) As current flows from B to A through resistor, B is at higher potential.
2V
35. i1 =
G i1 E
On shunting resistance S,
galvanometer S G
E
R + R¢ + As i1 = i2
GS S+G
E E = R + G R + R¢ + GS G+S GS R + R¢ + =R+G G+S G2 R¢ = G+S
Þ
P
31. (a) As voltmeter is ideal, it has infinite resistance, therefore current is zero. (b) V = E  ir Þ E = 5.0 V (c) Reading of voltmeter Þ V = 5.0 V
E
i2 =
– +
VP  VQ = 50 + 3.0 i VQ = 100  (50 + 60) =  10 V
R'
i2
3.0W
50 V 2.0W
the
R
P 0.5 (a) P = Vi Þ V = = = 5.0 V i 1.0 (b) E = V  iR = 5  2 = 3 V (c) It is clear from figure that positive terminal of X is towards left. 150  50 30. i = = 20 A 3+2
150 V
2V
E R+G
R=2.0W E B
Q
2V
4W
R
i
A
2V
4W
2V
2V
29. i2 R = 2 W < 5 W Clearly X is doing negative work.
– + i
…(i) …(ii)
32. V1 = E  i1r Þ E  1.5 r = 8.4 V2 = E + i2r Þ E + 3.5 r = 9.4 On solving, we get r = 0.2 W E = 8.7 V
1W
4V 0.5W
36. I
A
A I1 I2
I2 =
r V I= R+ r R
r V
B
with
12 Þ Þ
r V = R + r IR R IR  V 5 ´ 2500  100 = = r V 100 100 r= ´ 2500 = 20.16 W 12400
39. S =
ig i  ig
(G + R) S
R G
A
37. R=
60 V
300W
400W V
Let R be the resistance of voltmeter As reading of voltmeter is 30 V, 1 1 1 + = Þ R = 1200 W R 400 300 If voltmeter is connected across 300 W resistor,
B
i  ig ig
S G
20  103
´ 0.005  20 103 = 79.995 W L1  L2 0.52  0.4 40. r = R= ´ 5 = 1.5 W L2 0.4 =
41. Let R be the resistance of voltmeter V A
100W
i 3W
60 V 3.4 V 300W V
Effective resistance of 300 W resistor and voltmeter 300 ´ 1200 R¢ = = 240 W 300 + 1200 60 i= 400 + 240 60 = A 640 3 = A 32 \Reading of voltmeter, 3 V = iR ¢ = ´ 240 32 = 22.5 V R¢ 38. V2 = V, R1 + R2 ¢
R2 ¢ =
V2 =
rR2 120 = r + R2 3
= 40 W 40 120 60 + 40
= 48 V
100 R 100 + R 100 R =5 + 100 + R 3.4 i= = 0.04 100 R 5+ 100 + R 4R 0.2 + = 3.4 100 + R
Re = 3 + 2 +
400W
Þ
Þ R = 400 W Reading of voltmeter, 100 ´ 400 100 R V =i´ = 0.04 ´ 100 + R 100 + 400 = 3.2 V If the voltmeter had been ideal, Reading of voltmeter 100 = ´ 3.4 = 3.24 V 105 L R 42. 1 = 1 L2 R2 L1 8 (L1 + L2 = 40 cm) Þ = 40  L1 12 Þ
L1 = 16 cm
from A.
13 43. S =
ig i  ig
Maximum power dissipated by the circuit 2 P ¢max = Imax Re 3 = 15 ´ ´ 2.4 = 54 W 2
(G + R) R=
Þ
i  ig ig
S G
20  0.0224 ´ 0.0250  9.36 0.0244 = 12.94 W E 44. (a) i = RV + r =
Rv V i r
47. Total power of the circuit, P = P1 + P2 + P3 = 40 + 60 + 75 = 175 W V2 V2 As P = Þ R= R P (120)2 = = 82.3 W 175 48. Thermal power generated in the battery R
E
V = iRV = (b)
(c)
E RV RV + r
i r
E
r 1 = 1% = RV + r 100
P1 = i2r = i ( E  V ) = 0.6 W Power development in the battery by electric forces P2 = IE = 2.6 W
RV = 99r = 99 ´ 0.45 = 44.55 W RV V = E RV + r
As RV decreases, V decreases, decreasing accuracy of voltmeter. 45. (a) When ammeter is connected E IA = RA + R + r When ammeter is removed R + R+ r E = A I= IA R+ r R+ r I (b) A = 99% I R+ r 99 = RA + R + r 100 1 1 RA = ( R + r) = (3.8 + 0.45) Þ 99 99 RA = 0.043 W R+ r IA (c) As , as RA increases, I A = I RA + R + r decreases, decreasing the accuracy of ammeter. r max 36 46. Imax = = = 15 A R 2.4 For the given circuit 1 3 Re = R + R = R 2 2
49. The given circuit can be considered as the sum of the circuit as shown. 2W 35 A 16
2W
21/6A 2W
2/16A
14/6A
+
3W
7V
5/16A
3W 1V
2W 2A
3 A 16
ß
2W 1A 3W
1V
7V
P1 = 7 ´ 2 = 14 W, P2 =  1 ´ 1 = 1 W E1  E2 12  6 50. (a) i = = = 0.5 A R1 + R2 4 + 8 \
(b) Power dissipated in R1 = I 2 R1 = 1 W and power dissipated in R = I 2 R2 = 2 W (c) Power of battery E1 = E1I = 12 ´ 0.5 = 6 W (supplied) Power of battery E2 = E2I =  6 ´ 0.6 =  3 W (absorbed)
14 E 12 51. I = = =2A R+ r 5+1
1 1 1 1 1 54. (a) = + + + R 4 6 14 4 4W
(a) P = EI = 12 ´ 2 = 24 W (b) P1 = I 2 R = 22 ´ 5 = 20 W (c) P2 = I 2r = 22 ´ 1 = 4 W
4W 4W 6W
8W
ÞA
2W
A
4W
8W
6W
B
4W
52. (a)
2W
4W
ß
1.60W I1 I3
I
B
4W
I2 2.40W
6W
4.80W
A
B
14W
4W
1 31 = R 42 42 R= 31 1 1 1 1 (b) = + + Re 2 R 2 R R
28.0V
1 1 1 1 = + + R R1 R2 R3 1 1 1 1 = + + R 1.60 2.40 4.80 R = 0.80 W V 28.0 (b) I1 = = = 17.5 A R1 1.60 V 28.0 = = 11.67 A I2 = R2 2.40 V 28.0 = = 5.83 A I3 = R3 4.80
R A
R
R
B R
ß 2R
(c) I = I1 + I2 + I3 = 35.0 A
P2 =
V 2 (28)2 = = 326.7 W R2 2.4
V 2 (28)2 P3 = = = 163.3 W R3 4.8 V2 R Resistor with least resistance will dissipate maximum power. V2 53. (a) P = Þ V = PR R = 5 ´ 15 ´ 103 = 2.74 ´ 102 (f) As, P =
= 274 V V (120)2 (b) P = = = 1.6 W R 9 ´ 103 2
R
R
2R
A
(d) As all the resistance connected in parallel, voltage across each resistor is 28.0 V. V 2 (28)2 (e) P1 = = = 490 W R1 1.6
R R
B A
R
R
R
Þ
B
R
Wheatstone bridge is balanced R Þ Re = 2 i2+ i3 4W 2W
(c) i1+ i2 A
i2
i3
i3 3W
i1
1W
2W
3W
i1 i + i 1 2 B i1
1W
ß 4W 2W
A
3W
1W
3W
1W
2W B
15 ß
2.4W 2W
Þ
2W
6W
A
B
2W ß 6.4
2W
i3– i4 (d)
i4
2W
10W
2W
B
2W
4W
5W
11W
B
i2
R2 R3 R1 + R2 + R3 R1 R3 RB = R1 + R2 + R3
10W
10W Þ 10W 5W B
10W 10W
A 10W
10W
RC =
R
R
B
R
R
R
R
R
Þ R
R
R
A 10W
R
R
R
R
R
R
(f)
25W B
R2 R3 R1 + R2 + R3
R
R
10W
ß 5W
A
10W
RA =
i 1 + i 2 + i3 B
ß
A
B
O 1W
By StarDelta Method 10W
5W 5W
5W
ÞA
8W
B
i1 10W
10W
C
1 W 1W 2
D
A
i3
i4 5W
10W
10W D
ß 5.8W
5W
i2
10W
B
2W 4W
B
A
8W
B Þ A
8.5R
1.52W
10W
5W
i3
i1 + i2 + i3 A
(e) A
B
2W
C
8W 2W
6W
A
Þ A A
2W 4W
R
R
R
B
A B
ß
ß 2R
R
10W
R
5W A
ß 4.17W A
R
R
B
2R
Þ
5W
R
R A
B
B
2R
A
ß 3R Þ A
2R
B
A
6 W 5
2R
B
B
As circuit is symmetrical about perpendicular bisector of AB, lying on it are at same potential.
16 F
D
B
A
B
2i1+i2 i1 1W i1 2W
i1 B
2W
E C
F
R
1W
A
R2 3 R = = 0.6 R1 5 R
Þ
R2 = 0.6 R1
B
2W
Þ 5W
5W
1W
56. F
D
A 8W
2W
15W 6W
40W E ß
B
1W
15W
ß 0.71W A
B
2W
A
Clearly C and D, E and F are at same potential. B
15W
20W 30W
2W
1W
R
R
\
1W
2W
D
C,E
15W 8W
6W
40W
20W
30W D,F
30W 46W
2W
8W
A
C
B
ß
F
55. A
E
R
D
ß A
B
C
R2 = 3R
2W
2W
R
ß
2W
2W
R
2W
1W
F
R
A
E
C
A
i1
D
R
Þ
(g)
R
B
E
Let R be the resistance of each conductor, and R1 be the effective resistance between A and F in first case then, \ R1 = 5 R If R2 be effective resistance between A and F in second case then,
50W
Here, C and E , D and F are at same potential. Re = 23.3 W r
57. (a) a
r
r
r
r b
r
A
2 r 3
2r a
5 r 8
r
r b a
r
ß 5 r 3 B r
r
b
17 b
(b)
r
r
b
r
2r
r
Þ
a
o
r
a
r
r
Þ
r
a Þ a
b
ß r/ 3
4 r 3
2r ß
B
b
b r
r
r
a
r
r
(c)
ß
r
r
r
r b
r
a Þ b
a
r
r
r
r
As Wheatstone bridge is balanced. r 58. Re = 2 i1–i3
r
(b)
r
b
i3
2i1+i2 i1 a i i1 2
ß 2r a Þb
i3
b
i1
i3
i3
a
i1
i2
i1–i3 2r
ß r
As Wheatstone bridge is balanced b
r r r r
r r
(d)
r
Þ
a r
r
r 4
Þ
r r
a
r
r r
r
r
r
r
r
b
Þ
a r
(e)
r a
8r/3
r r
r
a
r
b r
b
2r
Þ a
2r
r
2r/ 3
8r/ 3 ß a
r
4/5r b
Objective Questions (Level1) 1. When ammeter is connected in series Re = R + RA Hence, net current decreases. So RA should be very low. 2. Amount of charge entering per second from one face is equal to the amount of charge leaving per second at the other, hence I is constant. Again, I = not constant. vd = neA eF As t vd = m
E=
Þ 3. R =
mvd = not constant et
V I
Þ
[R] =
[V] [ML2T 3I1 ] = [I] [I] = [ ML2T 3I2 ]
1 4. s = r As unit of resistivity is ohmm and unit of s is ohm 1 m 1. 5. Fact.
r
b
r
r
r r
Þ a b
r
r ß 2r/3
r r
2r 2r
r
b
18 6. E = I ( R + r ) Case I E = 0.5 (3.75 + r ) Case II
I1 + I2 + I3 = 0 6  V0 3  V0 2  V0 + + =0 6 3 2 Þ 6  V0 + 2 (3  V0 ) + 3(2  V0 ) = 0
E = 0.4 (4.75 + r ) On solving r = 0.25 W, E = 2 V 5 I 50 7. = Þ I = Ig 2 I g 20 S=
Ig I  Ig
GÞG =
I  Ig
S
Ig
I I 13. vd = = neA ne pr 2 vd ¢ =
V0 = 3 V
v v 2I = d = nep (2r )2 2 2
14. Voltmeter has higher resistance than ammeter. Again higher the range of voltmeter, higher will be its resistance. 1 R2 15. I2 = I 1 1 1 + + R1 R2 R3
3 ´ 12 2 = 18 W
= 1 I 50 Ig G S= G= I  Ig 49
8. I g = 2% I =
R1
V2 9. P = R
I1 2
P + DP =
R2=20W
I
V R + DR
R3=15W
As R µ l DR =  10% R V2 V2 æ 1 ö DP = =ç  1÷ P 0.9 R R è 0.9 ø 10 = P » 11% P 9 10. Potential difference between any two points is zero. l l 11. r= 1 2 R l2 75  60 = ´ 10 60 = 2.5 W 12. (b) By applying KCL at O A
I1
A
I2
I3
æ 1 I 1 1 ö ÷ =  çç + R1 I2 R2 è R2 R3 ÷ø 0.8 1ö æ1 = ç + ÷ 0.3 ´ 20 è 20 15 ø 1 = 60 R1 = 60 W E 16. (d) I1 = , V1 = I1 RV RA + RV Þ
A
V
I1 B I2
E
3W
6W O V0 2W I3 C
= E  I1 RA If resistance is connected in parallel with voltmeter,
19 R A
Pe =
V
19. As bulb A is in series with entire circuit.
I2
20. I =
E
I2 =
and
V2 = n 2P Re
E > I1 RRV RA + R + RV
Vab = E2  Ir2 = 0 18 3´1 = 0 R+3
V2 = E  I2 RA < V1
17. Before connectivity resistance is parallel with ammeter A
V
E1 + E2 18 = R + r1 + r2 R + 3
R=3 W
Þ R 21. I2 = I R + RV I
I1
A
I1
I2 E
I1 =
= E  I1 RA After connecting resistance in parallel to the ammeter.
R + 2500 = 125 R 2500 R= W » 100 W 24
RA
V
22.
I2
E
E , RA + RV 2 1 Reading of ammeter = I2 2 E 1 = > I1 RA + 2 RV 2 2E V = I2 RV = < 2V1 RA + 2 RV I2 =
18. Re =
V
V R = I RV R + RV 100 R = 5 2500 R + 2500
E , V1 = I1 RV RA + RV
A
R
R/10
R/10
R/10
R/10
R/10
R/10
R/10
R/10
R/10
R/10
ß R/5 R/5 R/5
R n2
R/ 25
Þ
R/5 R/5
R/n R/n
23. R/n 2
P=
V R
R1 20 1 = = R2 80 4
…(i) R1 + 15 40 2 = = R2 60 3
20 R1 15 2 + = R2 R2 3 15 2 1 5 =  = R2 3 4 12
30. (a) VAB = kL = 0.2 ´ 100 = 20 mV RAB VAB = E RAB + R RAB 0.02 = ´2 Þ RAB + 490
R2 = 36 W, R R1 = 2 = 9W 4
Þ
24. (b) As V1 =
25. (d)
VAB =  I (25 + 15) 1 =  ´ 40 »  4 V 9
E RPB
V , R1 = R2 2 RV ´ 100 = 50 100 + RV
RAB + 490 = 100 RAB 490 RAB = » 4.9 W 99
Þ
RV = 100 W 2 = = 0.4 W + r 4+1
31. (c) When key is open,
VAB = IRAB = 1.6 W VPB 1.6 K = = = 0.016 V/cm L 100 E 1.2 L= 1 = = 75 cm K 0.016
R
2R
2R
R
3R
3R
Þ
I1
I1 A
26. (d) VAB = 3 ´ 2 + 3 + 1 ´ 4  2 + 6 ´ 1 = 17 V 1A
2E 3R
I1 =
A E
3/2R ß
E
2A 3R
2W 3A
2W
2W A
27. (c) Ee =
2W 1W
1W
28. (a) V =
A
B 3V
E
2V
When key is closed
E1r2 + E2r1 =2V r1 + r2 rr re = 1 2 = 0.5 W r1 + r2
For maximum power R = re and
I1
Pmax =
R
2R 2R/3
2R
R
I2
Ee2 (2)2 = =2W 4re 4 ´ 0.5
Þ I2
A
A E
ß 4R/3
R E R+ r
æE ö æ 2.2 ö r = ç  1÷ R = ç  1÷5 èV ø è 1.8 ø 10 = W 9 E1  E2 29. (d) I = R1 + R2 + r1 + r2 10  5 1 = = A 25 + 15 + 2.5 + 2.5 9
I2 A
3E 4R I1 8 = I2 9
I2 = \
2R/3
E
21 1 I 32. (b) S = G = 34 ´ 3663 33 I  Ig I 34 = 111 W Ig
a b
r r
33. (b)
r
r r
A r
r r
B Þ A r
r
B r
r
= 2.5W 36. (c) Let RAD = RBC = x
2r
r
1W
2r/3 B ÞA
A
B
r
5/3r ß 5/11r
1W
r
r
2r/3
11 ´ 1.5 5 = 3.3 W
r
r
r
r r
r r
B ÞA
r r
r r
B
r r
r
A
r/2
2r
A
1W
B
1W
P R B
2r
R R
R Q
Þ
R R1 = =3 W 6 l2 R2 = Þ R2 = 15 W l1 + l2
R
R/2
R
Þ R/3
R Q
ß
P
As the circuit is symmetrical about perpendicular bisector of AB, all points lying on it are at same potential. L1 R L1 + L2
1W
P
2r
B ÞA
1W
1W 1W
38.
r
r ß 2r
35. (c) R1 =
1W
Þ 2 R2 + RR0 + R2 + RR0 = 2 RR0 + R02 Þ 3 R2 = R02 R Þ R= 0 3
r A
1W
Clearly x < 1 as 1 W resistor is in parallel with some combination. Now RAB = x + 1 + x = 2x + 1 As x < 1 1 < RAB < 3 R ( R + R0 ) 37. (d) RAB = R + = R0 2 R + R0
5 r 11
r
1W
r
r=
34. (b)
1W
1W
ß
5/3r
Re =
Hence R1 and R2 are in parallel R1 R2 Re = R1 + R2
2r
r
60° R1
R2
5 6 R = 25W Ü Q
5 11 R
R
22 4 R1 = ´ pr = 2 = R2 2pr 4 4 R3 = ´2 r = 2pr p 1 1 1 1 = + + Re R1 R2 R3 1 1 p 4+ p = + + = 2 2 4 4 4 Re = W 4+ p
39. Wheatstone bridge is balanced. A
P R
R
R
R 2R/3
2R/3
R
R
R R
R
Ü R R
R
R
R
B R
2R/3
2R/3
R
\
R
B ßA R
R
R
Þ 4R/3 4R/3
4R/3
7R/3 Þ
7R/3
4R/3
42. (d) Points C and D are shorted hence the portion above line CD can be removed.
B
7 R 6 L1 1 40. (d) R1 = R= R L1 + L2 12 Þ
Re =
C
D
=3 W
B
A
30°
O
ß
C
A B
D
B
A
ß R2 =
L2 11 = R = 33 W L1 + L2 12
B
A
A
41. (a) Resistance per unit length of wire 4 = 2 pr
R
Ü B
R
B
R
R
R R
R R/2
B R
R/2 R/2
R/2 R2
R
R
A
R/2
A
R/2
R/2
B
Þ
R/2
R ß 3R/2
B A B
A
R R
R
A
R3
R/2
43. (b) As AB is line of symmetry, we can fold the network about AB.
= 2.75 W
R1
Ü A
R1 and R2 are in parallel, 3 ´ 33 R1 R2 Re = = R1 + R2 3 + 33
R/2
R
R/2
B
23
JEE Corner Assertion and Reason 1. (d) V = IR, If V = 0 either I = 0 or R = 0
P=
2. (b) As all the resistors are in parallel potential difference is same, hence V2 is maximum if R is minimum. P= R I 2t r 3. (b) dH = I 2dRt = dH A I is same everywhere, hence portion having less area is more heated. I Again J = A J A > J B. \ Reason is also correct but does not explain assertion. 4. (b) Both assertion and reason are correct but reason does not explain the cause of decrease in voltmeter reading. 5. (b) As RA < RV , more current passes through ammeter when positions of ammeter and voltmeter are interchanged and potential difference across voltmeter becomes less that emf of cell. 6. (c) During charging current inside the battery flows from positive terminal to negative terminal. Reason is false while assertion is true. E 7. (d) I = is maximum when R is zero R+ r hence reason is false.
E2 R is maximum at R = r. ( R + r )2
V V2 ,P= both I and P are inversly R R proportional to R hence both decrease with increase in R which increases with temperature. According to Ohm’s law V µ I not V = IR. As R can be variable also.
8. (c) I =
9. (d) Drift velocity is average velocity of all the electrons but velocity of all electrons is not constant. L 10. (a) R = r A m r= 2 ne t with increase in temperature, electron collide more frequently, i.e., t decreases, increasing r and hence R. E r + E2r1 11. (d) E = 1 2 \ E1 < E < E2 r1 + r2 If E1 < E2 rr r = 1 2 , r < r1, r < r2 r1 + r2 12. (d)
R1 L1 = R2 L2
Hence there is no effect of one while measuring using meter bridge.
Objective Questions (Level2) E  E1 1.5  1.3 1. (b) I = 2 = r1 + r2 r1 + r2 0.2 = r1 + r2 V = E1 + Ir1 Þ
1.45 = 1.3 +
0.2 r1 r1 + r2
r1 0.15 = Þ 0.2r1 = 0.15r1 + 0.15r2 r1 + r2 0.2 0.05r1 = 0.15 r2 r1 = 3 r2
…(i)
2. (c) Let R = Resistance of voltmeter, ER = 198 V V1 = R1 + R V2 =
ER ER = = 180 V R2 + R 2 R1 + R 2 R1 + R 198 11 = = R1 + R 180 10
20 R1 + 10 R = 11 R1 + 11 R 9 R1 = R
…(i) …(ii)
24 From Eq. (i), ER = 198 R1 + R 10 R1 E = 198 ´ = 220 V 9 R1 3. (b) P = I 2 R As R is same for all bulbs and maximum current passes through bulb A, it will glow most brightly. V 4. (c) R + RA = =5W I R = 5  RA < 5 W L1  L2 10 5. (a) r = R= ´ 132.40 L2 60 » 22.1 W 6. (b) Current through R when S is open. E1 + E2 I1 = R + r1 + r2 Current throughR when S is closed E1 I2 = R + r1 DI = I2  I1 E1 + E2 E1 = R + r1 R + r1 + r2 E1r2  E2( R + r1 ) = ( R + r1 )( R + r1 + r2 ) DI = + ve if E1r2 > E2( R + r1 ) 7. (a) VA = IR 2 I 3 I x
B
A
35 =1 \Required ratio = 35 L  L2 xy 9. (a) r = 1 R= R L2 y R  20 t  10 10. (d) = 40  20 30  20 R = t + 10 E 10 I= = R t + 10 dq 10 = dt t + 10 30 10 30 q=ò dt = 10[log e( t + 10)] 10 10 t + 10
Þ
= 10 log e 2 11. (b) Let l1 length is kept fixed and l2 is stretched, l l R1 = r 1 , R2 = r 2 A A Initial resistance, …(i) R = R1 + R2 3 Now full is stretched times, ie, 2 3 l2 ¢ = ( l1 + l2 )  l1 2 1 = ( l1 + 3 l2 ) 2 A ¢l ¢ 2 Al2 A2 ¢ = 2 2 = l2 l1 + 3 l2 1 ( l1 + 3 l2 )2 2 R2 ¢ = r 2 Al2
y 1 I 3
R2 ¢ = r
C
2 I ´ 1.5 R = IR 3 1 VC = I ´ 3 R = IR 3 VA = VB = VC
Now,
VB =
\
8. (d) Current through 15 W resistor 30 = =2A 15 VBC = (2 + 5) ´ 5 = 35 V Voltage drop across R = 100  (30 + 55) = 35 V
Þ Þ 12. (b)
( l1 + 3 l2 )2 4 Al2
R¢ = 4 R R1 + R2 ¢ = 4 ( R1 + R) ( l + 3 l2 )2 l1 + 1 = 4( l1 + l2 ) 4l2 l2 1 = l1 7 l2 1 = l1 + L2 8
l1 X = Þ l1 = 40 cm R 100  l1
If R¢ = 8 W
25 l1 ¢ X = R¢ 100  l1 ¢ l1 ¢ = 60 cm l1 ¢  l1 = 20 cm
Þ 13. (c) I1 =
when, k1 is open and k2 is closed, net resistance is 100 ´ 100 Rnet = r + R1 + = ( r + R1 + 50) 100 + 100 I0 E …(ii) \ = 2 r + R1 + 50
0.1 I 0.1 + 9.9
The above two equations are satisfied if r = 0 and R1 = 50 W.
10 mA
9W A I1
16. (b) 20 W, 100 W and 25 W resistors are in parallel.
I2 0.1W
I
0.9W
4W +
A
B
4W
C
20W
25W
Þ
V
But I1 = 10 mA 10 I= ´ 10 mA = 1000 mA 0.1 = 1 kA
6W
6W – 4A
R = 20 W V = IR = 80 V
14. (d) Effective emf of two cells E r + E2r1 2 ´ 6 + 4 ´ 2 E= 12 = r1 + r2 2+6 20 = = 2.5 V 8 RAB = 4 ´ 4 = 16 W RAB 16 VAB = E0 = ´ 12 R + RAB 4 + 16
17. (a) Hence, points A and C, B and D are at same potential. V R
A
R
R
B
C A
V k = AB = 2.4 V/m L
The equivalent circuit is given by
Now, E = kl
V
E 2.5 25 L= = = k 2.4 24
R R
A, C
15. When k1 and k2 both are closed, the resistance R1 is short circuited. Therefore net resistance is
B, D
R
A
k1 E, r
I
ß V R/3
A
100 ´ 100 = r + 50 100 + 100 E I0 = r + 50
A
Rnet = r + \
D
E, r
= 9.6 V
Þ
10W
100W
E, r
…(i)
26 E
=1A R +r 3 R V = I ´ =3 V 3 Ig r 18. (c) S = G , G = r, S = I  Ig 4 I=
1 1 (I  Ig ) Þ Ig = I 4 5 = 0.006 A 10 5 19. (d) I1 = = A 14 7
21. (a) Voltage sensitivity of voltmeter 1 µ Resistance of voltmeter Vs1 R +G = 2 \ Vs2 R1 + G R2 + 50 30 = 20 2950 + 50
Ig =
I1
8W
B
R2 + G =
30 ´ 3000 = 4500 20 R2 = 4450 W
Þ 22. (b) For x = 0
6W
VAB = E E k1 = L EL1 E0 = k1L1 = L
P 4W I
I2
3W A 10 V
10 A I2 = 7 VP  VA = I 2 ´ 4 40 A = 7 40 A VP  VB = I1 ´ 8 = 7 VA  A B = 0 Another method R R As, 1 = 3 , VB = VA R2 R4 20. (b) For series connection V1 R = 1 V2 R2 R1 3 \ = R2 2 L rL Now, R1 = r 1 = 12 , A1 pr1 L rL R2 = d 2 = 22 A2 pr2 R1 L1 æ r2 ö = ´ç ÷ R2 L2 çè r1 ÷ø
\ r1 = r2
r2 1 = r1 2
For x = x (say) RAB E RAB + x
VAB = k2 =
RABE ( RAB + x ) L
E0 = k2L2 =
RABEL2 ( RAB + x ) L
…(ii)
From Eqs. (i) and (ii), R + L2 L1 = AB ( RAB + x ) 20 =
Þ
10 ´ 30 10 + x
x =5W
Þ
23. (d) To obtain null point similar terminal of both the batteries should be connected. 24. (c) Wheatstone bridge is balanced. 20W 1.4A
4W 15W
Þ
I1 20W
4W
50W
10W
1.4A I2
2
2 ´6 R2L1 = =2 R1L2 3 ´1
…(i)
50W
\
10W
R2 I1 = ´I R1 + R2 60 = ´ 1.4 84 =1A
27 25. (b)
L1  L2 R L2 L2r R= L1  L2 490 ´ 10 = = 490 W 10
29. (c)
r=
E,B,H R A
B E
Þ
R
A,C,F
10V F
Þ
R R
D
C
R
50W G 10W
H
R
D
More than One Correct Options
G
10V
I1
R
R R
I
Þ
If connected in series 1 1 1 = + P P1 P2
I2
I3
R
Ü
R
Þ t = t1 + t2 If connected in parallel P = P1 + P2 tt t= 12 Þ t1 + t2
10V
10V
10 I1 = = 2.5 A 2 ´2 26. (b) Effective resistance of voltmeter 3 kW resistor, 3 ´6 R1 = = 2 kW 2+6 R1 2 V1 = E = ´ 10 = 5 V R1 + R2 4 27. (d) P1 = P2 = P3, Clearly R2 = R3 R2 R1
and
31. E =
E1r2 + E2r1 6 ´ 3 + 5 ´ 2 = r1 + r2 2+3
= 5.6 V As there is no load. V = E = 5.6 V If E1 = E2, I = 0 E  E2 6  5 I= 1 = = 0.2 A r1 + r2 2+3 32. Let V = Potential difference between T1 and T2.
i R3
\
H = P1t1 = P2t2 H H t1 = , t2 = P1 P2
30.
ß
i2 = i3 =
i 2
T1
A
B T2
I2
2
i 1 P1 = i2 R1, P2 = æç ö÷ ´ R2 = i2 R2 è2ø 4 1 2 P3 = i R3 4 R2 = 4 R1, R3 = 4 R1 R1 : R2 : R3 = 1 : 4 : 4 \ E 2 28. As E = kL1 Þ k = = = 250 L1 500 1 V/cm = 250 1 V = kL2 = ´ 490 cm 250 = 1.96 V
I1
C
V RA + RB V I2 = RC
I1 =
Now,.
I A = I B = I1 IC = I2 Also, V = IC RC = I1( RA + RB ) = I A RA + ID RB I B I1 RC = = IC I2 RA + RB
28 33. As R1 ¹ R2
Ie = If S2 is closed
B
A
V1 ¹ V2 V3 = V1 + V2
2W a
R2
V1 = V2 L R=r A
b
39. a
r
L2 = 2L1 R1 = R2 A2 = 2 A1
d
d
b
d
2r
r
b r
a r
ß b
2r
c
b
a 5r 3 ß 5r 8
2r
c ß
r
a
d
r/ 2
r 2r 3
r 2r
c ß a
ß
b
d r
2r
b
r c ß
r
r
ß
d r
c ß a
r
b
r
b
r
r
c ß a
35. If E > 18 V current will flow from B to A and viceversa.
r
r
b
r
r
a
r
r
b
1 (For constant current) Also, vd µ A 1 vd 2 = vd1 Þ vd1 = 2 vd 2 2 Again, vd µ E \ E1 = 2 E2
If S1 is closed
>I
10 V
I C
a
36. V = kl If Jockey is shifted towards right, I and hence k will decreases as k µ I. Hence L will increase. If E1 is increased, k will increase, hence L will decrease. If E2 is increased L will increase as V will increase. If r is closed V will decrease hence L will decrease. E E , Initially, I = 37. Ie = Re + re R+ r
r 2
I =6A From b to a. Vc  Va = 2 ´ 6 = 12 V
But and \
R+
>I
38. Vb  Va =  10 + 2I = 2 V
34. As R1 = R2 R1
E
Ie =
V3
Also,
R +r 2
V2
V1
I
E
r
d
29
Match the Columns 1. By applying KCL at e i1 + i2 + i3 + i4 = 0
S R3
b
R1
R2
V1
V2
2W a
1W i1
e i4
i3 1W
E
c
2W d
2  Ve 4  Ve 6  Ve 4  Ve + + + =0 1 2 1 2 Ve = 4 V, I1 =  2 A, i2 = 0 , i3 = 2 A, i4 = 0 2. Current is same at every point and A1 < A2 i J = 횧 J1 > J 2 A i vd = 횧 vd1 > vd 2 neA R r r= = 횧 r1 > r2 L A V k = 횧 k1 > k2 L 3. When switch S is closed V1 decreases, V2 increases, \ Current through R1 decreases and through R2 increases.
4. [R] =
[V] [ML2T 3A 1 ] = [I] [A]
= [ML2T 3A 2 ] [W ] [ML2T 2 ] [V] = = [ q] [AT] = [ML2T 3A 1 ] 2 3 2 2 [ R][ A ] [ML T A ][L ] [ s] = = [L] [L] = [ML3T 3A 2 ] 1 [ s ]= = [M1L3T 3A 2 ] [r ] EA  EB 5. I = =1A R + rA + rA 4V 1W
1V
1W
1W
VA = EA  IrA = 3 V VB = EB + IrB = 2 V PA = IVA = 3 W PB = IVB = 2 W
21
Electrostatics Introductory Exercise 21.1
1. No, because charged body can attract an uncharged by inducing charge on it. 2. Yes. 3. On clearing, a phonograph record becomes charged by friction.
4. No. of electrons in 3 g mole of hydrogen atom = 3 ´ 6.022 ´ 1023 \ q = ne = 3 ´ 6.022 ´ 1023 ´ 1.6 ´ 1019 = 2.9 ´ 105 C
Introductory Exercise 21.2 1.
1 q1q2 1 e2 × 2 = × 4pe0 r 4pe0 r 2 Gm1m2 Fg = r2 Fe e2 = Fg 4pe0 × Gm1m2 Fe =
=
2.
3. Let us find net force on charge at A. A
9 ´ 109 ´ (1.6 ´ 1019 )2
B
6.67 ´ 1011 ´ 9.11 ´ 1031 ´ 1.67 ´ 1027 FAB =
= 2.27 ´ 1039 1 q1q2 F= × 4pe0 r 2 qq e0 = 1 2 2 4p Fr [ e0 ] = =
q
Net force on charge at A FA = FAB cos 30° + FAC cos 30° =
3 q2 4pe0 × a 2 FAB cos 30°
[IT ]2 [MLT ][L]
q C
1 q2 1 q2 × 2 FAC = × 4pe0 a 4pe0 a 2
[ qe ]2 [ F ][ r ]2 2
q
FAC
FAC cos 30°
2
= [ M1 L3 T 4 I 2 ]
FAB
60°
SI units of e0 = C 2 N1m 2. FAC sin 30°
A
FAB sin 30°
31 ®
®
4. F OA = F OC ®
and
®
F OB =  F OD A q
B q
–q O q C
q D
Hence, net force on charge at centre is zero. 5. No. In case of induction while charge comes closer and like charge moves further from the source. + + + + +
– – – –
+ + + +
The cause of attraction is more attractive force due to small distance. But if electrostatic force becomes independent of distance, attractive force will become equal to repulsive force, hence net force becomes zero. 6. When the charged glass rod is brought near the metal sphere, negative induces on the portion of sphere near the charge, hence it get attracted. But when the sphere touches the rod it becomes positively charged due to conduction and gets repelled by the rod. 7. Yes as qmin = e Fmin =
1 e2 × 4pe0 r 2
8. No. Electrostatic force is independent of presence or absence of other charges. ®
®
^
^
9. F21 =  F12 = (  4 i + 3 j) N.
Introductory Exercise 21.3 1. False. E =
1 q × 4pe0 r 2
2. VA > VB as electric lines of force move from higher potential to lower potential. 3. False. Positively charged particle moves in the direction of electric field while negatively charged particle moves opposite to the direction of electric field. 4. False. Direction of motion can be different from direction of force. s 5. E = Þ s = Ee0 = 3.0 ´ 8.85 ´ 1012 e0 = 2.655 ´ 1011 C/m 2 6. q1 and q3 are positively charged as lines of force are directed away from q1 and q3. q2 is negatively charged because electric field lines are towards q2. 7. If a charge q is placed at A also net field at centre will be zero.
A
E q
B q
O q D
q C
Hence net field at O is same as produced by A done but in opposite direction,02 i.e., 1 q E= × 4pe a 2 8. Net field at the centre (O) of wire is zero. If a small length of the wire is cutoff, net field will be equal to the field due to cutoff portion, i.e., 1 dq dE = × O 4pe0 R2 q R dl 1 2p R = × 4pe0 R2 q dl = 8p 2e0 R3
32 1 q ® × 3 r E= 4pe0 r ®
9.
=
9 ´ 109 ´ 2 ´ 106 (3 2 + 42 )3/ 2
^
^
^
^
(3 i + 4 j) =  144 (3 i + 4 j) N/C
Introductory Exercise 21.4 1. Gain in KE = loss of PE æ1 1ö 1 1 mv2 = × q1q2 ç  ÷ ÷ çr 2 4pe0 è 1 r2 ø 1 ´ 104v2 2 1 1 ö =  1 ´ 106 ´ 2 ´ 106 ´ 9 ´ 109 æç ÷ è 1 0.5 ø 2
v = 360 v = 6 10 ms 1 2. W = q ( VA  VB ) 6 æ 1  1 ´ 106 1  1 ´ 10 ö ÷ = 2 ´ 106 ç × × ç 4pe0 ÷ 1 4pe0 2 è ø 3 =  9 ´ 10 J = 9 m J
3. Whenever work is done by electric force, potential energy is decreased. W =  DU U2 = U1  W =  8.6 ´ 108 J qq 4. No. As U = 1 2 4pe0r If there are three particles 1 é q1q2 q2q3 q3q1 ù U= ×ê + + ú 4pe0 êë r12 r23 r31 úû Here U may be zero. In case of more than two particles PE of systems may same as if they were separated by infinite distance but not in case of two particles.
Introductory Exercise 21.5 1. Vba =
Wa ® b q
= 12 ´ 102 = 1200 V
2. l = a x (a) SI Units of l = C/m l a= x C/ m Hence SI unit of a = = C/m 2. m L
Þ
L x dx a × 4pe0 ò0 x + d L dx ù a é L = × ê dx  d ò ú 0x + d 4pe0 êë ò0 úû a = × [[ x ]L0  d [ln ( x + d )]L0 ] 4pe0 L + dù a é = × ê L  d ln ú 4pe0 ë d û
3. Consider an elementary portion of length dx at a distance x fro my centre O of the rod.
P
P
d
x d
(b) Consider an elementary portion of rod at a distance x from origin having length dx. Electric potential at P due to this element. dV =
l dx 1 × 4pe0 x + d
Net electric potential at P L 1 l dx V =ò × 0 4 pe 0 x + d
d O 2l
x
d–x
Electric potential at P due to this element, l dx 1 dV = × 4pe0 d 2 + x 2 V=
l l ×ò 4pe0 l
dx d2 + x 2
33 l é 1 x ù l × sin d úû  l 4pe0 êë q x = ´ 2 sin 1 4pe0 × 2l d q x sin 1 V= 4pe0 l d =
4. Consider the cone to be made up of large number of elementary rings. O q
l
L
Total potential at O Q sin q L QL sin q V= dl = 2pe0 RL ò0 2pe0 RL Q [L sin q = R] = 2pe0 L
dl
x
Charge on the elementary ring; Q dQ = s dA = × 2px dl p RL 2Ql sin q or dQ = dl RL Potential at O due to this ring 1 dQ dV = × 4pe0 l Q sin q = dl 2p e0 RL
U = qV Qq = 2pe0 L
R
Consider one such ring of radius x and thickness dl. Let q be the semivertical angle of cone and R be the radius of cone.
Introductory Exercise 21.6 1. (a) V = a ( x 2  y2 ) æ ¶v ^ ¶v ^ö ^ ^ i= j÷÷ =  2ax i + 2 y j E =  çç ¶ x ¶ y ø è
V is constant hence E is zero. For x >4 V is decreasing at constant rate, hence E is positive. (50  100) dV 3. E = == 10 V/m dr 5 0
(b) V = axy æ ¶v ^ ¶v ^ö ^ ^ i= j÷÷ =  a ( y i + x j) E =  çç ¶ x ¶ y ø è
True.
2. From x =  2 to x = 0 & x = 2 to x = 4 V is increasing uniformly.
® ®
4. (a) VP  VD = E × l = 0 ® ®
(b) VP  VC = E × l = 20 ´ 1 ´ cos 0° = 20 V
–2
O
2
4
Hence, E is uniform and negative From x = 0 to x = 2
A
B
D
1m C
x
1m
® E = 20V/m
(c) VB  VD =  20 ´ 1 =  20 V (d) VC  VD =  20 ´ 1 =  20 V
34
Introductory Exercise 21.7 ®
® E
B+q
=
x Q A
2l Q x
q
F29
1. F1 = qE towards right F2 = qE towards left Net torque about q, t = qE (2l  x ) sin q + qEx sin q = q (2l) E sin q = pE sin q ®
®
®
t = p´ E
2.
E1 =
1 q × 4pe0 ( y2 + a 2 )2
E2 =
1 q × 4pe0 ( y2 + a 2 )2
E3 =
1 2q × 4pe0 y2
q 4pe0
é 2 cos q cos q ù ^  2 ê 2  2 új 2 y +a y + a 2 úû êë y ù^ 2q é 1 y =ê új 4pe0 êë y2 ( y2 + a 2 )3/ 2 úû 2 2 3/ 2  y3 ù ^ 2q é ( y + a ) =ê 2 2 új 4pe0 êë y ( y + a 2 )3/ 2 úû
F1 2l
^
E =  ( E3  E1 cos q  E2 cos q) j
2 3/ 2 ù é æ ö ê y3 ç1 + q ÷  y3 ú 2÷ ç ú^ y ø 2q ê è =új ê 4pe0 ê y2 ( y2 + a 2 )3/ 2 ú ú ê úû êë As y >> a ù é 3æ 3 q2 ö ÷  y3 ú ê y çç1 + 2÷ 2y ø 2q ê ú^ è E=× új 4pe0 ê y5 ú ê úû êë 2 3 qa ^ j E=4pe0 y4
Net field at P
Introductory Exercise 21.8 1. (a) Charge q is completely the hemisphere hence flux through hemisphere is zero.
(b) Charge inside the sphere is q hence flux through hemisphere f=
q e0
(c) As charge q is at the surface, net flux through hemisphere f=
q 2 e0
2. When charge is at any of the vertex, net flux through the cube, q f= 8e0 If charge q is at D, flux through three faces containing D is zero and the flux f is divided equal among other three faces, hence
fEFGH = and
1 q = f 2pe0
fAEHD = 0
3. True. As electric field is uniform, flux entering the cube will be equal to flux leaving it. q \ fnet = 0 Þ fnet = e0 Þ q =0 4. (a) As net charge inside hemisphere is zero, E 1
2
f1 + f2 = 0 But E is parallel to surface 2.
35 \ f2 = 0 Hence, f1 = 0 (b) Again, f1 + f2 = 0
ÐPOQ = 2q = 120° E
1
2
f2 = E ´ pR2 = p R2E \ f1 =  f2 =  p R2E R/2 1 5. cos q = = , q = 60° R 2
P
2p R 3 Charge inside sphere, q q 2p q= 0 ´ RÞ 0 2pR 3 3 \Flux through the sphere q q f= = 0 e0 3 e0 \Length of arc PQ =
6. Net charge inside the cube = 0. \Net flux through the cube = 0.
Q q
R
R — 2
O
Introductory Exercise 21.9 Q
1 q + q + qB × =0 4pe0 2R '+
3q
+
B –2q
–(q
1. VB =
q
q+Q ) q' + +Q –(q +Q) q+Q
A +2q –q q
C B
R
R
2R
A
3R
2R
VB = q B =  2q Total charge inside a conducting sphere appears on its outer surface, \Charge on outer surface of A = 2q and charge on outer surface of B = 2q  2q = 0 2. Let q¢ = charge on sphere B and charge f flows from sphere C to A.
1 æ q¢ + q + Q 2q  Q ö ×ç + ÷ =0 4pe0 è 2R 3R ø
…(i) Þ 3 q¢ + q + f = 0 Again, VP = VC 2q  Q ù 1 éq + Q 1 3 q + q¢ q ×ê + + × ú= 4pe0 ë R 2R 3 R û 4pe0 3R 6( q + Q ) + 3 q¢ + 2(2q  Q ) = 2(3 q + q¢ ) 4q + 4q + q¢ = 0 On solving
36 5 24 Q=q, q¢ = q 11 11 A
B
Charge on 0 inner surface
 (q + Q ) 6 =q 11
3.
2q + q' C
–(q')
C
q'
 (q¢ + q + Q ) 18 = q 11
–2q
B A + 2q –q A
q
Charge on q + Q = 6 q¢ + q + Q 3q + q = 9 q outer surface 18 11 11 =q 11 A Charge on  q inner surface
B  2q
Charge on + 2q outer surface

4q 3
C 4 + q 3 2 + q 3
AIEEE Corner Subjective Questions (Level1) 1. F =
1 q (Q  q) × 4pe0 r2
4.
For maximum force dF 1 Q  2q = × =0 dq 4pe0 r2 Q q= 2 1 2 d 2F = × <0 dq2 4pe0 r 2 Q Hence F is maximum at q = . 2 2. Minimum possible charge on a particle = e. 9 19 2 1 e 2 9 ´ 10 ´ (1.6 ´ 10 ) \Fmin = × 2= 2 2 4pe0 r (1 ´ 10 ) 3.
= 2.3 ´ 1024 N 1 q1q2 Fe = × 4pe0 r 2 Gm1m2 Fg = r2 q1q2 Fe = Fg 4pe0 Gm1m2 =
(3.2 ´ 1019 )2 ´ 9 ´ 109 6.67 ´ 1011 ´ (6.64 ´ 1027 )2
= 3.1 ´ 1035
…(i) …(ii)
F1 =
1 q1q2 × 4pe0 r 2
F2 =
1 q2 × 4pe0 r 2
…(i) …(ii)
[As both the spheres are identical, find charge on both the spheres will be equal] q  q2 q= 1 2 q1  q2 = 2q Þ From Eq. (ii), q2 = 4pe0 r 2F2 =
(50 ´ 102 )2 ´ 0.036 9 ´ 109
= 1012
q = 106 C = 1 mC From Eq. (i), (50 ´ 102 )2 ´ 0.108 q1q2 = 4pe0r 2F1 = 9 ´ 109 = 3 ´ 1012 Also, q1 + q2 = 2q = 2 ´ 106 On solving q1 = ± 3 mC and q2 = + 1 mC
37 5. (a) F1 =
q1 Q 1 × 4pe0 (3 a / 2)2 +a/2 +a O ® Q ® q2 F2 F1
–a –a1
1 4q1 Q × 4pe0 9a 2 qQ 1 F2 = × 2 4pe0 ( a / 2)2 1 4q2 Q = × 4pe0 a 2 F1 =
FAO …(i)
For net force on Q to be zero. FAB = FAO 1 4q2 1 qq Q × = × 4pe0 L2 4pe0 L2 9 Þ q= Q 4 4 Q= q Þ 9 4 As Q is negative Þ q = q 9 (b) PE of the system 1 é 4q2 qQ 4qQ ù U= ×ê + + ú 4pe0 êë L x L  x úû
…(ii)
For net force on Q to be zero F1 = F2 or q1 = 9 q2 1 4q1 Q (b) F1 = × 4pe0 25a 2 –a q1
O
+Q + ® F2
F1 =
+a 2 Q ® F1 q2
1 4q2 Q × 4pe0 9a 2
For net force on Q to be zero. F1 + F2 = 0 q1 25 Þ = q2 9
x
O
Hence, equilibrium is unstable. N sin 60°
O N R
60°
Fe
R
N cos 60°
–x
Q
B +4q
Let the charge Q is planed at a distance x from A (+ Q charge) 1 qQ FOA = × 4pe0 x 2 4q Q 1 FOB = × 4pe0 ( x  x )2 For net force on Q to be zero. FOA = FOB 4q Q 1 qQ 1 × 2 = × 4pe0 x 4pe0 ( L  x )2 ( L  x )2 = (2x )2 L x= 3 Force on A,
1 é 4q2 4qQ 8qQ ù ×ê ú =0 4pe0 êë L 3L 3 L úû
=
7. FBD of af placed at left can be given by
6. (a) In order to make net force on charge at A and B zero, Q must have negative sign. A +q
1 4q2 × 4pe0 L2 1 qQ = × 4pe0 x 2 1 qq Q = × 4pe0 L2
FAB =
A
R
B
mg
DABD is equilateral As beads are in equilibrium mg = N sin 60° Fe = N cos 60° Fe = cot 60° mg q2 = 4pe0 R2 mg cot 60° 4pe0 R2 mg 3 6pe0mg = 2R 3
q=
38 Here, dE sin f components of field will cancel each other. Hence, Net field at O 1 l p/ 2 E = ò dE cos f = × ò cos f df 4pe0 R  p/ 2 1 2l = × 4pe0 R
8. As ball are in equilibrium O a T
T
T cosa
13. Consider elementary portion of the rod of length dl at a distance l from the centre O of the rod.
a Fe
Fe
T sina
dE sin f
r mg
dl
Fe = T sin a mg = T cos a Fe = mg tan a q2 = 4pe0r 2 tan a r = 2 l sin a 2 q = 16pe0 l2 sin 2 a tan a q = 3.3 ´ 108 C.
l
mg
Here,
O
dl
=
9 ´ 10 ´ (  8.0 ´ 10 ) ((1.2)2 + (1.6)2 )3/ 2 ^
^
\ ^
(1.2 i  1.6 j )
^
12. Consider an elementary portion on the ring of length dl subtending angle df at centre ‘O’ of the ring. Charge on this portion,
dE sin f
dE cos f dE cos f dE O
df
dE sin f f R
dl
\
dE cos f
dl
dq = l dl = l Rdf 1 dq 1 l df × 2 = dE = 4pe0 R 4pe0 R
dE sin f
Q dl L 1 dq dE = × 4pe0 ( a sec f)2 Q dl 1 = × 4pe0 La 2 sec 2 f dq = l dl =
=  18 2 (1.2 i  1.6 j ) N/C.
dE
dE cos f P
Charge on this portion
10. See Q.7. Introductory Exercise 21.3. q ® 1 11. E= ×1 r 4pe0 r 3 9
f Q
dE
9. Same as Q.7. Introductory Exercise 21.3.
9
dE
df
Now, Þ \
l = a tan f dl = a sec 2 f df 1 Q df dE = × 4pe0 La
Net Electric field at P. E = ò dE cos f [ dE sin f components will cancel each other as rod in symmetrical about P.] 1 Q q = × cos f df 4pe0 La ò q 1 2Q sin q = × La 4pe0 L L But sin q = = 2 2 4a + L2 L 2 a 2 + æç ö÷ è2ø 2Q 1 E= × \ 4pe0 a 4a 2 + L2
39 14. (a) As shown in figure, direction of electric field at P will be along + ve yaxis.
Clearly resultant field is along angle bisector of field towards 9 and 10.
y
6E1
6E1
–Q
6E1
E
E1
E2
E 6E1
x
P
6E1 6E1
Q
Hence time shown by clock in the direction of electric field is 9 : 30. F  eE 16. (a) a = = m m  1.6 ´ 1019 ´ 1 ´ 103 = 9.1 ´ 1031
(b) Positive xaxis. y E2
+Q
P
E
x
E1
+Q
(c) Positive yaxis. E E1
E2
–Q
+Q
1 q × 4pe0 R2 11 E11
u 12 E12
10 E10
1 E1 E2
2
E3 3 E4
E8 7
E7 E 6 6
E
q
9 E9 8
(b) v = u + at 5 ´ 106 t= = 2.8 ´ 108 = 28 ns. 1.76 ´ 1014 (c) Dk = work done by electric field. = F × x =  eEx =  1.6 ´ 1019 ´ 1 ´ 103 ´ 8 ´ 103 =  1.28 ´ 1018 J Loss of KE = 1.28 ´ 1018 J 25 ms 1 17. Here, ux = u cos 45° = 2
P
15. Let E1 =
=  1.76 ´ 1014 ms 2 u = 5.00 ´ 108 cm/s = 5 ´ 106 ms 1 v=0 v2  u 2 = 2as (5 ´ 106 )2 s= = 1.4 ´ 102 = 1.4 cm 2 ´ 1.7 ´ 1014
E5 5
4
Resultant fields of two opposite charges can be shown as given in figure.
u y = u sin 45° =
25 ms 1 2
a x = qE = 2 ´ 106 ´ 2 ´ 107 = 40 ms 1 a y =  10 ms 1 1 2 y = u yt + t ay y=
25 t  5t 2 2
40 at the end of motion, t = T and y = 0 5 s T= \ 2
Case I. In between two charges : let potential is zero at a distance x from q1 towards q2. x
Also at the end of motion, x=R 1 x = ux t + a x t 2 \ 2 2 25 5 æ 5 ö R= ´ + 20 ´ ç ÷ 2 2 è 2ø = 312.5 m m 2 sin 2q 18. (a) R= qE qER sin 2q = mu 2 1.6 ´ 1019 ´ 720 ´ 1.27 ´ 103 = 1.67 ´ 1027 ´ (9.55 ´ 103 )2 = 0.96 2q = 88° or 92° q = 44° or 46° 2mh sin q T= 2E 1 ´ 1.67 ´ 1031 2 1.6 ´ 1019 ´ 720
=
= 1.95 ´ 1011 s ®
®
19. (a) a = 
V= =
q2
q2 1 q1 1 × + × =0 4pe0 x 4pe0 100  x 1 2 ´ 10 × x 4pe0
6

6 1 3 ´ 10 × =0 4pe0 100  x
200  2x = 3 x x = 20 cm
Þ
Case II. Consider the potential is zero at a distance x from charge q, on its left. x
100 cm
q1
\ V= =
2 ´ 9.55 ´ 103 ´
100–x
q1
q2
q2 1 q1 1 × + × =0 4pe0 x 4pe0 100 + x 1 2 ´ 10 × x 4pe0
6
+
6 1  3 ´ 10 × =0 4pe0 100 + x
200 + 2x = 3 x x = 200 cm 21. Let us first find the potential at a point on the perpendicular bisector of a line charge. Consider a line of carrying a line charge density l having length L.
19
1.6 ´ 10 ´ 120 ^ eE =j m 9.1 ´ 1031
dl
^
=  2.1 ´ 1013 i m/s 2 Dx 2 ´ 10 4 (b) t = = = ´ 107 s 5 ux 1.5 ´ 10 3
l
q
vy = u y + a y t 4 = 3.0 ´ 106 ´ 2.1 ´ 1013 ´ ´ 107 3 = 0.2 ´ 106 m/s ®
^
r
^
v = (1.5 ´ 105 ) i + (0.2 ´ 106 ) j
20. Absolute potential can be zero at two points on the xaxis. One in between the charges and other on the left of charge a1 (smaller in magnitude).
O q1
100cm q2
f
L
Consider an elementary portion of length dl on the rod. Charge on this portion dq = l dl l dl 1 dV = × \ 4pe0 r sec f Now,
X
l = r tan f dl = r sec 2 f df
41 \ \
l sec f df 4pe0 q l V = ò dV = × ò sec 2 f df 4pe0 q l = [lnsec q + tan q]q q 4pe0
dV =
l é ½ sec q + tan q ½ù ½ú = ê ln ½ 4pe0 êë ½ sec q  tan q ½úû 2l lnsec q + tan q = 4pe0 In the given condition q = 60° Potential due to one side 2l V1 = V2 = V3 = × lnsec 60° + tan 60° 4pe0 2l = × ln2 + 3  4pe0
Hence, particle at B is faster than that at A. 24. Centre of circle is equidistant from every point on its periphery, 1 q Hence, × , V0 = 4pe0 R where q = Q1 + Q2 =  SQ 1 5Q \ × V0 = 4pe0 R 1 q Similarly, Vp = × 2 4pe0 R + Z2 =
1 × 4pe0
SQ 2
R + Z2
25. Initial PE 1 q1q2 × 4pe0 r1 1 q1q2 Uf = × 4pe0 r2 Ui =
Work done by electric force W =  DU =  (Uf  Ui ) O
Total potential at O V = 3 V1 = =
6l × ln2 + 3  4pe0
Q × ln2 + 3  2pe0 a ® ®
22. (a) V2  V1 =  E× d =  250 ´ 20 ´ 102 =  50 V W = DV = q( V2  V1 ) = 12 ´ 106 ´  50 =  0.6 mJ (b) V2  V1 =  50 V 23. By work energy theorem W =DK 1 1 q( V1  V2 ) = mv22  mv12 2 2  5 ´ 106 (20  800) 1 = ´ 2 ´ 104( V22  (5)2 ) 2 v22 = 55 v2 = 55 = 7.42 ms 1 When a particle is released in electric field it moves in such a way that, it decreases its PE and increases KE
æ1 1ö 1 × q1q2 ç  ÷ ÷ çr 4pe0 è 2 r1 ø Þ W =  9 ´ 109 ´ 2.4 ´ 106 ´ (  4.3 ´ 106 ) 1 ö æ 1 ç ÷ è 0.25 2 0.15 ø =
W =  0.356 mJ 1 é q1q2 q2q3 q3q1 ù 26. (a) U = ×ê + + ú 4pe0 êë r12 r23 r31 úû é 4 ´ 109 ´ (  3 ´ 109 ) = 9 ´ 109 ê 0.2 êë (  3 ´ 109 ) ´ (2 ´ 109 ) + 0.1 +
4 ´ 109 ´ 2 ´ 109 ù ú 0.1 úû
U = 9 ´ 108 [  6  6 + 8] =  360 nJ (b) Let the distance of q3 from q1 is x cm. Then qq qq ù 1 é q1q2 U= ×ê + 2 3 + 3 1ú =0 4pe0 êë 0.2 0.2  x x úû é 4 ´ 109 ´ (  3 ´ 109 ) Þ 9 ´ 109 ê 20 ´ 100 2 êë (  3 ´ 109 ) ´ 2 ´ 109 + (20  x ) ´ 102
42 +
2 ´ 10
9
9
´ 4 ´ 10 ù ú =0 úû
Þ
6 6 8 + =0 10 20  x x
Þ
x = 6.43 cm
^
^
^
V =  (5 i  3 j)  (5 k ) = 0 ^
(b) r = 4 i + 3 k ^
^
V =  (5 i  3 j)  (4 i + 3 j) =  20 kV ®
^
29. E = 400 j V/m ®
^
^
(a) r = 20 j cm = (0.2 j) m ® ®
V =  E× r =  80 V ®
^
(b) r = (  0.3 j) m ® ®
V =  E× r = 120 V ®
V =0 ^
30. E = 20 i N/C ®
^
^
(a) r = (4 i + 2 j ) m
= 4.07 ´ 105 Vm. q (b) f = Þ q = e0f e0
® ®
V =  E× r =  80 V ®
^
^
(b) r = ( 2 i + 3 j ) m
= 8.85 ´ 1012 ´ 780 = 6.903 ´ 109 q = 6.903 nC
® ®
V =  E× r =  40 V 31. (a) [ A ] =
[ ML2 T 3 I1 ] [V ] = [ xy + yz + zx ] [L2 ]
(c) No. Net flux through a closed surface does not depend on position of charge.
= [ ML0 T 3 I1 ] æ ¶v ^ ¶v ^ ¶v ^ ö ® (b) E =  Ñ V =  çç i+ j+ k÷ ¶y ¶z ÷ø è ¶x ^
^
® 4 æ3 ^ ^ö 36. E = ç E0 i + E0 j÷ 5 5 è ø ®
^
S = 0.2 j m 2 =
^
=  A [( y + z ) i + ( z + x ) j + ( x + y) k ] (c) at (1m, 1m, 1m)
V  0 =  (40 + 60) V =  100
q = e0f = 8.8 ´ 1012 ´ 360 = 3.18 ´ 109 C = 3.186 nC 3.60 ´ 106 q 36. (a) f = =e0 8.85 ´ 1012
^
(c) r = (0.15 k ) ®
^
¶v =  ( Ay  2 Bx ) ¶x ¶V Ey =  ( Ax + C ) ¶y ¶V Ez = =0 ¶Z (b) For E = 0 Ex = 0 and Ey = 0 Hence, Ey = 0 Ax + C = 0 C x=A Ex = 0 C Ay  2 B æç  ö÷ = 0 è Aø 2 BC y=A2 2 BC ö æ C Hence, E is zero at ç  , ÷. A2 ø è A q 34. f = e0
^
^
^
33. (a) Ex = 
(a) r = 5 k
®
^
=  20 ( i + j + k )
Þ Þ
® ®
^
^
® ®
28. V =  E× r
^
^
32. VB  V0 =  E× r
27. Let Q be the third charge 1 é q2 qQ qQ ù U= ×ê + + ú =0 4pe0 êë d d d úû q Q=2
®
^
E =  10 ( 2 i + 2 j + 2 k )
x ´ 102
\
® ®
f = E× S =
1^ 2 jm 5
4 Nm 2/C 25
43 4 ´ 2.0 ´ 103 Nm 2/C 25 = 320 Nm 2/C
1 p Þa= 2 3 R = b tan a = 3 b
cos a =
=
®
37. E =
E0 x ^ i x1 = 0 l ®
E1 = 0 x2 = a ® E a^ E2 = 0 i l Flux entering the surface f1 = 0 Flux leaving the surface E a3 f2 = E2a 2 = 0 l 5 ´ 103 ´ (1 ´ 102 )3 = 2 ´ 102 = 0.25 Nm 2/C q Net flux, f = e0 q f2  f1 = e0
Hence proved ® ® ^ ^ ® ^ ^ ^ 39. E =  B i + C j  D k, S1 =  L2 i, S2 =  L2 j, ® ® ® ^ ^ ^ S4 =  L2 k, S4 =  L2 j, S3 =  L2 i, ® ^ S6 =  L2 k \
®® ®® f1 = E × S1 = BL2, f2 = E × S2 = CL2, ®® ®® f3 = E × S3 =  BL2, f4 = E × S4 =  CL2, ®® ®® f5 = E × S5 =  DL2, f6 = E × S6 = DL2
(b) f = f1 + f2 + f3 + f4 + f5 + f6 = 0 40. W = 2p (1  cos f)
q
a
–q
q = e0 ( f2  f1 ) = 8.85 ´ 1012 ´ 0.25 = 2.21 ´ 1012 C = 2.21 pC 38. Consider the charge is placed at vertex of the cone of height b and radius R.
R a Q
Let a be the semivertical angle of the cone, then solid angle subtended by the cone. W = 2p (1  cos a ) Flux passing through cone W f= × ftotal 4p 1 But (Given) f = ftotal 4 W=p 2p (1  cos a ) = p 1 1  cos a = 2
f1 = f2 =
2p (1  cos a ) q W × cot ftotal = 4p 4p e0 1 = (1  cos a ) 2e0
Total flux through the ring f = f1 + f2 q q æç = (1  cos a ) = 1e0 e0 ç è
ö ÷ 2 2÷ R +l ø l
41. From the given equation, y
a q O
45°
a
radius of hemisphere = a and its centre is at ( a, 0, 0)
x
44 æ q A + q B + qC ö ç ÷ b c ø è ö s æ a2 ç =  b + c÷ ç ÷ e0 è b ø 1 æ q A + q B + qC ö VC = ç ÷ c 4pe0 è ø
1 VB = 4pe0
1 ö æ W = 2p (1  cos a ) = 2p ç1 ÷ 2ø è 1 ö æ 2p ç1 ÷ W 2ø q f= ftotal = è × 4p 4p e0 q æ 1 ö f= ç1 ÷ 2 e0 è 2ø
=
2
2
42. q1 = s (4pr ), q2 = s (4p R ) q2
s e0
æ a 2  b2 + c 2 ö ç ÷ ç ÷ c è ø
44. (a) As charge Q is placed at the centre of the sphere, charge  Q will appear on the inner surface and Q on its outer surface.
q1
+Q
r
A
B
–Q
R Q
Q 4 p ( r + R2 ) 1 æ q1 q2 ö VA = ×ç + ÷ 4pe0 è r Rø
But, q1 + q2 = Q Þ s =
2
2 2 1 æ s (4pr ) s (4p R ) ö ÷ = ×ç + ÷ r R 4pe0 çè ø
1 Q ( r + R) × 4pe0 r 2 + R2 1 æ q1 + q2 ö 1 Q VB = ×ç × ÷= 4pe0 è R ø 4pe0 R =
2
2
Hence,
sin =
and
sout
Q
4pa 2 Q = 4pa 2
(b) Entire charge inside the sphere appears on its outer surface, hence sin = 
Q+q Q and sout = 4pa 2 4pa 2
(c) In case (a) E=
43. q A = s (4pa ), q B =  s (4pb )
1 Q × 4pe0 x 2
+Q
s
–Q
–s s
Q
a
x b
A B C
c
qC = s (4pc 2 ) 1 æ q A q B qC ö VA = + + ç ÷ 4pe0 è a b c ø s = ( a + b + c) e0
In case (b) E = E1 + E2 E1 = Field due to charge Q. E2 = Field due to charge on shell. 1 Q E= × 4pe0 x 2 for x < a As field due to shell is zero for x < a.
45 and
E=
1 Q+q , for x > a × 4pe0 x 2
(d) Let Q A be the charge on inner sphere. –Q
45. Let Q be the charge on the shell B, b —q c –q
C
QA
b —q –q c –b —q c
R C
q
3R
A
B a
A
B
b
1 é q + Q  qù + ê ú =0 c û 4pe0 ë b æ b  cö Q = qç ÷ è c ø
VB =
Charge distribution on different surfaces is shown in figure. 46. (a) Let E1 and E2 be the electric field at P due to inner shell and outer shell respectively.
P r
47. (a) At r = R 1 é Q  2Q 3Q ù V= ×ê + + ú 4pe0 ë R 2R 3 Rû 1 Q = × 4pe0 R At r = 3 R V=
+2Q B
R 3R
1 é QA Q ù VA = ×ê =0 4pe0 ë R 3 R úû Q QA = 3
A –Q
1 é Q  2Q + 3Q ù ×ê ú 4pe0 ë 3R û 1 2Q = × 4pe9 3 R
(b) Let E1, E 2 and E 3 be the electric fields at 5 r = R due to shells A, B and C 2 respectively. C
1 2Q and E2 = 0 Now, E1 = × 4pe0 r 1 2Q \ E = E1 + E2 = E1 = × 4pe0 r 1 é 2Q Q ù (b) VA = × 4pe0 êë R 3 R úû 1 é 2Q  Q ù VB = ×ê ú 4pe0 ë 3 R û 1 é 2Q 2Q ù VA  VB = × 4pe0 êë R 3 R úû 1 4Q = × 4pe0 3 R
(c) Whenever two concentric conducting spheres are joined by a conducting wire entire charge flows to the outer sphere. \
Q A = 0, Q B = 0
B
3R A R 2R
\
–2Q 3Q Q
1 Q × 4pe0 æ 5 ö 2 ç R÷ è2 ø 1 4Q = × 4pe0 25 R 1 2Q E2 = × 4pe0 æ 5 ö 2 ç R÷ è2 ø E1 =
(outwards)
46 1 8Q = × 4pe0 25 R
(inward)
E3 = 0 5 Net field at r = R 2 E = E2 + E1 =
1 4Q × 4pe0 25 R
(inward)
(c) Total electrostatic energy of system is the sum of selfenergy of three shell and the energy of all possible pairs i.e., 2 1 é Q 2 (  2Q ) (3Q )2 U= ×ê + + 4pe0 êë 2 R 2 ´ 2 R 2 ´ 3 R
Q (  2Q ) (  2Q ) ´ 3Q Q ´ 3Q ù + + ú 2R 3R 3R û 1 Q U= × 4pe0 R +
(d) Let q charge flows from innermost shell to outermost shell on connecting them with a conducting wire. 3Q +q –2Q 2R Q2
R
3R
1 é Q  q  2Q 3Q + q ù ×ê + + ú 4pe0 ë R 2R 3R û 1 3Q  2q = × 4pe0 3R
VA =
1 é Q  q  2Q + 3Q + q ù VB = ×ê ú 4pe0 ë 3R û 1 2Q = × 4pe0 3 R But VA = VB 1 3Q  2q 1 2Q × = × 4pe0 3R 4pe0 3 R Q Þ q= 2 \Charge on innermost shell = Q  q =
Q 2
and charge on outermost shell= 3Q + q =
7Q 2
1 3Q  2q × 4pe0 3R 1 2Q = × 4pe0 3 R
and
VA =
(c) In this case E1 =
1 × 4pe0
Q 2
5 2 R æç Rö÷ è2 ø 2Q 1 (outward) = × 4pe0 25 R 1 2Q E2 = × 4pe0 æ 5 ö 2 ç R÷ è2 ø 1 8Q (inward) = × 4pe0 25 R 5 Net electric field at r = R 2 1 6Q (inward) E = E2  E1 = × 4pe0 25 R
Objective Questions (Level1) 1. f = E × A Units of f = N/C ´ m 2 = N  m 2/C or V/m ´ m 2 = V  m 2. Net force F ¢ = mg  qE qE g¢ = g m l T ¢ = 2p >T g1
3. Electric lines of force terminate at negative charge. 1 q2 4. F = × 4pe0 l2 Initial PE Ui = Find PE
1 æ q2 q2 q2 ö ÷ = 3 Fl ×ç + + 4pe0 çè l l l ÷ø
47 Uf =
1 æ q2 q2 q2 ö 3 ÷ = Fl ×ç + + 4pe0 çè 2l 2l 2l ÷ø 2 3 W = Uf  Ui =  Fl 2
W = Uf  Ui = =
5. KE = qV
V1 : V2 : V3 =
Þ
q2 [4  2 2 ] J 4pe0a
8. Potential at point P
1 mv2 = qV 2 2qV v= m q1V1 q2V2 q3V3 V1 : V2 : V3 = : : m1 m2 m3
Þ
2R P
R
e ´1 e ´2 2e ´ 4 : : m 2m 4m
3R
1 q1 + q2 × 4pe0 3R
V=
V1 : V : V3 = 1 : 1 : 2 6. V =
2 q2 q2 + [4  2 ] 4pe0 a 4pe0 a
1 q × 4pe0 r
9 ´ 109 ´ 3 ´ 106 = 9000 3R R =1m
Þ
9. As distance of every point of ring from axis is same. kq , But x = 2 R V= R2 + x 2
4r
= V¢ = 1 7. Ui = 4pe0
1 2q V × = 4pe0 4r 2
10. For equilibrium, w
é q2 ù q2 ´4 + ´ 2ú ê2a êë a úû –q
+q
r g m
–q
+q
A –q
+q
=
+q
–q
 q2 [4  2 ] 4pe0 a 2
2
2
ù 1 é q q q × ê´2 + ´2 ´ 2ú 4pe0 êë a a 2 úû 2 q2 =4pe0 a
Uf =
kq 3R
30°
B
q sin
mg
Fe
mg cos q
mg sin 30° = Fe 1 q2 mg sin 30° = × 4pe0 r 2 r=q
1 4pe0 mg sin 30°
= 2.0 ´ 106
91 ´ 109 0.1 ´ 10 ´
» 20 cm
1 2
48 11. Net force on C = 0
13. Data is not sufficient.
A
B
14. If the charges have opposite sign, electric field is zero on the left of smaller charge. 15. Net field is only due to charge on C. Aq
O
C
D
FCO
FCD
F
qC
FCA
FCB E
2
(2 2  1) Q 2
FCB =
1 × 4pe0
FCD =
2 2 1 (2 2  1) Q × 4pe0 a2 2
FCA =
1 (2 2  1) Q × 4pe0 2a 2
FCO =
2 1 2 (2 2  1) Q × 4pe0 a2
D
17. Charge distribution is shown in figure. +8Q –2Q +2Q –4Q +4Q
(2 2  1)Q é (2 2  1)Q (2 2  1)Q + ê 2 a2 2 êë (2 2  1)Q 2q ù + + ú =0 1 úû 2
7Q 4 s 12. E = e0
q
16. On touching two spheres, equal charge will appear on both the spheres and for a given total charge, force between two spheres is maximum if charges on them are equal.
Net force on C F = FCA + FCO + FCB cos 45° + FCO cos 45° =
q
1 q q E= × = 2 4pe0 (2a ) 16pe0 a 2
a2
2
qB
q=
F = eF =
18. V = es e0
Acceleration of proton se F a= = m me0 1 s = ut + t 2 a u =0 t=
=
25 me0 25 = a se 2 ´ 0.1 ´ 1.67 ´ 1027 ´ 8.8 ´ 1012
= 2 2 ms
2.21 ´ 109 ´ 1.6 ´ 1019
1 q × 4pe0 r
If drops coalesce, total volume remains conserved, 4 4 pR3 = 1000 ´ pr 3 3 3 R = 10r 1 1000q V¢ = × = 10V 4pe0 10q 19. VA =
1 éq Qù × + ú 4pe0 êë r Rû
B
q A R
r
49 1 é q + Qù ×ê ú 4pe0 ë R û q é1 1 ù VA  VB = × 4pe0 êë r R úû VB =
O R
P
r
VA  VB µ q \ If q is doubled, VA  VB will become double. 20. Charge distribution is shown in figure.
1 q 1 1 3q × = × × 4pe0 R + r 2 4pe0 2 R 1 3° = R + r 4R
2Q –3Q 3Q –Q Q
4R = 3 R + 3r R r= 3 25. Net charge on any dipole is zero. 26. For net force to be zero.
®® ^ ^ ^ 21. f = E × S = (5 i + 2 j) × ( i ) = 5 Vm. 22. FDA = FDC =
q
1 Qq × 4pe0 a 2
T
T cos q
q qE
q
A Q
7 sin q
B mg
Q
mg cos q qE or T sin q = qE Þ T = sin q V1 1 q 27. E1 = × = 4pe0 a a V2 1 q E2 = = b 4pe0 b2 T cos q = mg Þ T =
q
FDC
Q C
D
FDB FDA
FDB =
1 q2 × 2 4pe0 2a
Net force on charge at D ® F0 = FDB + FDA cos 45° + FDB cos 45° = 0 Þ
1 q 4pe0 × 2 4 a
Q Qù éq + ê2 + ú =0 2 2û ë q = 2 2Q
23. As VB = 0, Total charge inside B must be zero and hence charge on its outer surface is zero and on its inner surface is  q. 1 24. Vp = V0
But
Þ
E1 = E2 V1 V2 = a b V1 a = V2 b
28. Electric field on equatorial lines of dipole is opposite to dipole moment. 29. Potential difference between two concentric spheres is independent of charge on outer sphere. q 1 30. E = × 2 4pe0 r
50 1 q V= × = Er 4pe0 r V 3000 r =½ ½ = =6m ½ E ½ 500 6 ´ (  3000) q = 4pe0rV = =  2 mC 9 ´ 109 31. F1 = F2
qq 1 q1q2 1 × = × 1 2 4pe0 r12 4p Ke0 r22 r 50 r2 = 1 = = 10 5 m K 5
1 q VR E= × = 2 4pe0 r 2 r 36. When outer sphere is earthed field between the region of two spheres in nonzero and is zero in all other regions. ®® 37. W = F × s = qEs cos q W 4 E= = = 20 N/C qs cos q 0.2 ´ 2 ´ cos 60° 38. V1 =
1 éê Q × 4pe0 ê R ë
» 22.3 m
ù ú d 2 + R2 úû Q
Q
32. Electric field at a distance r from infinite line charge l E= 2pe0r
–Q
R
R
dV =  E dr V2
òV
1
b
l 1 × ln 2pe0 2
W = q( V2  V1 ) =
ql 1 ln 2pe0 2
34. r = (4  1)2 + (2  2)2 + (0  4)2 = 5 m V=
1 q 9 ´ 10 ´ 2 ´ 10 × = 4pe0 r 5
2
ù 1 éê Q Q ú + 2 2 ê R ú . d + R 4pe0 ë û é ù 1 ê 2Q 2Q ú V1  V2 = 2 2ú ê R . d + R 4pe0 ë û ù é Q ê 1 1 ú V1  V2 = + 2 ê R . d + R2 úû 4pe ë V2 =
33. As negative charge is at less distance from the line charge, it is attracted towards the line charge. 9
d
1
a
V2  V1 =
Þ
r
dV =  ò E dr
8
= 36 V
(b) and (c) are wrong. 1 q 35. V = × 4pe0 R
0
39. Electric field inside a hollow sphere is always zero. ®® ®® 40. W = F × r = q E × r ^
^
^
^
= q ( E1 i + E2 j)  ( a i + b j) = q ( aE1 + bE2 )
At a distance r from the centre,
JEE Corner Assertion and Reason 1. Negative charge always moved towards increasing potential. On moving from A to B potential energy of negative charge decreases hence its KE increases. 1 q1q2 2. U = × 4pe0 r
If q1 and q2 have opposite sign, U decreases with decrease in r. dU F=Þ work done by conservative force dr always decreases PE. dV 3. E = =  (10) = 10 V/m along xaxis. dr 1 q 4. V = × 4pe0 R
51 Inside the solid sphere. 1 qr E= × 4pe0 R3 R at r = 2 1 q V E= × = 4pe0 2 R2 2 R
^
^
^
^
^
^
VA =  (4 i + 4 j) × (4 i ) =  16 V VB =  (4 i + 4 j) × (4 i ) =  16 V VA = VB Hence, Assertion is false. 7. In the line going A and B, the energy of third charge is minimum at centre.
Assertion is correct. Reason is false as electric field inside the sphere is directly proportional to distance from centre but not outside it. 5. Gauss theorem is valid only for closed surface but electric flux can be obtained for any surface. 6. Let V0 = Potential at origin,
8. Dipole has both negative and positive charges hence work done is not positive. 9. Charge outside a closed surface can produce electric field but cannot produce flux. 1 qx a is maximum at x = 10. E = × 2 2 3/ 2 4pe0 ( x + a ) 2 1 q is maximum at x = 0. But V = × 4pe0 a 2 + x 2
Objective Questions (Level2) 1. Electrostatic force always acts along the line joining the two charges, hence net torque on charge + 2q is always zero. U
30° +Q
+2q
R
As net torque is zero angular momentum of charge remains conserved. Initial angular momentum Li = m ( V sin 30° ) R When the separation between the charges become minimum, direction of motion of charge + 2q become perpendicular to the line joining the charges. V'
r
+2q
® ^ ^ ^ ® 2. v1 = v j, v 2 = 2v cos 30° i + 2v sin 30° j ^
As velocity along yaxis is unchanged, electric field along xaxis is zero. For motion along xaxis, vx2  ux2 = 2a x ( x  x0 ) ( 3v)2  0 3v2 = 2a 2a 3 mv2 Fx = ma x = 2a ® 3 mv2 ^ F = i 2a ® ® F =eE
ax =
Þ Also,
® 3 mv2 ^ E =i 2ea Rate of work done by electric field at B ® ® æ 3 mv ^ö ^ ^ P = F× v = ç i ÷ × ( 3v i + v j) 2 a è ø
+1q
=
\find angular momentum Lf = mv¢ r =
mvr 2
By conservation of angular momentum 3 Li = Lf Þ r = R 2
^
= 3 i+vj
3 3 mv3 2a
3. Electric field is always possible, hence a must be positive and b must be negative. +q a
–Q b
52 4. The system can be assumed as a combination of three identical dipoles as shown in figure. Here, P1 = P2 = P3 = Q (2a )
2
Uf =
2
1 q q × = 4pe0 2 ´ 3 a 24pe0 a
Heat produced = Ui = Uf =
® P2
=
q2 8pe0a
kq2 2a
7. Let Q charge flows to C ® P1
–q
D
60° 60°
4a
Q q
® P3
Net dipole moment of the system P = P1 + P2 cos 60° + P3 cos 60° = 2 p = 4 Qa Electric field on equatorial lines of short dipole is given by P 1 E= × 4pe0 x 3 Qa 1 4 Qa = × = 4pe0 x 3 pe0x 3 5. Potential at centre will be same as potential at the surface of inner shell i.e., 10 V. 6. Initial charge distribution is shown in figure, Initial energy of system +q
–q +q a
2q
2 1 é q2 (  q) q2 ×ê + + 4pe0 êë 2a 2 ´ 2a 2 ´ 3 a
+
q (  q) q ( q) q (  q) ù + + ú 2a 3a 3a û
5a 2 = 48pe0 a When switch S is closed, entire charge flows to the outer surface of outer shell,
A
B
1 é q + Q (  q) ù ×ê + ú =0 4pe0 ë 3 a 4a û q Q=4  qù 1 é q Q \ VA = × + + 4pe0 êë 2a 3 a 4a úû 1 é q ù = ê ú 4pe0 êë s a úû 1 q × VA  VC = 4pe0 s a kq = sa VC =
1 q × 4pe0 R
1 3q × 4pe0 2 R 1 q \ VC  VS = × 4pe0 2 R 4 p dR 3 d R 2 1 3 = × = 4 pe0 2R s E0 and
3a
2a
C
8. VS =
S
Ui =
q
3a
VC =
9. As particle comes to rest, force must be repulsive, hence it is positively charged. Again on moving down its KE first increases than decreases, PE will first decrease than increase.
53 10. (1) is correct as the points having zero potential are close to Q2, Q2 < Q1. Again as potential near Q1 is positive, Q1 is positive, hence (2) is correct. At point A and B potential is zero not field, hence they may or may not be equilibrium point. Hence (3) is wrong. At point C potential is minimum, Q positive charge placed at this point will have unstable equilibrium but a negative charge will be in stable equilibrium at this position. Hence, (4) is wrong. 11. V1 is always negative and V2 is always positive. 12. Electric field between the two points is positive near q1 and negative near q2, hence q1 is positive and q2 is negative. Again neutral point is closer to q1, hence q1 < q2. 13. Electric field due to a conductor does not depend on position of charge inside it. ® ^ ^ 14. E = 400 cos 45° i + 4000 sin 45° j ^
^
= 200 2 ( i + j)
qE l m
v= At point B T = qE +
17. Velocity of particle at any instant V O
E
(x, 0)
qE t m \ L = mvr = qE x0t Hence, angular momentum of the particle increases with time. V = at =
18. By work energy theorem W = DK Þ Þ Þ
1 mv2 2 æ 1 Q 1 3Q ö 1 ÷ =  mv2 qç × × ç 4pe R 4pe 2 R ÷ 2 0 0 ø è Qq u= 4pe0mR q ( VS  VC ) = 0 
19. Potential at the centre of negatively charged ring
® ® ® ® V A  V B =  E × rAB
–Q ^
mv2 = 2 qE r
^
^
^
=  200 2 ( i + j) × (2 j  3 i ) ´ 10
2
+Q 2R
= 2 2 V » 2.8 V 15. Potential difference between two concentric spherical shells does not depend on charge of outer sphere. Hence, V A ¢  VB ¢ = VA  VB But VB ¢ = 0 \ VA ¢ = VA  VB. 16. By work energy theorem, ® E
A
l q l(1–cosq)
B
Work done by electric field = charge is KE 1 qE l (1  cos q) = mv2  0 2
R
R 1
Ö3R
2
1 æ Q Q ö V1 = ×ç + ÷ 4pe0 è R 2Rø Q =8pe0 R Potential at the centre of positively charged ring Q ö 1 æQ V2 = ×ç ÷ 4pe0 è R 2 R ø Q = 8pe0 R Kinetic energy required = Work done required Q = q ( V2  V1 ) = 4pe0 R
54 20. Ex = Ey = 
Vx 2  Vx1 x2  x1 Vy 3  Vy1 y3  y1 ^
=
16  4 = 3 V/m 2 2
=
12  4 =  4 V/m 4 2
^
^
2
24. T  mg cos q  qE sin q =
E
^
\E = Ex i + Ey j = (3 i  4 j) V/m. q
21. Consider a point P ( x, y) where potential is zero. Now, VP
q
2Q
Q (+3a,0)
O
ö  2Q 1 æç Q ÷ + 2 2 2 2 ÷ 4pe0 ç (3 a  x ) + y ( ) 3 x + a + y ø è =0 ( x + 3 a )2 + y2 = 4 [(3 a  x )2 + y2 ] Þ 3 x 2 + 3 y2  30ax + 27a 2 = 0 Þ Þ x 2 + y2  10ax + 9a 2 = 0 The equation represents a circle with radius VP =
2
10a ö 2 = æç ÷  9a = 4a è 2 ø 10 and centre at æç a, 0ö÷ = (5a, 0) è2 ø Clearly points x = a and x = 9a lie on this circle. 22. Work done = qEy = Charge in KE 1 Hence, K f = mv2 + qEy 2 All other statements are correct. 23. Electrostatic force of attraction provides necessary centripetal force. mv2 lq ie, = r 2pe0r Þ T=
V=
T qE
mg
P(x,y)
(–3a,0)
mv l
lq 2pe0m
2pe0m 2pr m = 2pr = 2pr V 2 K lq lq
Tension will be minimum when velocity is minimum. Minimum possible in the string is zero. mv2 ie, =  ( mg cos q + qE sin q) l Diff. both sides w.r.t. q 2mv dv …(i) = mg sin q  qE sin q l dq For minima or maxima dv qE = 0 Þ q = tan 1 dq mg 1 qE or p + tan mg Differentiating Eq. (i) again, 2 2mv d 2v 2m æ dv ö + × ç ÷ = mg sin q + qE sin q 2 l dq l è dq ø d 2v qE \ 2 = + ve for q = tan 1 mg dq 1 qE and –ve for q = p + tan mg 25. q A = s (4pa 2 ), q B =  s (4pb2 ) and qC =  s (4pc 2 ) 1 æ q A + q B qC ö VB = ×ç + ÷ b c ø 4pe0 è ù s é a2 ê 2  b + cú e0 êë b úû 2 2 2 ù q 1 éq q q2 q2 26. Ui = ×ê + ´ 2ú 4pe0 êë a a a a 2a ûú  2 q2 = 4pe0a =
Uf =
1 q2 × 4pe0 a
W = Uf  Ui =
1 q2 × ( 2 + 1) 4pe0 a
55 27. q A = s (4pa 2 ), q B =  s (4pb2 ), qC = s (4pc 2 ) Given, VA = VC 1 æ q A q B qC ö 1 æ q A + q B + qC ö ×ç ×ç + + ÷ ÷= b c ø 4pe0 è c 4pe0 è a ø a 2  b2 Þ ab+ c= +c c Þ a+ b=c 28. Potential at minimum at midpoint in the region between two charges, and is always positive. 1 q2 29. Ui = × =U 4pe0 r Uf =
1 q2 × ´3 = 3 U 4pe0 r
\W = Uf  Ui = 2 U 30. Loss of KE = Gain in PE 1 1 qQ mv2 = × 2 4pe0 r 1 rµ 2 v 31. When the spheres are in air
On dividing Eq. (ii) by Eq. (i), Fe ¢ mg  FB = Fe mg FB 1 0.8 1 =1 =1 = K mg 1.6 2 K =2 é ù 1 êq q 32. VP = × ´2 ´ 2ú ú 4pe0 ê a a 2 + b2 ë û 2 2 é 2q ê a + b  a ùú = × 4pe0 ê a a 2 + b2 ú û ë 2 1/ 2 ù é æ ö ê a ç1 + b ÷  aú ú a 2 ÷ø 2q ê çè 2q b2 = = × 3 ú ê 4pe0 ê a a 2 + b2 ú 4pe0 a ú ê úû êë [As b << a] 1 5q 33. In any case electric field at origin is × 4pe0 r 2 1 5q along xaxis and along yaxis. × 4pe0 r 2 2
O
34. < u > =
q T Tcosq q
Tsinq
Fe
Fe mg
mg
T cos q = mg T sin q = Fe …(i) Fe = mg tan q \ When the spheres are immersed in liquid O
1 1 æ 1 q ö ÷ e0 E2 = e0 ç × 2 2 çè 4pe0 R2 ÷ø é 9 ´ 109 ´ 1 ´ 109 ù ú 1 ê 9 = e0 ê ú 2 ê 12 ú êë úû e0 3 J/m = 2
35. If Q is initial charge on B 1 Q then, VA  VB = × =V 4pe0 b Now, if A is earthed, let charge q moves on A from ground, then Q
q
q
T' T'
T'cosq FB
q
F'e mg
\
F'e
Tsinq mg
F1 ¢ = T ¢sin q g  FB = T ¢cos q Fe ¢ = ( mg  FB ) tan q
1 æ q Qö ×ç + ÷ =0 4pe0 è a bø a q= Q b
VA = …(ii)
56 1 q+Q VB = × 4pe0 b 1 Qæ a a = × ç1  ö÷ = V æç1  ö÷ 4pe0 b è bø bø è ® æ ¶v ^ 36. E =  çç i+ è ¶x æ2 ^ =ç i+ è 1 ^
^
¶v ^ ¶v ^ ö j+ k÷ ¶y ¶z ÷ø 2 ^ 2 ^ö j+ k÷ 1 1 ø ^
= 2 ( i + j + k ) N/C If VP is potential at P, then ®® VP  V0 =  E × r ^
^
^
^
^
^
VP  10 =  2( i + j + k ) × ( i + j + k ) =  6
42. Let charge q¢ flows through the switch to the ground, then 1 é Q  q¢ Q ù ×ê ú =0 4pe0 êë r 2 r úû 1 q¢ = Q 2 43. After n steps 1 1 q¢ = n Q and q = n  1 Q 2 2 1 æ q¢ qö ÷ =0 ×ç + \ VA = 4pe0 çè r 2 r ÷ø 1 q¢ + q × VB = 4pe0 2 r =
VP = 4 V 37. On touching two spheres, charge is equally divided among them, then due to induction a q charge æç  ö÷ appears on the earthed sphere. 2 è ø 38. Negative charge will induce on the conductor near P. ì ï 0 for r < rA ï kQ P ï 39. í for rP < r < rB ï r ï k(Q A  Q B ) for r > r B ï r î AsQ B > Q A E is –ve for r > rB. ® æ ¶v ^ ¶v ^ö 40. E =  çç i+ j÷ ¶y ÷ø è ¶x ^
^
= k ( y i + x j) ® E = k y2 + x 2 = kr 41. Let charge on outer shell becomes q. B 2r
A r
S2
S1
1 2
n+1
é Q ù ê ú êë 4pe0r úû
44. Consider a spherical Gaussian surface of radius r ( < R) and concentric with the sphere,
O
1 æQ + qö ÷ =0 ×ç 4pe0 çè 2 r ÷ø q = Q
R
Charge on a small sphere of radius r dq = d dV = 4pr 2d dr æ r3 ö = r pd0 ç r 2  ÷ dr ç R ÷ø è Total charge inside the Gaussian surface, r æ r3 ö q = 4p d0 ò ç r 2  ÷ dr 0 ç R ÷ø è é r3 r4 ù = 4pd0 ê ú êë 3 4 R úû 1 q d ér r2 ù × 2 = 0ê \E = ú 4pe0 r e0 êë 3 4 R úû 45. Total charge inside the surface. é R3 R3 ù 1 3 Q = 4pd0 ê ú = pd0 R r úû 3 êë 3 E=
VB =
r
d R3 1 Q × 2 = 0 2 4pe0 r 12e0r
ér r2 ù ú ê êë 3 4 R úû For maximum intensity of electric field
46. E =
d0 e0
57 é1  r ù = 0 ê3 2Rú ë û 2 Þ r= R 3 d d 2E =  0 =  ve, 2 Re0 dr 2 dE d0 = dr e0
q1 =  Q Now, if A is earthed q1 q2
2 hence E is maximum at r = R. 3 é 2l æ 2 R ö 2 ù ê ç ÷ ú r 3 ø ú r0 R 47. Emax = 0 ê 3  è = e0 ê 3 4 R ú qe0 ê ú êë úû
1 æ q2 q1 ö ×ç + ÷ =0 4pe0 è a bø a a q2 =  q1 = Q b b
qA =
48. Potential difference between two concentric spheres do not depend on the charge on outer sphere. 49. When outer sphere B is earthed q1 B
Q
1 æ Q + q1 ö ×ç ÷ =0 4pe0 è b ø
VB =
A
50. When connected by conducting wires, entire charge from inner sphere flows to the outer sphere, ie, a q3 = q1 + q2 æç  1ö÷Q èb ø ab = Q b
More than One Correct Options 1. Before earthing the surface B, 1 æ qA qB ö VA = ×ç + ÷ =2V 4pe0 è R 2 R ø 1 æ qA + qB ö 3 VB = ×ç ÷= V 4pe0 è 2 R ø 2 qA 1 Þ = qB 2 On earthing the sphere B, qA ¢ = qA 1 qA ¢ + qB ¢ × =0 VB = 4pe0 2R Þ
qB ¢ =  qA ¢ qA ¢ = 1 qB ¢
As potential difference does not depend on charge on outer sphere, V V A ¢  VB ¢ = VA  VB = 2 1 VA ¢ = V 2
2. For the motion of particle y
E
m x
ux = 0, vx = v, a x =
qE , a y =  g, m
x0 = 0, y0 = 0 x = x 0 + ux t + x=
1 ax t 2 2
qE 2 t 2m
…(i)
y = y0 + u y t + = ut 
1 a yt2 2
1 2 gt 2
At the end of motion t = T , y = 0, x = R
…(ii)
58 From Eq. (ii),
4. For all charges to be in equilibrium, force experienced by either charge must be zero ie., force due to other two charges must be equal and opposite.
1 0 = æçu  gT ö÷ T 2 è ø 2u 2 ´ 10 T= = =2s g 10
q1
From Eq. (i),
q2
q3
qE 2 T 2m 1 ´ 103 ´ 104 = ´ 4 = 10 m 2´g
Hence all the charges must be collinear, charges q1, and q3 must have same sign and q2 must have opposite sign, q2 must have maximum magnitude. Such on equilibrium is always unstable.
Now, v2y  u 2y = 2a y ( y  y0 ) At highest point (i.e., y = H), vy = 0 0  (10)2 =  2 ´ 10 ( H  0) H =5m
5. Flux through any closed surface depends only on charge inside the surface but electric field at any point on the surface depends on charges inside as well as outside the surface.
R=
Q2
3. Let R be the radius of the sphere 1 q × V1 = 4pe0 R + r1 q ´ 109 ´ q
= 100 ( R + S ) ´ 102 1 q × V2 = 4pe0 R + r2
Þ
Q1
…(i)
9
9 ´ 10 ´ q
Þ
( R + 10) ´ 102
= 75
…(ii)
6. As net charge on an electric dipole is zero, net flux through the sphere is zero. But electric field at any point due to a dipole cannot be zero.
On solving, R = 10 cm,
7. Gauss’s law gives total electric field and flux due to all charges.
5 50 ´ 109 C = ´ 1010 C 3 3 Electric field on surface, 5 9 ´ 109 ´ ´ 109 1 q 3 E= × = 4pe0 R2 (10 ´ 102 )
8. If two concentric spheres carry equal and opposite charges, Electric field is nonzero only in the region between two sphere and potential is is zero only outside both the spheres.
q=
= 1500 V/m Potential at surface, 1 q V= × 4pe0 R
5 ´ 109 3 10 ´ 102
9 ´ 109 ´
= 150 V Potential at Centre 3 VC = VS = 225 V 2
9. As force on the rod due to electric field is towards right, force on the rod due to hinge must be left. The equilibrium is clearly neutral. 10. If moved along perpendicular bisector, for all identical charges, electrostatic potential energy is maximum at mid point and if moved along the line joining the particles, electrostatic potential energy is minimum at the midpoint.
59
Match the Columns 1. (a ® s), (b ® q), (c ® r), (d ® p). If charge at B is removed B
® E
® E
at r = 2 R ® E
® E F
Vout
C
A
11 V 8 1 q = × 4pe0 r
=
1 q V × = 4pe0 2 R 2 1 qr Ein = × 3 4pe0 R
V2 =
D
E
R 2 1 q V V E1 = × = = 2 4pe0 2 R 2R 2 1 q Eout = × 4pe0 r 2
at r = Enet = ED cos 30° + EE cos 30° = 3E If charge at C is removed Enet = ED cos 60° + Ef cos 60° =E If charge at D is removed ® ® ® E net = 0 and E B =  E E and
® ® EE =  EE
If charge at B and C both are removed, Enet = EE + ED cos 60° + EF cos 60° = 2E 2. (a ® q), (b ® p), (c ® s), (d ® r). ®® V =  E× r If
® ^ r = 4 i, V =  8 V,
If
® ^ r = 4i, V = 8V
If
® ^ r = 4 j, V =  16 V,
If
® ^ r =  4 j, V = 16 V
3. For a solid sphere q 1 Vin = × (3 R2  r 2 ) 4pe0 2 R3 R at r= 2 q æ R2 ö 1 ç 3 R2 ÷ V1 = × 3 ç 4pe0 2 R è 4 ÷ø
(Q R = 1 m)
at r = 2 R E2 =
1 q V V × = = 2 4pe0 (2 R) 4R 4
\(a ® s), (b ® q), (c ® q), (d ® p). 4. (a ® r), (b ® q), (c ® q), (d ® s) 5. (a ® p), (b ® q), (c ® r), (d ® s) For a spherical shell, ì 0 for r < R E = ïí Kq ï r 2 for r ³ R î ì Kq for r £ R ï V = ïí R ï Kq for r ³ R ï r î For a solid sphere, ì Kqr ï 3 for r £ R E = ïí R ï Kq for r ³ R ï r2 î ì Kq 2 2 ï 2 (2 R  r ) for r £ R ï 2 R V =í Kq ï for r ³ R ï r2 î
22
Capacitors Introductory Exercise 22.1
1. C =
q [ AT ] = [C ] = V [ML2 T 3A 1 ] 1
2
4
7mC
3mC
–7mC
3mC
2
= [M L T A ] 2. False. Charge will flow if there is potential difference between the conductors. It does not depend on amount of charge present. 3. Consider the charge distribution shown in figure. 10–q
q
1
2
E4 E2
–q 3
(q – 4) 4
Hence, if q1 and q2 be charge on two plates then. q1
q2
q3
q4
E3 P
E1
Electric field at point P EP = E1 + E3  E2  E4 10  q q 4 q q = + 2e0 A 2e0 A 2e0 A 2e0 A But P lies inside conductor \ EP = 0 Þ 10  q  q + q  q + 4 = 0 Þ q = 7 mC Hence, the charge distribution is shown in figure. Sortcut Method Entire charge resides on outer surface of conductor and will be divided equally on two outer surfaces.
q1 + q2 = 3 mC 2 q  q2 = 7 mC q2 = 1 2 q  q1 q3 = 2 =  7 mC 2
q1 = q4 =
q1
q2
q3
q4
61 4. Charge distribution is shown in figure. q + q2 q =q1 = q4 = 1 2 2 q  q2 5q q2 = 1 = 2 2 q2  q1 5q q3 = =2 2 \ Charge on capacitor = Charge on inner side of positive plate.
5q 2 e A C= 0 d q 5q d V= = C 2e0 A q=
and \
Introductory Exercise 22.2 1. All the capacitors are in parallel
10V
1mF
q = CeV =
2mF
C1
C2
2 ´ 1200 = 800 mC 3
C1
C2
V1
V2
3mF C3
q1 = C1V = 1 ´ 10 = 10 mC q2 = C2V = 2 ´ 10 = 20 mC q3 = C3V = 3 ´ 10 = 30 mC 2. Potential difference across the plates of capacitor V = 10 V q = CV = 4 ´ 10 = 40 mC
V
800 q (b) V1 = = = 800 V 1 C1 800 q = = 400 V V2 = 2 C2 C1,V1
3. In the steady state capacitor behaves as open circuit. 2W
4W A
C2,V2
B I
I 6W
30V
I=
30 =3 A 6+4
Potential difference across the capacitor, VAB = 4 ´ I = 4 ´ 3 = 12 V \Charge on capacitor q = CVAB = 2 ´ 12 = 24 mC 1 1 1 1 1 4. (a) = + = + Ce C1 C2 1 2 2 Ce = mC 3
Now, if they are connected in parallel, æ C V + C2V2 ö ÷ Common potential, ç V = 1 1 ç C1 + C2 ÷ø è 1 ´ 800 ´ 2 ´ 400 1600 V = = 1+2 3 1600 3200 q1 = C1V = mC, q2 = C2V = mC 3 3 5. Common potential C V + C2V2 V= 1 2 C1 + C2 But V = 20, V2 = 0, V1 = 100 V, C1 = 100 mC 100 ´ 100 + C2 ´ 0 \ = 20 400 + C2 Þ
C2 = 400 mC
62
Introductory Exercise 22.3 1. Let q be the final charge on the capacitor, work done by battery C
R
q=
Þ
q0 (1  e  t / t ) 2
\At time t, q0 (1  e  t / t ) 2 q Charge on C2 = q0  q = 0 (1 + e t / t ) 2 Charge on C1 = q =
4. Let q be the charge on capacitor at any instant t
V
W = qV Energy stored in the capacitor 1 U = qV 2 \Energy dissipated as heat 1 H = U  W = qV = U 2 2. We have I = I0 e  t / t I0 1 = I0 e  t / t Þ e  t / t = 2 2 t = t ln 2 = 0.693 t t = 0.693 time constant. 3. Let capacitor C1 is initially charged and C2 is uncharged. R
q0 –q
+q
C
R
–q
I
E
By Kirchhoff’s voltage law q + IR = E C dq CE  q = dt RC q t dt dq òq0 CE  q = ò0 RC Þ q0 = CE (1  e  t / t ) + q0e  t / t where, t = RC 5. (a) When the switch is just closed, Capacitors behave like short circuit.
C1–(q0–q)
E
S
I q
C1
–q C2
At any instant, let charge on C2 be q, charge on C1 at that instant = q0  q By Kirchhoff’s voltage law, ( q0  q) q  IR  = 0 C C dq q0  2q Þ = dt RC q t dt dq Þ ò 0 q0  2q = ò 0 RC Þ
[ln ( q0  2q)]q0 1 = [ t ]t0 2 RC
R2 C2 R1
R3
\Initial current Ii =
E R1
(b) After a long time, i.e., in steady state, both the capacitors behaves open circuit, E If = R1 + R3 6. (a) Immediately after closing the switch, capacitor behaves as short circuit,
63 I2
A I1
I E
R1
R2 C
S
B
I1 =
\
E E and I2 = R1 R2
(c) Potential difference across the capacitors in the steady state, V =E \Energy stored in the capacitor 1 U = CE2 2 (d) After the switch is open R2
R1
E
C
(b) In the steady state, capacitor behaves as open circuit, E I1 = , I2 = 0 \ R1
Re = R1 + R2 t = R3C = ( R1 + R2 ) C
AIEEE Corner Subjective Questions (Level1) 1. C =
3 e0 A Cd 1 ´ 1 ´ 10 = ÞA= e0 d 8.85 ´ 1012
= 1.13 ´ 108 m 2 e0 A1 e0 A2 2. C1 = and C2 = d d If connected in parallel e A e A C = C1 + C2 = 0 1 + 2 2 d d e0( A1 + A2 ) e0 A = = d d where, A = A1 + A2 = effective area. Hence proved. 3. The arrangement can be considered as the combination of three different capacitors as shown in figure, where
A 2 = k3e0 A C3 = 2d / 2 2d k3e0
Therefore, the effective capacitance, C2C3 C = C1 + C2 + C3 =
e0 A æ k1 k1k3 ö ÷ ç + ç 2d è 2 k2 + k3 ÷ø
4. (a) Let the spheres A and B carry charges q and  q respectively, q
–q
a
b d
A C2 C1 C3
A 2 = k1e0 A C1 = 2d 4d A k2e0 2 = k2e0 A C2 = 2d / 2 2d k1e0
\
B
1 é q qù × 4pe0 êë a d úû 1 é q qù ×  + VB = 4pe0 ëê b d ûú VA =
Potential difference between the spheres, q é1 1 2ù V = VA  VB = + 4pe0 êë a b d úû 4pe0 q C= = V 1 + 12 a b d Hence proved.
64 (b) If d ® ¥ 4pe0 4pe0 × ab = 1 1 a+b + a b If two isolated spheres of radii a and b are connected in series, then, C1C2 C¢ = C1 + C2 C=
Let effective capacitance between A and B C AB = x As the network is infinite, C PQ = C AB = x Equivalent circuit is shown in figure, 2C
RAB = C +
B C
C
C
C A
B
6. q = CV = 7.28 ´ 25 = 182 mC 6 q 0.148 ´ 10 7. (a) V = = = 604 V 12 C 245 ´ 10 e A Cd (b) C = 0 Þ A = e0 d
C C 2C
C
C
A
B C 2C/3
5C/3 A
=
B A
B
(b) C C
C
C
C A
B C
B
C
C
C 4C/3 A
A
B
B C/3
(c) 2C A C B
P
C
2C
C
8.85 ´ 1012
2C
8. (a) E0 = 3.20 ´ 105 V/m E = 2.50 ´ 105 V/m 3.20 ´ 105 E k= 0 = = 1.28 E 2.50 ´ 105 (b) Electric field between capacitor is given by s E= e0 Þ
C Q
2C
245 ´ 1012 ´ 0.328 ´ 103
= 9.08 ´ 103 m 2 = 90.8 cm 2 6 q 0.148 ´ 10 (c) s = = = 16.3 mC/m 2 A 9.08 ´ 103
C
A
2Cx =x 2C + x
Þ 2C 2 + Cx + 2Cx = 2Cx + x 2 Þ x 2  Cx  2C 2 = 0 On solving, x = 2C or  C But x cannot be negative, Hence, x = 2C
5. (a)
C
Q
B
C¢ = C
\ Hence proved.
x
C
where, C1 = 4pe0a, C2 = 4pe0b 4pe0 × ab C¢ = \ a+b
A
P
A
¥
the
plates
s = e0E = 8.8 ´ 1012 ´ 3.20 ´ 105 = 2.832 ´ 106 C/m 2 = 2.832 mC/m 2
9. (a) q1 = C1V = 4 ´ 660 = 2640 mC q2 = C2V = 6 ´ 660 = 3960 mC
of
65 As C1 and C2 are connected in parallel, V1 = V2 = V = 660 V C1 = 4.00mF
C2 = 6.00mF
660 V
(b) When unlike plates of capacitors are connected to each other, Common potential C V  C1V1 6 ´ 660  4 ´ 660 V= 2 2 = C1 + C2 6 = 220 V q1 = C1V = 4 ´ 220 = 880 mC q2 = C2V = 6 ´ 220 = 1320 mC V 400 10. E = = = 8 ´ 104 V/m d 5 ´ 103 Energy density, 1 1 u = e0E2 = ´ 8.85 ´ 1012 ´ ( 8 ´ 104 )2 2 2 = 2.03 ´ 102 J/m 3 = 20.3 mJ/m 3 11. Dielectric strength = maximum possible electric field V V E= Þd= d E 5500 = = 3.4 ´ 104 m 1.6 ´ 107 ke A Cd C= 0 ÞA= d ke0 =
1.25 ´ 109 ´ 3.4 ´ 104 3.6 ´ 8.85 ´ 1012
= 1.3 ´ 102 m 2 = 0.013 m 2 12. Let C P and CS be the effective capacitance of parallel and series combination respectively. For parallel combination, U P = 0.19 J 1 U P = C PV 2 2 2 ´ 0.1 2U C P = 2P = Þ = 0.05 F V (2)2 = 50 mF
For series combination, US = 1.6 ´ 102 J = 0.016 J 1 US = C S V 2 2 2US 2 ´ 0.016 CS = 2 = Þ = 0.008 F (2)2 V = 8 mF Now, C P = C1 + C2 = 5 mF or C2 = (5  C1 ) mF 1 1 1 1 and = + = CS C1 C2 8 1 1 1 + = Þ C1 S  C1 8 On solving, C1 = 40 mF, C2 = 10 mF or viceversa. 13. In the given circuit, q q E + =5 VA  VB = C1 C2 A
Þ Þ \
+q
C1 –q
E
+q
C2
–q B
q q  10 + =5 106 2 ´ 106 q = 10 ´ 106 C = 10 mC q q = 10 V, V2 = =5V V1 = C1 C2
14. (a) In order to increase voltage range n times, ncapacitors must be connected in series. Hence, to increase voltage range to 500V, 5 capacitors must be connected in series. Now, effective capacitance of series combination, 10 CS = Cn = = 2 pF 5 Hence, no parallel grouping of such units is required. Hence, a series grouping of 5 such capacitors will have effective capacitance 2 pF and can withstand 500 V. (b) If n capacitors are connected in series and m such units are connected in parallel, Ve = nV mC Ce = n Here, V = 100 V Ve = 300 V V n = e =3 \ V
66 C = 10 pF Ce = 20 pF
nCe 3 ´ 20 = =6 C 10 Hence, the required arrangement is shown in figure. m=
15. Case I. V1 =
C2 V = 60 V C1 + C2
A
B
V1
V2
100 V
V2 = Þ Þ
C1 V = 40 V C1 + C2 C1 2 = C2 3 3 C2 = C1 2
Case II. V1 =
C2 = 10 V C1 + C2 + 2
2mF
Þ
C1 + 2 90 = =9 C2 10
Þ
C1 + 2 = 9C2
Þ
C1 + 2 = 9 ´
3 C1 2
25 4 C1 = 2 Þ C1 = mF 2 25 = 0.16 mF 3 C2 = C1 = 0.24 mF 2 16. (a) q = CV = 10 ´ 12 = 120 mC e A (b) C = 0 d If separation is doubled, capacitance will become half. i.e., C C¢ = 2 C q¢ = E¢ V = V = 60 mC 2 e0 A pe0r 2 (c) C = = d d If r is doubled, C will become four times, i.e., C ¢ = 4C q¢ = C ¢ V = 480 mC 17. Heat produced = Energy stored in the capacitor 1 1 H = CV 2 = ´ 450 ´ 106 ´ (295)2 2 2 = 19.58 J e A 8.85 ´ 1012 ´ 2 18. (a) C = 0 = d 5 ´ 103 Þ
= 3.54 ´ 106 F = 3.54 mF (b) q = CV = 3.54 ´ 109 ´ 10000 = 35.4 ´ 106 = 35.4 mC V 10000 (c) E = = = 2 ´ 106 V/m d 5 ´ 103 19. Given, C1
A V1
C2
C3
B V2
100 V
V2 =
C1 + 2 = 90 V C1 + C2 + C
V
C1 = 8.4 mF, C2 = 8.2 mF C3 = 4.2 mF, V = 36 V
67 (a) Effective capacitance, 1 1 1 1 = + + Ce C1 C2 C3 1 1 1 = + + Þ Ce = 2.09 mF 8.4 8.2 4.2 q = CeV = 2.09 ´ 36 = 75.2 mC As combination is series, charge on each capacitor is same, i.e., 75.2 mC. 1 1 (b) U = qV = ´ 75.2 ´ 36 ´ 106 2 2 = 1.35 ´ 103 J = 1.35 mJ (c) Common potential, C V + C2V2 + C3V3 V= 1 1 = 10.85 V C1 + C2 + C3 1 (d) U ¢ = (C1 + C2 + C3 )V 2 2 1 = ´ ( 8.4 + 8.2 + 4.2) ´ (10.85)2 ´ 106 2 = 1.22 ´ 103 J = 1.22 mJ 20. The Given circuit can be considered as the sum of three circuits as shown 3mF +12 –12 +6 5V
6mF
+4
+2 2mF
4mF
–2
–6
–4
3mF +12 –12 +
6mF
–24 +12 2mF
+24
+24 –12
–24 4mF
10V + –4 +
3mF +4
5V
+8/3 –8/3 –8 6mF 2mF +8
3mF +20mC –20mC 6mF
4 +4 — 3
4mF
5V
–10mC 5V
4 –4 — 3
+50 — mC 3
2mF –50 — mC +10mC 3 10V
+40 — mC 3 4mF –40 — mC 3
(Charge is shown in mC). Hence, charge on 6 mF capacitor = 10 mC 40 and Charge on 4 mF capacitor = mC 3 21. (a)
C1=8.4mF
C3=4.2mF
a C2=4.2mF b C5=8.4mF
C4=4.2mF 8.4mF
a 4.2mF
2.1mF
b 8.4mF 8.4mF a
2.52mF a
b
6.3mF b 8.4mF
(b) Charge supplied by the source of emf q = CV = 2.52 ´ 106 ´ 220 = 554.4 mC q1 = q5 = q = 554.4 mC 4.2 q2 = q 4.2 + 2.1 4.2 = ´ 554.4 mC = 369.6 mC 6.3 2.1 2.1 and q3 = q4 = q= ´ 554.4 mC 4.2 + 2.1 6.3 = 184.8 mC q1 554.4 V1 = = = 66 V = V5 C1 8.4 q 369.6 V2 = 2 = = 88 V C2 4.2 q 184.8 V3 = V4 = 3 = = 44 V C3 4.2 22. Let C1 and C2 be the capacitances of A and B respectively. ke A ke A C1 = 1 0 1 , C2 = 1 0 2 \ d1 d2 C2 Now, V1 = V C1 + C2 C2 130 13 …(i) = = Þ C1 + C2 230 23
68
Þ
C1 V2 = V C1 + C2 C1 10 = C1 + C2 23
C2=2mF 4mF
…(ii)
From Eqs. (i) and (ii), C1 10 = C2 13 If dielectric slab of C1 is replaced by one for which k = 5 then, 5e A 5 C1 ¢ = 0 1 = C1 d1 2 V2 ¢ C1 ¢ 5C1 50 \ = = = V1 ¢ C2 2C2 26 50 V2 ¢ + V1 ¢ = 230 26 Also, V1 ¢ + V2 ¢ = 230 50 V1 ¢ = V1 ¢ 26 V1 ¢ = 78.68 V and V2 = 151.32 V 23. In this case
2mF 20V
C3=4mF
+ 3mF
C1=3mF
2mF 2mF
2mF 20V 3mF
3mF
6mF 20V
20V
3mF
6mF C1,V1
C2
C3
Common potential, C1V1 V= C2C3 C1 + C2 + C3 1 ´ 110 V= 1 + 1.2 110 = 2.2 = 50 V Charge flown through connecting wires, C2C3 = V C2 + C3 = 1.2 ´ 50 = 60 mC 24. (a) Hence, effective capacitance across the battery is 3 mF.
(b) q = CV = 3 ´ 20 = 60 mC (c) Potential difference across C1 6 V1 = ´ 20 = 10 V 6+6 q1 = C1V1 = 3 ´ 10 = 30 mC (d) Potential difference across C2 6 V2 = ´ 20 = 10 V 6+6 q2 = C2V2 = 2 ´ 10 = 20 mC (e) Potential difference across C3 4 V3 = ´ V2 = 5 V 4+4 q3 = C3V3 = 4 ´ 2 = 20 mC 25. (a) When switch S2 is open, C1 and C3 are in series, C2 and C4 are in series their effective capacitances are in parallel with each other. Hence, C1C3 q1 = q3 = V C1 + C3
69 1 ´3 ´ 12 = 9 mC 1+3 C2C4 q2 = q4 = C2 + C4 2 ´4 = ´ 12 = 16 mC 2+4
C2C3Q C1C2 + C2C3 + C3C1 C1C2C3V = C1C2 + C2C3 + C3C1
q=
=
\
(b) When S2 is closed, C1 is in parallel with C2 and C3 is in parallel with C4. C1
C3 S2
C2 V1
C4 V2
B
Therefore, C3 + C4 V C1 + C2 + C3 + C4 7 = ´ 12 = 8.4 V 10 C1 + C2 V V3 = V4 = C1 + C2 + C3 + C4 3 = ´ 12 = 3.6 V 10 q1 = C1V1 = 1 ´ 8.4 = 8.4 mC q2 = C2q2 = 2 ´ 8.4 = 16.8 mC q3 = C3V3 = 3 ´ 3.6 = 10.8 mC q4 = C4V4 = 4 ´ 3.6 = 14.4 mC V1 = V2 =
26. Initial charge on C1 Q = C1V0 Now, if switch S is thrown to right. Let charge q flows from C1 to C2 and C3. By Kirchhoff’s voltage law, q C2 –q
Q–q –(Q–q)
C1
q C3 –q
q q Qq + =0 C2 C3 C1 æ 1 1 1 ö Q ÷= q çç + + ÷ è C1 C2 C3 ø C1
C12 (C2 + C3 ) V C1C2 + C2C3 + C3C1 C1C2C3V q2 = q3 = q = C1C2 + C2C3 + C3C1 q1 = Q  q =
e0 A e AV , q = CV = 0 d d e0 A (a) C ¢ = , q¢ = q 2d (As battery is disconnected) q¢ V¢ = =2V C¢ 1 2 e0 AV 2 (b) Vi = V = 2C 2d 1 1 e A Uf = C ¢ V ¢2 = × 0 (2V )2 2 2 2d e AV 2 = 0 d e0 AV 2 (c) W = Uf  Ui = 2d
27. C =
28. In the steady state, capacitor behaves as open circuit, R1
I1
P I2
A
S
IC
B
E1
E2
R2
C
D
E1 = 1 mA and IC = 0 R1 + R2 E1 R2 VPD = I2 R2 = R1 + R2
I1 = I2 =
When switch is shifted to B, At this instant, E1 R2 VPD = R1 + R2 V E1 I2 = PD = = 1 mA R2 R1 + R2
70 E1 R2 E2 + E + VPD R1 + R2 I1 = 2 = R1 R1 ( R1 + R2 ) E2 + E1 R2 = R1 = 2 mA IC = I1 + I2 = 1 + 2 = 3 mA
q2 = C2V = 3 ´ 18 = 54 mC After closing the switch, q1 ¢ = C1V1 = 6 ´ 12 = 72 mC q2 ¢ = C2V2 = 3 ´ 6 = 18 mC Dq1 = 18 mC, Dq2 =  36 mC V 18 30. (a) I = = =2A R1 + R2 9
29. (a) When switch S is open, no current pass through the circuit,
V=18.0V I
V=18.0V
C1= 6mF
R1 = 6W
q
–
S
R1
C1
a
S
+
a
b
R2 = 3W
b
q
+ C2= 3mF –
R2
C2
q= Hence,
Now,
Vb  0 = 0 Vb = 0 18  Va = 0 Þ Va = 18 V Þ Va  Vb = 18 V (b) a is at higher potential. (c) When switch S is closed,
and
C1C2 V = 2 ´ 18 = 36 mC C1 + C2
Va  0 = IR2 Þ Va = 6 V q 36 = = 12 V Vb  0 = 3 C2
Va  Vb =  6 V (b) b is at higher potential. (c) When switch S is closed, in steady state, V =18V
V=18.0V I
q1 + –
R1
C1
R1
a
I
a
S
b
b R2
C2
S
R2
C1
+ q2
–
C2
I
I=
V =2A R1 + R2
Vb  0 = IR2 = 2 ´ 3 Vb = 6 V (d) q1 = C1V = 6 ´ 18 = 108 mC
Va  vb = 6 V q1 = C1V1 = 6 ´ 12 = 72 mC q2 = C2V2 = 3 ´ 6 = 18 mC Charge flown through S = q1  q2 = 72  18 = 54 mC
71 31. E
I
R
A
F
I–I1
I1 q
R
V
C
C
G
B
R
E1 R2 + E2 R1 V = 2 R1 + R2 R 3R Re = R + = 2 2 q = q0 (1  e  t / t ) CE q0 = 2 3 RC t= 2 CE q= (1  e 2t / 3 RC ) \ 2 dq E 2t / 3 RC (b) I1 = = e dt 3 R q In loop EDBA + I1 R  I2 R = 0 C q I2 = + I1 RC E E 2t / 3 RC = (1  e 2t / 3 RC ) + e 2R 3R E = (3  e 2t / 3 RC ) 6R Ee =
(a) Consider the circuit as combination of two cells of emf E and OV. C
R
F I1
R
I2
A
D
B I
I E
F
R
V
C
3R/2
I
V/2
Objective Questions (Level 1) 2
Q is independent of d. 2e0 A q 2. C = V
5. Incorrect diagram.
1. F =
On connecting the plates V becomes zero. 3. The system can assumed to a parallel combination of two spherical conductors. C = C1 + C2 = 4pe0a + 4pe0b = 4pe0 ( a + b) q 4. V = C On connecting in series q¢ = q = Charge on any capacitor C C¢ = n nq \ V¢ = = nV C
6. Charge on capacitor of capacitance 2C A
M
60V
C
C
B
N
A 60V
C
B
2C
C C = V = 30 C 2 2 q VMN = = 30 V C 7. For equilibrium, qE = mg V 4 q = pr 3rg d 3 r3 Vµ q
qE
mg
C — 2
72 3
V2 æ r2 ö q = çç ÷÷ ´ 1 V1 è r1 ø q2
R R
V2 = 4 V
Þ
8. Electric field between the plates is uniform but in all other regions it is zero.
R
R
R
9. Initially the capacitor offers zero resistance. 1W E 12V
E 4E = 3 R + R 7R 4 But potential difference across capacitor, V = IR 4E 10 = R Þ 7R Þ E = 17.5 V I=
6W
3W 1W 4A 2W
12V
i=
12 =4 A 1+2
10. q = CV = CE 11. In the steady state, capacitor behaves as open circuit. the equivalent diagram is given by 3 —R 4
R
I
12. As all the capacitors are connected in series potential difference across each capacitor is E 10 V= = = 2.5 V 4 4 VA  VN = 3V = 7.5 V VA = 7.5 V VN  VB = 2.5 V VB =  2.5 V 13. Heat produced = Loss of energy C1C2 = ( V1  V2 )2 2 (C1 + C2 ) =
E
2 ´ 106 ´ 2 ´ 106 2 (2 + 2) ´ 106
(100  0)
= 5 ´ 103 J = 5 mJ
3R
q = q0e  t / h I = I0 e  t / h P = I 2 R = I02e 2t / h R = P0e 2t / h h h¢ = Þ 2 C V + C2V2 E 15. Common potential = 1 1 = 2 C1 + C2 14.
R
R
E
16. VA  VB = 6 + 3 ´ 2 
9 + 3 ´ 3 = 12 V 1
73 17. In the steady state, current through battery 12V
I
2W
4W
B
O
I
C
C
R
C/2
S
C B
C
3C/2
3C/14
22. 1mF
y
2mF
2mF
19. For the motion of electron mu 2 sin 2q R= =l eE mu 2 sin 2 q and H= =d 2eE Dividing Eq. (ii) by Eq. (i), 4d tan q = l V 2Ve0 20. V = Ed Þ d = = E 6 2 ´ 5 ´ 8.85 ´ 1012 = 107 = 8.85 ´ 104 = 0.88 mm
…(i)
1mF
x
y
…(ii) 2mF
2 — mF 3
x
y 2mF
23. C1 =
21. R
1mF
1mF
x
18. C2 and C3 are in parallel Hence, V2 = V3 Again Kirchhoff’s junction rule  q1 + q2 + q3 = 0 Þ q1 = q2 + q3
Q
C B
6W
Potential difference across the capacitor, 3 VAB = 6 ´ = 9 N 2 q = CVAB = 2 ´ 9 = 18 mC \
P
C
A
12 3 I= = A 6+2 2
A
C
C
2mF A
C A
S
B
P and Q are at same potential, hence capacitor connected between them have no effect on equivalent capacitance.
x
y 8 — mF 3
k1e0 A k2e0 A + 2d 2d
( k1 + k2 ) e0 A (Parallel grouping) 2d 1 d d (Series grouping) = + C2 2k1e0 A 2k2e0 A 2k1k2 e0 A C2 = k1 + k2 d =
C1 ( k1 + k2 )2 (2 + 3 )2 25 = = = 4k1k2 4 ´ 2 ´ 3 24 C2
74 1
24.
d
C
A
2
A P C
C
Q C
CC
C
B A C — 2
C — 2
C
Q
d
C
3
C
d 4
C
2d
C
C
5 C
B
B
C
d
C
B
6
A
Capacitance of all other capacitance is same, e A i.e., C = 0 but that of formed by plates 4 d C as distance between these two and 5 is 2 plates is 2d. The equivalent circuit is shown in figure.
2C B
25. Cases (a), (b) and Wheatstone bridge.
(c)
are
balanced
26. The given arrangement can be considered as the combination of three capacitors as shown in figure.
2
A
C
C — 5 24
1
C1
2
A
C3
C
3
6 C
C
B
3 C — 3
C
B
k1e0 A 2d A k2e0 2 = k2e0 A C2 = d/2 d A k3e0 2 = k3e0 A C3 = d/2 d
4
5
C2
Hence,
A
C
P
B A
C
C1 =
Effective capacitance, C2C3 e A ék k2k3 ù C = C1 + = 0 ê 1 + ú C1 + C2 d ë2 k2 + k3 û 27. Here, plate 1 is connected to plate 5 and plate 3 is connected to plate 6.
C
A
B
C 4C — 7 A
B C
4C — 3
C
A
B C
11C — 7
\Ceq =
11e0 A 11 11 C= = ´ 7 mF = 11 mF 7 7d 7
75
JEE Corner Assertion and Reason 1. Capacitance =
q is constant for a given V
capacitor. 2. Reason correctly explains the assertion. 1 3. U = qV , W = qV 2 4. For discharging of capacitor q = q0e  t / t q dq =  0 e t / t t dt q =  0 e t / t RC Hence, more is the resistance, less will be the slope. 5. Charge on two capacitors will be same only if both the capacitors are initially uncharged.
6. As potential difference across both the capacitors is same, charge will not flow through the switch. 7. C and R2 are shorted. 8. Time constant for the circuit, t = RC 9. In series, charge remains same q2 1 and U= ÞU µ 2C C 10. In series charge remains same q q V1 = , V2 = \ C1 C2 On inserting dielectric slab between the plates of the capacitor, C2 increases and hence, V2 decreases. So more charge flows to C2.
Objective Questions (Level 2) ì 4Q ^ i ïï e0 A ï 2Q ^ 1. E = íi ï e0 A ï 4Q ^ ï e Ai î 0
for x < d
But, I01 = I02 Þ
for d < x < 2d Also, for 2d < x < 3 d
2. Let E0 = external electric field and E = electric field due to sheet E1 = E0  E = 8 \ E2 = E0 + E = 12 s Þ E = 2 V/m Þ =2 2e0 s = 4e0 3. When the switch is just closed, capacitors behave like short circuit, no current pass through either 6 W or 5 W resistor. 4. For charging of capacitor I = I0 e  t / t
t t V t ln I = ln R RC ln I = log I0 
V1 V2 = R1 R2
1 1 > R1C1 R2C2
Þ R2C2 > R1C1 As only two parameters can be different, C1 = C2 R2 > R1 and V2 > V1 5. Charge on capacitor at the given instant. q CE q= 0 = 2 2 Heat produced = Energy stored in capacitor q2 CE2 = = 2C 8 Heat liberated inside the battery, r = ´ Total heat produced r + 2r =
CE2 24
76 6. Capacitor is not inside any loop. E  E0 7. I = R + R0 q VBA =  E C E
R
12. VA =  VB ie, VA  VP = VP  VB q q Þ = C123 Cn Þ
I
Þ +q C –q
E B
A
R0
E0
q E C ( E  E0 ) RC q = IRC + R + R0
 E + IR =
8. C =
Cn = C123 1 1 1 1 = + + Cn C1 C2 C3
13. When connected with reverse polarity C1C2 H= ( V1 + V2 )2 2 (C1 + C2 ) C ´ 2C 25 = ´ ( V + 4V )2 = CV 2 2 (C + 2C ) 3 14.
H1 R2 R = = H2 R1 S 2mF
C1C2 C1 + C2
5W I1
e A C1 = C2 = 0 d e0 A C= 2d 2e0 A e A , C2 ¢ = 0 C1 ¢ = d 2d C1 ¢ C2 ¢ 2e0 A C¢ = = <C C1 ¢ + C2 ¢ 5d Re =
9.
R 3
RC t = ReC = 3 t / t q = q0(1  e ) = CV (1  e 3t / RC ) C1C2 10. Energy loss = ( V1  V2 )2 2 (C1 + C2 ) 2 ´4 = (100  50)2 ´ 106 2 ´ (2  14) = 1.7 ´ 103 J 11. q = q0e
[VP = 0]
 t / RC
I=
q dq = 0 e  t / RC dt RC
at t = 0 q0 = 10 RC q V0 = 0 = 10 ´ R = 10 ´ 10 = 100 V C I=
R I2
Also, H1 + H2 =
1 1 CV 2 = ´ 2 ´ 106 ´ (5)2 2 2
H1 + H2 = 25 mJ H2 = 25  10 = 15 mJ 10 10 R W = Þ R= 3 15 5 15. When current in the resistor is 1 A. q IR + =E C q 1 ´ 5 + = 10 2 q = 10 mC Þ When the switch is shifted to position 2. In steady state, charge on capacitor q¢ = 5 ´ 2 = 10 mC but with opposite polarity. \Total charge flown through 5 V battery, = q + q¢ = 20 mC Work done by the battery = 20 ´ 5 = 100 mJ Heat produced = W  DU But, DU = 0 \ H = W = 100 mJ
77 16. VA  VB = 
q q q +  =0 6 2 3 +
+
q
2mF – q
q2 =
R1
–
+ A
6mF
B
q
–
3mF
Hence, no charge will flow from A to B. 17. As potential difference across both the capacitors is same, they are in parallel. Hence, effective capacitance, 2e A C= 0 d e A 1 U = CV 2 = 0 V 2 2 d
2.5 ´ 0.5 ´ 30 2.5 + 0.5
= 12.5 mC q \ Vp  Va = 1 = 18 V C1 q Vp  Vb = 2 C3 12.5 = =5 V 2.5 Vb  Va = 13 V 23. As all the capacitors are identical, potential difference across each capacitor, 12V + –
B
C
18. Rate of charging decreases as it just charged. 19. Potential difference across capacitor = 6 V 1W
A
A
2W
B
C
C
I 5V
2V D
4W C
2mF
Þ
A C1 = 60mF C2 = 20mF
21. Common potential, C V  C1V1 3 ´ 100  1 ´ 100 V= 2 2 = C1 + C2 1+3 = 25 V
C1 1.0mF
a
q1 q2 2.5mF C3
C2 1.5mF q1 q2
b
30V
0.5mF C4
V0
C3 = 30mF C
B
q1 + q2 + q3 = 0 C1(C A  V0 ) + C2( VB  V0 ) + C3( VC  V0 ) = 0 C V + C2VB + C3VC V0 = 1 A C1 + C2 + C3
1 ´ 1.5 ´ 30 = 18 mC 1 + 1.5
P
E =3 V 4 VN  VB = 3 V VB =  3 V VA  VB = 12 V VA = 9 V
24. By Kirchhoff’s voltage law,
20. While charging Re = R Þ t = RC While discharging Re = 2 R Þ t = 2 RC
22. q1 =
N
V=
6V
3W
q = CV = 2 ´ 6 = 12 mC In loop ABCD, I ´1  2  5 = 0 Þ I = 7 A
C
V0 = Q
=
60 ´ 6 + 2 ´ 20 + 3 ´ 30 60 + 20 + 30 49 V 11
25. In the steady state, there will be no current in the circuit.
78 3mF
B 3mF
3mF
1mF
3mF
3mF 1mF
2mF
1mF 10V
2W
1mF
B
20W
10W
10V
10W
A
A
C1=6mF q1
C1=1mF q3
20W A
q2
10W
resistance. 30. Immediately after switch is closed, capacitor behaves like short circuit. V  t / RC V 31. i1 = , i2 = e  t / RC e R 2R 5t i1 1 6 RC = e i2 2 Increases with time. rd d 32. R = = A sA ke0 A d ke A d t = RC = ´ 0 sA d e0 8.85 ´ 1012 = = =6s 6 7.4 ´ 1012 C=
10V
C2 3 10 V1 = V= 10 = V C1 + C2 3+6 3
\ 26. I1 =
C2=3mF
B
C1C2 29. H = ( V1  V2 )2 is independent of 2 (C1 + C2 )
E1 18 = =3 A R1 + r1 5 + 1 E2 15 I2 = = = 2.5 A R2 + r2 4 + 2
I1 18V 1W
3mF A + – B q I1 5W
4W q
– D
+ 2mF
15V
C
In loop ABCD, q q  I2 R2 +  I1 R1 = 0 3 2 5q Þ = 3 ´ 5 + 2.5 ´ 4 Þ q = 30 mC 6 27. During discharging q = q0 e  t / t q0 = CE = 10 mC at t = 12 s, q = 10e 12/ 6 = 10e 2 = (0.37)210 mC C1C2 28. q = ( E1  E2 ) C1 + C2 C1 q ( E1  E2 ) Vap = = C2 C1 + C2 æ E  E2 ö ÷C =ç 1 çC + C ÷ 1 2ø è 1
Þ Þ
2W I2
I2
33. i = i0e t / t
Þ Þ Þ
 ln 4
i0 = i0e RC 2 ln 4 = ln 2 RC ln 4 = ln 2 RC RC = 2 2 2 R= = =4W C 0.5
34. Potential difference across each capacitor is equal, hence they are in parallel, charge on each capacitor q = CeV = 2 ´ 10 = 20 mC As plate C contributed to two capacitors, charge on plate, C = 2q = + 40 mC 35. Charge distribution on the plates of the capacitor is shown in figure Q/2
CV +
Q 2
Q/2
–CV +
\
Q¢ V¢ = = C
Q 2
CV +
C Q =V + 2C
Q 2
79 36. Let q be the charge on C2 (or charge flown through the switches at any instant of time) By Kirchhoff’s law S1
I
q=
C2q0 C1 + C2
t=
C1 + C2 C1C2 R æ C q0 ç1  e ç C1 è C1C2 C= C1 + C2
or + q
q0 – q –
–
q=
where,
+ C2
37. H =
q q q  0 =0 C2 C1 dq C2q0  (C1 + C2 ) q = dt C1C2 R q t dq dt ò0 C2q0  (C1 + C2 ) q = ò0 C1C2 R q 1 = [lnC2q0  (C1 + C2 ) q]0 C1 + C2 1 = t C1C2 R IR +
t æ  ö ç1  e t ÷ ÷ ç ø è
t ö RC ÷
÷ ø
C1C2 ( V1  V2 )2 2 (C1 + C2 ) =
æ q0 ö C1C2 ç ÷ 2 (C1 + C2 ) çè C1 ÷ø
2
C q02 C2q02 = 2C1 (C1 + C2 ) 2C12
=
38. Electric field in the gap will remain same. 39. Electric field inside the dielectric slab E V . E¢ = = k kd
More Than One Correct Options 1. Charge distribution is shown in figure EB = E1 + E2 + E3 + E4 Q Q Q Q = + + 4e0 A 4e0 A 4e0 A 4e0 A Q = 2e0 A +Q — 2
X
A
1
–Q — 2
+Q — 2
B
2
Y
2R
I2
2C
R
t ö æ q2 = 2CE ç1  e RC ÷ ÷ ç ø è t dq1 E  RC = e dt R t dq2 2E  2 RC = e dt R q0 q1 1 1 = Þ 1 = q2 2 q02 2
+Q — 2
4
EA = EC = E1 + E2  E3 + E4 but EA and EC have opposite direction.
t ö æ 2. q1 = CE ç1  e RC ÷ ÷ ç ø è
C
E
C
3
I1
t1 = t 2 = RC
3. V1 =
4 ´ 103 q = = 40V C 100 ´ 106
80 U µd Q Qd V= = C e0 A
Þ
V2 200W
V1
900W
V µd
Þ
6. When switch S is open
100W
C
A2
S
A1 C
I1 = 0 V1 40 2 = = A I2 = 900 900 45 2 80 V2 = I2 ´ 200 = ´ 200 = V 45 9 80 E = V1 + V2 = 40 + 9 440 = V 9 E 4. Initially I1 = 0,I2 = I = R C
I2
B
I1
I +
A
–
S
C +
– E
C ´ 2C 2 = C C + 2C 3 2 q1 = CE 3 When switch S is closed Ce ¢ = 2C q2 = 2CE Charge flown through the battery 4 Dq = q2  q1 = CE = positive 3 Ce =
7. Let charge q flows to C1 at it falls to the free end of the wire. A
E
C1
C1 –q
q
As the capacitor starts charging, I2 decreases and I1 increases, In the steady state E I1 = I = , I2 = 0 R At any instant P1 = I12 R, P2 = I22 R Steady state potential difference across the capacitor, E V= 2 1 CE2 U = CV 2 = 2 8 Q2 5. F = independent of d. 2e0 A Q independent of d. E= e0 A U=
Q 2 Q 2d = 2C 2e0 A
q2 + –
+
q3
2mF C2
3mF C3
–
q2–q + –
q3–q + –
C2
C3
By Kirchhoff’s voltage law, q2  q q3  q q + =0 C2 C3 C1 q2 q + 3 C2 C3 q= 1 1 q + + C1 C2 C3 V2 + V3 = 1 1 1 + + C1 C2 C3 150 + 120 q= = 180 mC 1 1 1 + + 2 3 1.5 q2 ¢ = q2  q = 150 ´ 2  180 = 120 mC
81 8. C =
e0 A d
I= Q 2 Q 2d = ÞU µd 2C 2e0 A Q Qd V= = Þ V µd C e0 A e A 1 C = 0 ÞC µ d d
U=
E=
Q Þ E is independent of d. e0 A
9. R = 1 + 2 = 3 W, C = 2 F q0 = CV0 = 2 ´ 6 = 12C At any instant æ t ö q = q0 ç e RC ÷ ÷ ç ø è t q0 RC dq = I= e dt RC at t = 0 q 12 I= 0 = =2 A RC 3 ´ 2
q0 e RC
 6 ln 2 6
=
12  1 ´ 6 2
1 =1 A 2 Potential difference across 1 W resistor Þ 1 ´1 = 1 V Potential difference across 2 W resistor Þ 1 ´2 = 2 V \ By Kirchhoff’s voltage law, potential difference across capacitors = 1 + 2 = 3 V. =2 ´
10. q = C1V1 = 1 ´ 10 = 10 mF 6mF 1mF 4mF
3mF
1mF
4mF
9mF
V1
V2
V3
E
E
q 10 = = 2.5 V 4 C2 q 10 = V V3 = 9 C3
V2 =
at t = 6 ln 2
Match the Columns 1. C1¢ = kC1 = 8 mF,
C C2 ¢ = 2 = 2 mF k C ¢C ¢ q¢ = 1 2 V = (1.6V ) mC C1 ¢ + C2 ¢ q = (2V ) mC
\q ¢ < q 2
q¢2 (1.6 V ) U2 ¢ = = = 0.64V 2, C2 ¢ 2 ´2 q2 (2)2 U2 = = = (1 V ) mC 2C2 2 ´ 2 U2 ¢ < U2
q¢ 1.6V = = 0.2V , C2 ¢ 8 q (2V ) V2 = = = 0.5V C2 4
1.6V q¢ , = e0 A e0 A 2V q V = = E2 = ke0 A 2 ´ e0 A e0 A E2 ¢ =
E2 ¢ > E2 ( a ® q), (b ® q), (c ® q), (s ® p). 2. Before switch S is closed, charge distribution is shown in figure (1). 4mF + –
S
40mC + – 3mF
20mC + – 2mF
V2 ¢ =
V2 ¢ < V
30V
Fig. 1 After switch S is closed, charge distribution is shown in figure (2).
82 C
4mF + –
S
+
120mC + – 3mF
S
60mC + – 2mF
2C
Fig. 2 (a ® s), (b ® p), (c ® q ), (d ® s). 3. (a ® q), (b ® p, r), (c ® q), ( d ® p, p) C1 C2 C3 (p) a
b
V V1 = V2 = V3 = 3 CV q1 = q2 = q3 = 3 C1 C2 a
b C3
V1 = V2 = V3 = V q1 = q2 = q3 = CV (r)
C2 a
b C1 C3
As combination is series, q1 = q2 q1 =1 Þ q2 U1 C2 8 = = U2 C1 9
2V V , V2 = V3 = 3 3 2CV CV q1 = , q2 = q3 + 3 3 V1 =
(s)
C1V1 + C2V2 V = 3 C1 + C2 1 1 U1 = C1V ¢2 =` CV 2 2 18 1 1 2 U2 = C2V ¢ = CV 2 2 9 C1C2 DU = ( V1  V2 )2 2 (C1 + C2 ) C ´C ( V  0 )2 = 2 (C1 + C2 ) 1 = CV 2 6 (a ® r), (b ® p), (c ® q). ke A e A 5. C1 = 0 + 0 2d 2d e0 A 3 e0 A = ( k + 1) = 2d 2d 1 d d = = + C2 2ke0 A 2e0 A d æ1 + kö = ÷ ç 2e0 A è k ø 2ke0 A 4e A Þ C2 = = 0 d (1 + k) 3d C1 9 = C2 8 V¢ =
30V
(q)
C1
(a ® s), (b ® s), (c ® s).
C2 a
b
6. Charge distribution is shown in figure. (a ® p), (b ® p, q), (c ® s), (d ® p, q, r). 4Q
C
3 2V V , V1 = V2 = V2 = 3 3 2CV CV , q1 = q3 = q2 = 3 3
4. Common potential
–
7Q
Q
–3Q 3Q
2Q
–2Q 2Q
7Q
0
0
7Q
23
Magnetics Introductory Exercise 23.1
1. [ Fe ] = [ Fm ] [ qE] = [ qvB] Þ
® Fm = qvB sin q
Þ é E ù = [v] = [L T 1 ] ê Bú ë û
® ® ® 2. F = q ( v ´ B ) ® ® ® ® \ F ^ v and F ^ B Because cross product of any two vectors is always perpendicular to both the vectors. ® ® ® 3. No. As Fm = q ( v ´ B )
If Fm = 0, either B = 0 or sin q = 0, i.e., q = 0 ® ® ® 4. F = q ( v ´ B ) ^
^
^
=  4 ´ 106 ´ 106 ´ 102 [(2 i  3 j + k ) ^
^
^
´ (2 i + 5 j  3 k )] ^
^
^
=  4 ´ 102(4 i + 8 j + 16 k ) ^
^
^
=  16 ( i + 2 j + 4 k ) ´ 102 N
Introductory Exercise 23.2 1. As magnetic field can exert force on charged particle, it can be accelerated in magnetic field but its speed cannot increases as magnetic force is always perpendicular to the direction of motion of charged particle. ® ® ® 2. Fm =  e ( v ´ B ) ® By Fleming’s left hand rule, B must be along positive zaxis. 3. As magnetic force provides necessary centripetal force to the particle to describe a circle. mv2 qvB = r mv r= Þ qB
(a)
r=
mv qB
Þ r µm Hence, electron will describe smaller circle. 2pr 2p m (b) T= = v qB 1 qB f = = T 2p m 1 Þ f µ m \electron have greater frequency. 4. Electrons are refocused on xaxis at a distance equal to pitch, i.e., n = p = vT 2p mv cos q = eB
84 mv 5. (a) If L ³ r = , qB
6. r =
O q × × × × p – q × 2 2 × × ×
× × × × × × × × × × × × × × q/2 × ×
× × × × × × ×
× × × × × × ×
× × × × × × ×
× × × × × × ×
× × × × × × ×
mv 2mk = eB eB 2m eV 1 2mV = = eB B e
For electron, q
r=
= 1.67 ´ 104 m = 0.0167 cm For proton r=
L × × × × × × × ×
(b) The particle will describe a semicircle. Hence, q = p L q (c) = cos l 2 L q = cos Þ q 2 2 R sin 2 L 1 = sin q Þ sin q = R 2 p Þ q= 6
31 ´ 100 1 2 ´ 9.1 ´ 10 0.2 1.6 ´ 1019
7. r =
mv = qB
27 ´ 100 1 2 ´ 1.67 ´ 10 19 0.2 1.6 ´ 10
= 7 ´ 103 m = 0.7 cm 2m k qB
\
rµ
\
rp : rd : ra =
m q
1 2 4 : : 1 1 2 = 1 : 2 :1
Introductory Exercise 23.3 1. Let at any instant ® ^ ^ ^ V = Vx i + Vy j + Vz k Now, Vx2 + Vy2 = V02 = constant qE and V2 = V0 f m ® V is minimum when V2 = 0 at and
f =
2. After one revolution, y = 0,
2p mv sin q qB
Hence, coordination of the particle, æ 2p mv sin q ö ÷ = ( x, b) = çç0, ÷ qB ø è ® ® ® ^ ^ ^ 3. F = i ( l ´ B ) = ilB [ i ´ ( j + k )] ® ( F ) = 2 ilB
mv0 qE
Vmin = V0
x = p = pitch of heating
=
^
^
^
^
^
^
^
^
^
4. No. as i ´ ( i + j + k ) = i ´ j + i ´ ( j ´ k ) ^
^
But i ´ j = 0 ^
^
^
^
^
^
^
\ i ´ (i + j + k) = i ´ ( j ´ k)
85
Introductory Exercise 23.4 1. Consider the disc to be made up of large number of elementary concentric rings. Consider one such ring of radius x and thickness dx. Charge on this ring
x
w qx dx
´ px 2 p R2 wq = 2 x 3dx R Magnetic moment of entire disc, \ wq R M = ò dM = 2 ò x 3dx R 0 w q é R4 ù 1 2 = 2 ê ú = w qR R êë 4 úû 4 dM = di ´ A =
® ® ® 2. M = i ´ [(OA ´ AB )] dq = dq =
q q × dA + ´ 2px dx p R2 pR2 2qx dx 2
R Current in this ring, dq w dq w qx dx di = + = T 2p pR2 \ Magnetic moment of this ring,
® ^ ^ OA = OA cos q j + OA sin q k ® ^ AB = AB i ®
^
^
^
M = i ´ OA × AB [(cos q j + sin q k ) ´ i ]
\
é æ 3 ^ 1 ^ ö ^ù = 4 ´ 0.2 ´ 0.1 ê ç j + k ÷ ´ iú 2 ÷ø êë çè 2 ûú ^
^
= (0.04 j  0.07 k ) Am 2
Introductory Exercise 23.5 1. (a) B1 = B2 = B3 = B4 m i = 0× [sin 45° + sin 45° ] 4p l / 2 2 1
i
45° l 2
3
2pr = 4l Þ r = B=
r=
pm 0 i = 24.7 mT (inward) 4l
i
x
2 m0 i 2. B = × 4p x (As P is lying near one end of conductor 1) B2 = 0 (Magnetic field on the axis of a current carrying conductor is zero) 2
4
m0 i
P
l
m 0 2 2i × 4p l Net magnetic field at the centre of the square, m 8 2i B = B1 + B2 + B3 + B4 = 0 × 4p l 2 2 m 0i = = 28.3 mT (inward) pl =
(b) If the conductor is converted into a circular loop, then
2l p
i
1
B = B1 m i = 0× 4p x By right hand thumb rule, direction of magnetic field at P is inward.
86 3. Magnetic field due to straight conductor at O i
O
m0 2i × 4p R Magnetic field at O due to circular loop m i B2 = 0 2R By right hand thumb rule, both the filds are acting inward. Hence, m i B = B1 + B2 = 0 (1 + p ) 2pR 4p ´ 107 ´ 7 æ 22 ö = ç1 + ÷ 7ø 2p ´ 10 ´ 102 è B1 =
60° m 0i m 0i (inward) B2 = × = 360° 2b 12b 60° m 0i m 0 (outward) B3 = = = 360° 2a 12a As B3 > B2, Net magnetic field at P, B = B3  B2 =
m 0i é 1 1 ù 12 êë a b úû
6. AB, AP and BP from Pythagorus triplet, hence ÐAPB = 90° i1 A X 5.0cm P ® B1
13.0cm
12.0 cm
= 58 ´ 106 T = 58 mT (inward). 4. B1 = B2 = 0 (Magnetic field on the axis of current carrying conductor is zero)
B X i2
1
® m 2i ^ B1 = 0 × 1 PB 4p r1
i
\ 3
® m 2i ^ B 2 = 0 × 2 AP 4p r2
R 2
B = B12 + B22
O
B3 = =
1 m 0i m 0i × = 4 2R 8R 4p ´ 107 ´ 5 8 ´ 3 ´ 102
= 2.62 ´ 105 T = 26.2 mT (inward). 5. B1 = B2 = 0 (Magnetic field on the axis of straight conductor is zero) 2
3
1 60° 4
®
B2
2
æi ö æ i1 ö ç ÷ + ç 2÷ çr ÷ çr ÷ è 2ø è 1ø
2
=
m0 2p
=
2 2 4p ´ 107 3 ö æ 3 ö ´ æç ÷ +ç ÷ 2p è 0.05 ø è 0.12 ø
= 1.3 ´ 105 T = 13 mT 7. t = NIAB cos q = 100 ´ 1.2 ´ 0.4 ´ 0.3 ´ 0.8 ´ cos 30° = 9.98 Nm Rotation will be clockwise as seen from above.
87
Introductory Exercise 23.6 1. By right hand thumb rule, direction of magnetic field due to conductor A, B, C and D are as shown in figure. C
A X
At point B B1 =
m 0 I1 m I × , B2 = 0 × 2 ¯ 4pe0 r2 4p r2
Net field at B B = B2  B1 =
® B ® D BA
2m
® BC ® BB
=
® B
m0 1 × ( I2  I1 ) 4p r2
107 ´ (3  2) = 0.67 ´ 104 T 3 ´ 103
= 67 mT
B X
D
BA = BB = BC = BD =
3. Consider the cylinder to be made up of large number of elementary hollow cylinders.
m 0 2I × 4p r
Here, I = 5 A
R r
r=
a 0.2 = = 0.14 2 2
\Net magnetic field at P
Consider one such cylinder of radius r and thickness dr. Current passing through this hollow cylinder, di = jdA = j ( 2pr dr ) 2p br 2dr (a) Total current inside the portion of radius r1,
B = ( BA + BD )2 + ( BB + BC )2 m0 4 2 I × 4p r 107 ´ 4 2 ´ 5 = = 20 ´ 106 T 0.2 / 2
=
r1
I1 = ò di = 2pbò r 2dr 0
= 20 mT Clearly resultant downward.
magnetic
field
is
2. At point A I2
B1
×× I1
m b q
r1
é r3 ù = 2 pb ê 1 ú êë 3 úû 0 2 = p br13 3 By ampere’s circuital law, ò B × dl = m 0i1 2 pr1
B2
B1 =
m 0 I1 × 4p r1
B2 = 0 (Magnetic field inside a current carrying hollow cylinder is zero) m I \ Ba = B1 + B2 = 0 × 1 4p r1 =
10
7
´1
1 ´ 103
 104 T
= 100 mT (upward)
2 B1 ´ 2 pr1 = m 0 æç p br13 ö÷ è3 ø m 0 br12 B1 = Þ 3 (b) Total current inside the cylinder R
i = 2 pbò r 2dr 0
2 p bR3 3 m 2 i m 0 bR3 B2 = 0 = 4p r2 3 r2 =
88
AIEEE Corner Subjective Questions (Level1) ® ^ ^ ^ 7. Let B = Bx i + By j + Bz k
1. Positive. By Flemings left hand rule. 2. Fm = evB sin q Fe Þ v= eB sin q =
® ® ® (a) F = q( v ´ B ) ^
1.6 ´ 1019 ´ 3.5 ´ 103 ´ sin 60°
^
Þ Bx =  0.175 T, Bz =  0.256 T (b) Cannot be determined by this information. ® ® ® (c) As F = q ( v ´ B )
3. Fm = qvB sin q = (2 ´ 1.6 ´ 1019 ) ´ 105 ´ 0.8 ´ 1 = 2.56 ´ 1014 N ® ® ® 4. (a) Fm = e ( v ´ B ) ^
® ® F ^B ^
=  1.6 ´ 1019 [(2.0 ´ 106 ) i + (3.0 ´ 106 ) j] ^
^
´ (0.03 i + 0.15 j)
®® Hence, B × F = 0 ® ^ 8. B = B i ® ^ (a) v = v j
^
=  (6.24 ´ 104 N) k
® ® ® ^ F = q ( v ´ B ) =  qvB k
® ® ® ^ (b) = Fm e ( v ´ B ) =  (6.24 ´ 104 N) k ® ^ (b) v = v j
® ® ® 5. Fm = e ( v ´ B ) ^
^
® ® ® ^ F = q ( v ´ B ) = qvB j
^
(6.4 ´ 1019 ) k =  1.6 ´ 1019[(2 i + 4 j) ^
^
´ ( Bx i + 3 Bx j)] ^
® ^ (c) v =  v i ® ® ® F = q( v ´ B ) = 0
^
6.4 ´ 1019 k =  1.6 ´ 1019[2Bx k ] 6.4 ´ 1019  3.2 ´ 1019
=  2.0 T
6. (a) As magnetic force always acts perpendicular to magnetic field, magnetic field must be along xaxis. F1 = qv1B sin q1 5 2 ´ 103 F1 Þ B= = qv1B sin q1 1 ´ 106 ´ 106 ´ 1 2 Þ or
B = 103 T ® ^ B = (103 T ) i
(b) F2 = qv2 B sin q2 = 1 ´ 106 ´ 106 ´ 103 ´ sin 90° = 103 N F2 = 1 mN
^
=  7.8 ´ 106 ´ 3.8 ´ 10 3( Bz i  Bx k )
= 9.46 ´ 106 m / s
Bx =
^
7.6 ´ 103 i  5.2 ´ 103 k
4.6 ´ 1015
® ^ ^ (d) v = v cos 45° i  v cos 45° k ® ® ® qvB ^ F = q( v ´ B ) = j 2 ® ^ ^ (e) v = v cos 45° j  v cos 45° k ® ® ® qvB ^ ^ F = q( v ´ B ) = ( j  k) 2 qvB ^ ^ =( j + k) 2 2m k 2 m eV mv 9. r = = = qB eB eB B=
Þ =
2mV r e
2 ´ 9.1 ´ 1031 ´ 2 ´ 103 1.6 ´ 1019
´ 0.180
89 = 0.36 ´ 104 T B = 3.6 ´ 104 T mv qBr 10. (a) r = Þv= qB m =
13. The component of velocity along the magnetic field (i.e., vx ) will remain unchanged and the proton will move in a helical path. z
1.6 ´ 1019 ´ 2.5 ´ 6.96 ´ 103 3.34 ´ 10
vy cos wt
= 8.33 ´ 105 ms 1 T pm (b) t = = 2 qB =
3.14 ´ 3.34 ´ 10
vy
vy sinwt
27
1.6 ´ 1019 ´ 2.5
= 2.62 ´ 108 s 1 (c) k = eV = mv2 2 mv2 Þ V= 2e 3.34 ´ 1027 ´ ( 8.33 ´ 105 )2 = 2 ´ 1.6 ´ 1019 = 7.26 ´ 103 V = 7.26 kV 11. (a)  q. As initially particle is neutral, charge on two particles must be equal and opposite. (b) The will collide after completing half rotation, i.e., T pm t= = 2 qB × × × × × × × × × × × × × ×
× × × × +q × –q × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × q × × × × ×
×B × × × × × × × × × × × × ×
10.0 12. Here, r = = 5.0 cm, 2 mv mv (a) r = ÞB= qB qr =
® B
27
9.1 ´ 1031 ´ 1.41 ´ 106 1.6 ´ 1019 ´ 5 ´ 102
= 1.6 ´ 104 T By Fleming’s left hand rule, direction of magnetic field must be inward. T pm (b) t = = 2 qB =
3.14 ´ 9.1 ´ 1031 1.6 ´ 1019 ´ 1.6 ´ 104
= 1.1 ´ 107 s
At any instant, Components of velocity of particle along Yaxis and Zaxis v¢y = vy cos q = vy cos wt and v¢z =  vz sin q = vz sin wt qB where, w= m ® ^ ^ ^ \ v = vx i + vy cos w t j  vz sin w t k 14. For the electron to hit the target, distance G S must be multiple of pitch, i.e., GS = np For minimum distance, n = 1 2p mv cos q Þ GS = p = qB Þ
p=
Þ
B=
2p 2 mk cos 60° qB
(mv = 2 mk)
2p 2 mk cos 60° qp
2 ´ 3.14 ´ 2 ´ 9.1 ´ 1031 ´ 2 ´ 1.6 ´ 1016 ´ = Þ
1.6 ´ 1019 ´ 0.1 B = 4.73 ´ 104 T
15. (a) From Question 5 (c) Introductory Exercise 23.2 L = sin q Þ L = R sin q R R R sin 60° = 2 mv0 mv L= = Þ 2qB 2qB0 (b) Now, L ¢ = 2.1 L = 1.05 R As L ¢ > R,
1 2
90 Particle will describe a semicircle and move out of the magnetic field moving in opposite direction, i.e., ^
v¢ =  v =  v0 i T pm t= = 2 qB0
and
® ^ ® ^ 16. v = (50 ms 1 ) i, B = (2.0 mT ) j As particle move with uniform velocity, ® ® ® ® F = q(E + v ´ B ) = 0 ® ® ® ^ E = B ´ v =  ( 0.1 N/C)k
Þ
17. If v be the speed of particle at point (0, y, z ) then by workenergy theorem, ® ^ B = –B j
z
E = EK. v vy q vx
mv R qB0 R v= m qB0 R cos q vx = v cos q = m qB0Z (Q R cos q = Z) = m 2qE0Z q2B02Z 2 Now, vz = v2  vx2 = m m2 ® ^ ^ ® 18. Given, E = E j , B = B k, ® ^ ^ v = v cos q j + v sin q k As protons are moving undeflected, ® ® ® ® F = 0Þ e (E + v ´ B ) = 0 qvB0 =
Þ
x
^
^
e ( E j  vB cos q j) = 0
E B cos q Now, if electric field is switched off 2p mv sin q 2p mE tan q p= = qB qB2 or
Z O
2
v=
(Component of velocity along magnetic field = vz = v sin q) 1 mv2 2 But work done by magnetic force is zero, hence, network done = work done by electric force = qEZ 1 qE0Z = mv2 \ 2 2qE0Z Þ v= m W = DK =
As the magnetic field is along Yaxis, particle will move in XZplane. The path of particle will be a cycloid. In this case, instantaneous centre of curvature of the particle will move along Xaxis. As magnetic force provides centripetal force to the particle, z vz
v q
q
vx
R X
19. F = I l B sin q F 0.13 I= = lB sin q 0.2 ´ 0.067 ´ sin 90° = 9.7 A Fm 20. For no tension in springs × ×I× ×I× × ××× ××× Fm = mg ××× ××× ××× ××× ××× ××× Þ I lB = mg ××× ××× mg 13.0 ´ 103 ´ 10 mg I= = lB 62.0 ´ 102 ´ 0.440
= 0.48 A By Fleming left hand rule, for magnetic force to act in upward direction, current in the wire must be towards right. 21. (a) FBD of metal bar is shown in figure, for metal to be in equilibrium, Fm + N = mg Fm Þ Fm = mg  N N I lB = m  N Þ mg V Þ lB = mg  N R R Þ V= ( mg  N ) lB
91 ® ® ^ ^ l 3 = cd =  (40 ´ 102 ) i + (40 ´ 102 m ) j
For largest voltage, N =0 R mg 25 ´ 750 ´ 103 ´ 9.8 V= = lB 50.0 ´ 102 ´ 0.450
® ® ® ^ F3 = I ( l 3 ´ B ) =  (0.04 N) k ® ® ^ ^ l 4 = da = (40 ´ 102 m ) i  (40 ´ 102 m ) k
= 817.5 V
® ® ® ^ ^ F4 = I ( l 4 ´ B ) = (0.04 N) i + (0.04 N) k
(b) If I lB > mg I lB  mg = ma I lB  mg V lB a= = g m Rm 817.5 ´ 50 ´ 102 ´ 0.45 =  9.8 2 ´ 750 ´ 103
® ^ 24. M = IA M ^
^
^
= (40 . 2 ´ 104 ) (0.60 i  0.80 j)Am 2 ® ^ ^ B = (0.25 T) i + (0.30 T) k
= 112.8 m/s 2 ^
® ® ® (a) t = M ´ B
22. I = 3.50 A, l =  (1.00 cm ) i Þ
^
= 0.20 ´ p( 8.0 ´ 102 )2(0.60 i  0.80 j)
^
l =  (1.00 ´ 102 m ) i
^
^
® ^ (a) B =  (0.65 T) j
^
^
^
= (  9.6 i  7.2 j + 8.0 k ) ´ 104 Nm.
® ® ® ^ Fm = I ( l ´ B ) =  (0.023 N) k
®® (b) U =  M× B =  (40.2 ´ 104 )(0.15) J
® ^ (b) B = + (0.56 T) k
»  6.0 ´ 104 J
® ® ® ^ Fm = I ( l ´ B ) = (0.0196 N) j ® ^ (c) B =  (0.33 T) i ® ® ® Fm = I ( l ´ B ) = 0 ® ® ^ (d) B = (0.33 T) i  (0.28 T ) k ® ® ® ^ Fm = I ( l ´ B ) =  (0.0098 N) j ® ^ ^ (e) B = + (0.74 T) j  (0.36 T ) k ® ® ® ^ ^ Fm = I ( l ´ B ) =  (0.0259 N) k + (0.0126 N) j ^
^
= (0.0126 N) j  (0.0259 N) K ® ^ 23. B = (0.020 T) j ® ® ^ l1 = ab =  (40.0 cm) j ^
=  (40.0 ´ 102 m ) j
25. Consider the wire is bent in the form of a loop of N turns, L Radius of loop, r= 2pN Magnetic dipole moment associated with the loop i L2 M = NiA = Ni ´ pr 2 = 4pN 2 2 iL B t = MB sin 90° = 4pN Clearly t is maximum, when N = 1 and the maximum torque is given by i L2B tm = 4p 26. Consider the disc to be made up of large number of elementary rings. Consider on such ring of radius x and thickness dx. Charge on this ring,
® ® ® F1 = I ( l1 ´ B ) = 0
x dx
® ® ^ l 2 = bc = (40.0 cm) k ^
=  (400 ´ 102 m ) k ® ® ® ^ F2 = I ( l 2 ´ B ) = (0.04 N) i
^
= (40.2 ´ 104 )(  0.24 i  0.18 j + 0.2 k )
dq =
q 2q ´ 2px dx = 2 x dx p R2 R
92 Current associated with this ring, dq w dq w q di = = = x dx T 2p pR2 Magnetic moment of this ring wq dM = px 2di = 2 x 3dx R Magnetic moment of entire disc, wq R 1 M = ò dM = 2 ò x 3dx = w qR2 4 R 0
y g
f
b
c h
z
…(i)
® ^ Mabcd =  i l2 k
Magnetic field at the centre of disc due to the elementary ring under consideration m di m 0w q2 dB = 0 = dx 2x 2pR2 Net magnetic field at the centre of the disc, m wq R m wq B = ò dB = 0 2 ò dx = 0 0 2pR 2pR M pR3 \ = B 2m 0 27. (a) By principle of conservation of energy, Gain in KE = Loss in PE KE =  PE cos q + ME 0.80 ´ 103 K f cos q = 1 =1 ME 0.02 ´ 52 ´ 103 10 = 13 10 q = cos 1 = 76.7° 13 10 (b) q = cos 1 = 76.7° 13 Entire KE will again get converted into PE
® ^ Mefgh = i l2 k ® ^ Madeh = i l2 j \Total magnetic moment of the closed path, ® ® ® ® ^ M = Mabcd + Mefgh + Madeh = i l2 j 31. Circuit is same as in Q.30 ® ^ ^ M = i l2 j = j ® ^ B =2 j ® ® ® t = M ´B =0 32. B1 =
m0 I × 4p r Here, B1 and B2 are perpendicular to each other, hence, 2 l
l
1
B = B12 + B22
= 1.5 ´ 1016 s 19 e 1.6 ´ 10 (b) i = = = 1.1 ´ 103 A T 1.5 ´ 1016
30. Suppose equal and opposite currents are flowing in sides a d and e h, so that three complete current carrying loops are formed,
m0 I × 4p r B2 =
28. DU = U2  U1 =  MB  ( + MB) =  2 MB =  2 ´ 1.45 ´ 0.835 =  2.42 J 11 2pr 2 ´ 3.14 ´ 5.3 ´ 10 29. (a) T = = v 2.2 ´ 106
= 1.1 mA (c) M = p r 2i = 3.14 ´ (5.3 ´ 1011 )2 ´ 1.1 ´ 103 = 9.3 ´ 1024 Am 2
x
e d
a
=
m 0 2I 107 ´ 2 ´ 5 × = 4p r 35 ´ 102
= 2.0 ´ 106 T = 2.0 mT 33. Clearly DBOC ~ DAOB r2 AD = \ r6 BC Þ
r2 = 2r = 100 mm
93 q a
A
C r2
O
2a
p q
I 1 a
r 4
and AD = 2BC = 200 mm r q = cos 1 = 45° BC 2 2I m0 I BBC = [sin 45° + sin 45° ] 4p r r m0 (outwards) = 4p m I BAD = 0 × (sin 45° + sin 45° ) 4p r2 =
2
3
q
l
D
a q
B
m0 2 I × 4p r2
(inwards)
Net magnetic field at O. 2 m 0I é 1 1 ù B = BBC  BAD = ê  ú 4p êë r1 r2 úû ù é 1 1 = 2 ´ 107 ´ 2 ê 3 3 ú 100 ´ 10 úû êë 50 ´ 10 6 (outwards) = 2 ´ 10 T = 2 mT 34. Let us consider a point P ( x, y) where magnetic field is zero. Clearly the point must lie either in 1st quadrant or in 3rd quadrant.
= B3 = B4 =
m0 I × 4p a 2
m0 I (sin 0 + sin 0) 4p 2a m I = 0× 4p 2a 2
Net magnetic field at P B = B1 + B2  ( B3 + B4 ) m I = 0× 4p 2a m I m I q 36. B = 2 ´ 0 ×  0 ´ =0 4p R 2 R 2p Þ q = 2 rad.
(inwards)
(outwards)
(inwards)
l
R q
l
37. (a) Consider a point P in between the two conductors at a distance x from conductor carrying current I1 (= 25.0 A),
l2
P(xy)
x l1
P
I2 = 75.0 A
I1 = 25.0 A r
B=
35.
m 0 2I1 m 0 2I2 × × =0 4p y 4p x
Þ
I1x = I2 y
Þ
æI ö y=ç 1÷x çI ÷ è 2ø
q = 45° B1 = B2 =
m0 I × (sin q + sin q) 4p a
Magnetic field at P m I m I B = 0 × 1  0 × 2 =0 4p x 4p r  x I1 I Þ = 2 x rx r  x I2 = Þ x I1 I1 25.0 Þ x= r= ´ 40 = 10 cm I1 + I2 100.0
94 (b) Consider a point Q lying on the left of the conductor carrying current I1 at a distance x from it.
x I2 = 75.0 A I1 = 25.0 A
m 0 I1 m 0 I2 × × =0 4p x 4p r + x I1 I = 2 x r+x I1 25.0 x= r= ´ 40 I2  I1 50.0
B= Þ Þ
=
2 ´ 0.0580 ´ 2.40 ´ 10
2
4p ´ 107 ´ 800
Þ I = 2.77 A (b) On the axis of coil, 2 NIA m B= 0× 2 4 p ( r + x 2 )3 / 2 æ r2 + x 2 ö BC ( r 2 + x 2 )3/ 2 ÷ = Þç 3 ç r2 ÷ B r è ø Þ x = 0.0184 m
3/ 2
=2
41. Let the current I2 ( = I ) upwards I2
I1
I3 I4
= 20 cm 2 N pr 2I m 38. B = 0 × 2 4 p ( r + x 2 )3 / 2 But, x = R m NI 4 2 Br B= 0 ÞN= m 0I 4 2r = Þ
4 2 ´ 6.39 ´ 104 ´ 6 ´ 102 4p ´ 107 ´ 2.5
N = 69
39. For magnetic field at the centre of loop to be zero, magnetic field due to straight conductor at centre of loop must be outward, hence I1 must be rightwards. At the centre of the loop
B =  B1 + B2  B3 + B4 m 2 = 0 × [  I1 + I2  I3 + I4 ] = 0 4p r I2 = I1 + I3  I4 = 10 + 8  20 = 2 A Negative sign indicates that current I is directed downwards. ® m I^ 42. B KLM =  0 i 4R
I
I2
I
K
D I1
=
40. (a) B =
N
I
R
B = B1  B2
M
L
m 0 2I1 m 0I2 × =0 4p D 2R pD I1 = I2 R
2 BR m 0 NI ÞI= m 0N 2R
® m I^ B KNM = 0 j 4R ® ® ® m I ^ ^ B = B KLM + B KNM = 0 (  i + j) 4R ® ® ® m Iqv ^ (a) F = q ( v ´ B ) =  0 k 4R ® ® ^ (b) l1 = l 2 =  2 R k ® ® ® ^ F1 = I ( l1 ´ B ) = 2 IRB i ® ® ® ^ F2 = I ( l 2 ´ B ) = 2 IRB i ® ® ® ^ F = F1 + F2 = 4 IRB i
95 ®®
43. (a) Length of each side
òbB × dl =  m 0I1 =  5.0 ´ 10
l
6
®®
Tm
òc B × dl = m 0( I2  I1 ) = 2.5 ´ 10
òdB × dl = m 0( I2 + I3  I1 ) = 5.0 ´ 10
q
46.
ÞI=
1 ®® 1 B × dl = ´ 3.83 ´ 107 m0 ò 4p ´ 107
= 0.3A (b) If we integrate around the curve in the opposite direction, the value of line integral will become negative, i.e.,  3.83 ´ 107 Tm. ®® 45. ò B × dl = m 0I As the path is taken counterclockwise ®® direction, ò B × dl will be positive if current is outwards and will be negative if current is inwards. ®® ò B × dl = 0 a
6
Tm
y
P1
x
r
a qq
(a) By Ampere’s circuital law ®® ò B × dl = m 0I
Tm
®®
a
2 pr l= n p q= n l pr a = cot q = cot q 2 n m i B = n ´ 0 × (2 sin q) 4p a m 2n 2 sin q = 0× 4p p r cot q p m 0 i n 2 sin 2 n = p 2p 2r cos n p m 0i n 2 sin 2 n (b) lim B = lim p n ®¥ n ®¥ 2p 2r cos n m 0i Þ lim = n ®0 2r ®® 44. ò B × dl = 3.83 ´ 107 Tm
6
P2
a r
Current density J=
I 2
=
2I pa 2
a pa 2  2p æç ö÷ è2ø Let us consider both the cavities are carrying equal and opposite currents with current density J. Let B1, B2 and B3 be magnetic fields due to complete cylinder, upper and lower cavity respectively. (a) At point P1 ® m 2I ^ m 2J ´ pa 2 ^ B1 =  0 × 1 i =  0 × i 4p r 4p r m I^ = 0 i pr 2 æaö 2 J ´ p ç ÷ ® m 2 I2 ^ m 0 è 2 ø ^i B2 = 0 × i= × a 4p 4p r  a r2 2 m 0I ^ =i aö æ 4p ç r  ÷ 2ø è ® m 0 2 I3 ^ m0 ^ B3 = × i= i aö 4p r + a æ 4p ç r + ÷ 2 2ø è ® ® ® ® B = B1 + B 2 + B 3 é ù ê 4 1 1 ú^ + + i ê a aú r+ ú ê r rêë 2 2 úû 2 2 ® m 0I é 2r  a ù ^ B = ê úi 4pr êë 4r 2  a 2 úû =
m 0I 4p
96 2
2
é 2r  a ù , towards left. ê 2 2ú êë 4r  a ûú (b) At point P2
® m I \ (B ) = 0 4pr
y x
q q
® B1
y
q B 2 sin ® B2
=
2 I3 a2 r2 + 4 m 0I
^
P2
®®
® B3 B2 cos q B3 cos q
B × dl = m 0l l òWXYZ ®® ®® ®® ®® Þ ò B × dl + ò B × dl + ò B × dl + ò B × dl = m 0l l W®X
X ®Y
Y ®Z
Z ®W
B l + 0 + B l + 0 = m 0l l 1 B = m0 l 2 In Fig. 2.
B2
B1
P
^
[sin q i  cos q j ]
2p 4r 2 + a 2
^
Q B2 Fig.2
^
[sin q i  cos q j ]
1 m0 l 2 B = B1  B2 = 0,
B1 = B2 =
é2 ù ê  2 cos q ú ^j 2 2 êr 4r + a úû ë r 2r but, cos q = = 2 2 a 4r + a 2 r2 + 4 ® m 0I é 2 ù^ 4r B = j \ ê  2 2ú 2p êë r 4r + a úû m I é 2r 2 + a 2 ù ^ = 0 ê 2 új 4pr êë 4r + a 2 úû ® m I é 2r 2 + a 2 ù (B ) = 0 ê 2 ú , upwards. 4pr êë 4r + a 2 úû m 0I 2p
At point Q, 1 m0 l 2 B = B1 + B2 = m 0 l ® ® ® ® ® m I dl ´ r m 0 q ( v ´ r ) 48. B = 0 × = × 4p 4p r3 r3 B1 = B2 =
y u o
47. Let us first find magnetic field due a current carrying infinite plate. Consider a rectangular amperian (WXYZ) as shown in Fig. 1.
B1
At point P,
® ® ® ® B = B1 + B 2 + B 3 =
z
A l Fig.1
B3 sin q
® m 2I ^ m I ^ B1 = 0 × 1 j = 0 j 4p r pr ® m0 2 I2 ^ ^ B2 = × [  sin q i  cos q j ] 2 4p a r2 + 4  m 0I ^ ^ = [sin q i + cos q j ] 2p 4r 2 + a 2 ® m B3 = 0 × 4p
w
x
loop
x
® ^ v = ( 8.00 ´ 106 ms 1 ) j ® ^ (a) r = (0.500 m ) i ® ® ® m 0 q( v ´ r ) B = × 4p r3
97 ^
=
^
107 ´ 6.00 ´ 106[( 8.00 ´ 106 j) ´ (0.500) i ]
(0.500) ® ^ Þ B =  (1.92 ´ 105 T ) k
3
Þ ® 107 ´ 4.00 ´ 106 [(2.00 ´ 105 ^i ) ´ (  0.300 ^j)] B1 = (0.300)3
® ^ (b) r =  (0.500 m ) j
^
=  ( 8.89 ´ 107 T ) k ® ® ® m q (v ´ r ) B 2 = 0 × 2 23 2 4p r2
® ® ® m 0 q( v ´ r ) B = × =0 4p r3 ® ^ (c) r = + (0.500 m ) k ® ® ® m 0 q( v ´ r ) ^ B = × = (1.92 ´ 105 T ) i 4p r3 ® ^ ^ (d) r =  (0.50 m) j + 0.500 m k ® ® ® m 0 q( v ´ r ) ^ B = × = (1.92 ´ 105 T ) i 4p r3 49. q =  4.80 mC =  4.80 ´ 106 C ® ^ v = (6.80 ´ 105 m / s ) i
Þ ® 107 ´ (  1.5 ´ 106 )[( 8.00 ´ 105 ^i ) ´ (  0.400 ^j)] B2 = (0.400)2 ^
=  ( 7.5 ´ 107 T ) k ® ® ® ^ B = B1 + B 2 =  (16.4 ´ 106 T ) k ^
=  (1.64 ´ 106 T ) k or
® ^ (a) r = (0.500 m ) i ® ® ® m 0 q( v ´ r ) B = × =0 4p r3 ® ^ (b) r = (0.500 m ) j ® ® ® m 0 q( v ´ r ) ^ B = × =  (1.3 ´ 106 T ) k 4p r3 ® ^ ^ (c) r = (0.500 m ) i + (0.500 m ) j
51. Magnetic force per unit length on the conductor AB, m 2I I f = 0× 1 2 4p r For equilibrium m f = g=lg l m 2I I …(i) Þ lg = 0 × 1 2 4p r Suppose wire AB is depressed by x, f
® ® ® m 0 q( v ´ r ) ^ B = × =  (1.31 ´ 106 T ) k 4p r3 ® ^ (d) r = (0.500 m ) k ® ® ® m 0 q( v ´ r ) ^ B = × = (1.31 ´ 106 T ) j 4p r3 ® ® ® m q (v ´ r ) 50. B1 = 0 × 1 13 2 4p r1 y q
A
I1
r
lg
C
I2
B
D
Net force on unit length of wire AB la = lg  f ¢ m 2I I m 2I I = 0× 1 2  0× 1 2 4p r 4p r  x m 0 2 I1I2 × x =× 4p r ( r  x ) If x << r
v v2
q2
m 0I1I2 x 2r 2 m II a =  0 1 22 x 2l r
la = 
0.300m 0.400m
B = 1.64 ´ 106 T (inwards)
x
…(ii)
98 General equation of SHM …(ii) a =  w2x Hence, motion of wire AB will be simple harmonic. From Eqs. (i) and (ii), m 0I1I2 w= 2 lr 2 2 lr 2 2p r T= = 2p = 2p w m 0I1I2 g 0.01 = 2 ´ 3.14 9.8
D I1
30 A
f CG
m 0 2I1I2 × 4p r f r I2 = m0 × 2I1 4p 4.00 ´ 105 ´ 2.50 ´ 102 = 107 ´ 2 ´ 0.600
= 8.33 A (b) As the wires repel each other, current must be in opposite directions. m 2I I 53. f CD = 0 × 1 2 4p r1
G I3
r2
r1 10 A
20 A
m 2I I = 0× 2 3 4p r2
f = f CD  f CG éI I ù mo × 2 I2 ê 1  3 ú 4p êë r1 r2 úû é 30 20 ù ´ 2 ´ 10 ê ú 2 3 10 5 102 úû ´ ´ êë
f = f CD  f CG = = 107
= 0.2 s 52. (a) f =
C I FCG 2 FCD 3 cm 5 cm
f = 12 ´ 104 N = 1.2 ´ 103 N/m F = f l = 1.2 ´ 103 ´ 25 ´ 102 = 3 ´ 104 N 54. Force per unit length on wire MN m 2I I f MN = 0 × 1 2 4p a m II L F = f MN ´ L = 0 1 2 2 pa Torque acting on the loop is zero because magnetic field is parallel to the area vector.
Objective Questions (Level 1) 1. Fact 2 pm 2. T = is independent of speed. qB 3. Outside the wire m 2I where, r is distance from the B= 0× 4p r centre. 4. The path will be parabola if force acting on the particle is constant in magnitude as well as in direction. m 2I 5. B = 0 × 4p r 4p rB m0 = 2I m ´ Wb / m 2 Units of m 0 = A = Wbm 1A 1 6. Fact
® ® ® 7. M = i A , where A = Area vector. 8. Force acting on a closed current carrying loop is always zero. 9. M = NIA ® ® ®® 10. a ^ B Þ a × B = 0 ^
^
^
^
^
^
( x i + j  k ) × (2 i + 3 j + 4 k ) = 2x + 3  4 = 0 Þ x = 0.5 12. A current carrying closed loop experiences a force magnetic field. mv P 13. r = , = qB qB
\ \
P = mv = momentum. 1 rµ q rp qa = Þ rP : ra = 2 : 1 ra q p
never
99
r
3 cm
3 cm
5 cm I
v=
Þ
2 eV m
Magnetic force, Fm = evB sin q Fm µ v Þ Fm µ v Hence, if potential difference is doubled, force will become 2 times. 19. Magnetic field at O due to P,
q q
6 cm
14. W = MB (cos q1  cos q2 ) Here, q1 = p, q2 = p  q W = MB (cos p  cos ( p  q)) =  MB (1  cos q) m0 I 15. BP = × (2 sin q) 4p r
P
P
Q I
I
5 cm
R/2 2
r = 5  3 = 4 cm = 4 ´ 102 m
R
3 sin q = 5 3 10 ´ 50 ´ 2 ´ 5 BP = 4 ´ 102
m 0 2I m I × = 0 4p R / 2 p R
(inwards)
Magnetic field at O due to Q, m m I 2I B2 = 0 × = 0 4p R / 2 p R
(inwards)
B1 =
7
\
= 1.5 ´ 104 T = 1.5 gauss. 16. Magnetic field on the axis of current carrying circular loop, 2M m …(i) B1 = 0 × 2 4 p ( r + x 2 )3 / 2 Magnetic field at the centre of current carrying circular loop, m 2M …(ii) B2 = 0 × 3 4p r From Eqs. (i) and (ii), B2 ( r 2 + x 2 )3/ 2 = B1 r3 =
(3 2 + 42 )3/ 2 33
125 27 125 B2 = ´ 54 = 150 mT Þ 27 ® ® ® ® ® 17. F = I( l ´ B ) = I ( ba ´ B ) =
® ® ® ® =  I ( ab ´ B ) = I ( B ´ ab) 18. Kinetic energy of electron, 1 K = mv2 = e V 2
R/2
O
2
Net magnetic field at O, B = B1 + B2 =
2 m 0I pR
20. As solved in Question 16, B2 æ x 2 + R2 ö ÷ =ç B1 çè R2 ÷ø Þ Þ
æ x 2 + R2 ö ç ÷ ç R2 ÷ è ø x 2 + R2
3/ 2
3/ 2
=8
=4 R2 x= 3 R
Þ
21. Component of velocity of particle along magnetic field, i.e., qE vy = t = aE t m is not constant, hence pitch is variable. 2 mK mv 22. r = = qB qB Now, R =
2 mK eB R¢ =
2m (2K ) e (3 R)
=
2 R 3
100 23. Same as question 1. Introductory exercise 23.6.
y
Note. Her diagram is wrong correct diagram should be
1
A x
2 4
C
I
O B x
24. r =
D
2 mK 2 mqV mv [K = qV ] = = qB qB qB r=
Þ
2 mV æ 1 ö ç ÷ q è Bø
25. Magnetic field due to a conductor of finite length. m I B = 0 × (sin a + sin b ) 4p r Here, a =  q2, b = q1 and r = a m I B = 0 (sin q1  sin q2 ) \ 2a 26. In case C, magnetic field of conductor 12 and 23 at O is inward while those of 34 and 41 at O is outward, hence net magnetic field at O in this case is zero. 2
I
3
O
1
4
® ® ® 27. dF = I ( dl ´ B ) ® ® But B dl at every point, ® hence, dF = 0. 28. B1 = B3 = 0 (Magnetic field on the axis of current carrying straight conductor is zero)
a b
3
x
®  1 æ m 0I ö ^ m I ^ B2 = ç ÷ k =  0 k, 4 è 2b ø 8b ® 1 æm I ö ^ m I ^ B3 = ç 0 ÷ k = 0 k 4 è 2a ø 8a ® ® ® ® ® B = B1 + B 2 + B 3 + B 4 =
m 0I é 1 1 ù ^ k 8 êë a b úû
29. Current associated with electron, q I= = ef T m I m ef B= 0 = 0 2R 2R 30. Same as question 1(a). Introductory Exercise 23.5. 31. At point 1, Magnetic field due to inner conductor is nonzero, but due to outer conductor is zero. Hence, B1 ¹ 0 At point 2, Magnetic field due to both the conductors is equal and opposite. Hence, B2 = 0 32. Apply Fleming’s left hand rule or right hand thumb rule. 33. Magnetic field due to straight conductors at O is zero because O lies on axis of both the conductors. f m 0I m 0If Hence, B = × = 2p 2x 4px 34. Inside a solid cylinder having uniform current density,
101 B=
m 0Ir 2 p R2
Here, r = R  x \
B=
r
m 0I ( R  x ) 2 p R2
35. Magnetic force is acting radially outward on the loop.
R
JEE Corner Assertion and Reason 1. For parabolic path, acceleration must be constant and should not be parallel or antiparallel to velocity. 2. By Fleming’s left hand rule. 3. Magnetic force on upper wire must be in upward direction, hence current should be in a direction opposite to that of wire 1. Reason is also correct but does not explain Assertion.
2 meV mv = qB eB
7. For equilibrium ® ® Fe + Fm = 0 Þ Þ
® ® ® q E =  q( v ´ B ) ® ® ® ® ® E = v ´B = B ´ v
® ® 8. Pm = Fm × v
4. t = MB sin a a = 90° \t = MB ¹ 0
® ® As Fm is always perpendicular to v , ® Pm = 0
5. F2 = I lBO x1 y
® B
1
2
r=
4
®® Again, Pe = Fe× v , may or may not be zero. 9. Reason correctly explains Assertion.
F2
F4
3 x1
x2
x
F4 = I lB0x2 F4 > F2 \ Hence, net force is along Xaxis. 6. Radii of both is different because mass of both is different
10. Magnetic force cannot change speed of particle as it is always perpendicular to the speed of the particle. v2 11. a = R but R also depends on v. F qvB a= m = \ m m Þ a µv
102
Objective Questions (Level 2) Charge on this cylinder,
1. For net torque to be zero.
T q
R
y O mg
IAB0 = mgR mgR mgR I= = AB0 pR2B0 mg = p RB0 2. As it is clear from diagram, l
I (– 4,0)
(2,0)
Effective length of wire, ® ^ l = (4 m ) i ® ® ® F = I( l ´ B) ® ® F I ® ® a = = ( l ´ B) m m =
2 ^ ^ ^ (4 i ´ (  0.02 k )) = 1.6 j m /s 2 0.1
3. Impulse = Change in momentum ò I lB dt = mv  0 lBò dq = mv mv m 2 gh dq = = lB lB 4. Consider the sphere to be made up of large number of hollow, coaxial cylinder of different height and radius. Consider one such cylinder of radius x, height y and thickness. Now, y = 2 R cos q, x = R sin q, dx = R cos q dq
x
q × (2p yx dx ) 4 pR3 3 = 3 q cos 2 q sin q dq Current associated with this cylinder, dq w dq 3 w q di = = = cos 2 q sin q dq T 2p 2p Magnetic moment associated with this cylinder, 3 qw dM = di A = cos 2 q sin q dq ´ px 2 2p 3 dM = R2wqA cos 2 q sin 3 q dq 2 0 3 M = ò dM = R2qò cos 2q sin 3 q dq p2 2 0 3 = R2w qò cos 2q (1  cos 2 q)sin q dq p /2 2 dq =
0
=
é cos 3 q cos 5 q ù 3 2 R wq ê ú 2 5 úû p/ 2 êë 3
=
1 2 R wq 5
5. As solved in question 5(c). Introductory Exercise 23.2. L = sin q R mV Here, L = d, R = qB qB d \ = sin q mV q V sin q or = m Bd 6. Force on portion AC will more compared to that on portion CB.
103 7. Consider an elementary portion of the wire carrying current I1 of length dx at a distance x from end B.
mv qB qB ( b  a ) v³ m
i.e.,
ba£
or
11. Consider an elementary portion of length dx at a distance x from the pivoted end. I1
I2
B
dx
dxe
a
2a x
x
Charge on this portion
Force on this portion dF = I1dx B m 2I I = 0 × 1 2 dx 4p a + x
q dx l Current associated with this portion dq qf di = = dx T l Magnetic moment of this portion pqf 2 dM = px 2di = x dx l pqf l 2 1 M= x dx = pqfl2 l ò0 3 dq =
Total force on wire AB 2 a dx m F = ò dF = 0 × 2I1I2 ò a a + x 4p m II = 0 1 2 ln 3 2p 8. Magnetic field line due to current carrying conductor is shown in figure. z
12. At x = 0, y = ± 2 m Effective length of wire ^
l = (4 m ) j y x
® ® ® ^ ^ \ Fm = I( l ´ B ) = 3(4 j ´ 5 k ) ^
= 60 i N 13. Effective length of wire, × × × × × × × × × × × × × × × ×
2 IA1 m B1 = 0 × 2 4p ( x1 + r12 )3/ 2
9.
2I × pr12 m = 0× 2 4p ( x1 + r12 )3/ 2 m 2I × pr22 B2 = 0 × 2 4p ( x2 + r22 )3/ 2 B1 r12 ( x22 + r22 )3/ 2 = B2 r22 ( x12 + r12 )3/ 2 But, and \
r1 = x1 tan q r2 = x2 tan q B1 =2 B2
10. b  a must be less than or equal to radius of circular path,
P
Q
× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×
I T
S 60°
g
3 4d R
3 a ´ cot 60° 4 a = 2 For equilibrium, I lB = Mg 2Mg Þ I= lB l = ST = 2 ´
104 14. For particle not collide with the solenoid, radius of path of particle £ half or radius of solenoid. mv r ³ qB 2 But B = m 0n i Þ
® ® ® 19. As E =  v ´ B Net force on the particle must be zero. 20. Consider an elementary portion of length dy at y  y on the wire. Force on this portion,
rqB m 0qr n i v> = 2m 2m
® ® dF = I ( dy ´ B )
16. Magnetic force cannot do work on charged particle, hence its energy will remain same, so that q remains same. Again, magnetic force is always along the string, it will never produce a torque hence, T will also remain same.
® ^ Here, dy =  dy j (Current is directed along
17. Let the xcoordinates of loops be as shown in figure,
Total force on the wire,
y
negative yaxis). ^
^
^
dF =  I { dy j (0.3 y i + 0.4 y j)} ^
=  2 ´ 103(  0.3 y dy k ) 1
^
F = ò dF =  2 ´ 103 ò (  0.3 y dy k ) 0
^
F = (3 ´ 104 k ) N ® ® ® 21. E =  v ´ B
1
O
then,
\
a
b
b+a
® rqB E  = vB = B m (5 ´ 102 )(20 ´ 106 )(0.1)2 = (20 ´ 109 )
x
F1 = Ia ( B0a )  0 = I a 2B0 F2 = Ia ( B0 ( b + a ))  Ia ( B0b) = I a 2B0 F1 = F2 ¹ 0
® \ E = 0.5 V/m
(1 mg = 109 kg )
22. Condition is shown in figure. ® B2 1
18. Consider an amperian loop of radius x ( b < x < c ), threaded by current the amperian loop, ××
A
2
® B1
a
C
B b
I¢ = I =
x 2  b2 c 2  b2
c2  x 2 c 2  b2
I \
I 2
I=
m 0I1 2 R1 m I B2 = 0 2 2 R2 B1 =
2
m 0I ¢ m 0I ( c  x ) = 2px 2px ( c 2  b2 )
B = B12 + B22 =
m0 2
2
æI ö æ I1 ö ç ÷ +ç 2÷ çR ÷ çR ÷ è 2ø è 1ø
2
105 2
® m 2I ^ B1 = 0 × j 4p a ® m 2I ^ B2 =  0 × i 4p a ® B3 = 0
ö æ 2 ÷ ç æ 5 2 ö 4p ´ 10 5 ÷ ç ç ÷ + = ´ ç 5 ´ 102 ÷ 2 ç 5 ´ 102 ÷ è ø ÷ ç 2 ø è 4p ´ 107 ´ 2 = = 4p ´ 105 T 2 ´ 102 7
23. Initially, net force on the particle is zero. Hence, E V= B Now, if electric field is switched off. mv E æ q = Sö r= = ç ÷ qB SB2 èm ø 24. For equilibrium, mg [f = magnetic force per unit length on f = l the conductors] m 0 2I1I2 Þ × =lg 4p r m 2I I r= 0× 1 2 Þ 4p l g =
® ® ® ® m 0i ^ ^ B = B1 + B 2 + B 3 = ( j  i) 2pa 26. Effective length, l = AC = 42 + 3 2 =5 m C
B
A
F = I lB = 2 ´ 5 ´ 2 = 20 N 27. At point P,
107 ´ 2 ´ 100 ´ 50 0.01 ´ 10
= 0.01 m Clearly, equilibrium of conductor B is unstable. ® ® ® 25. If B1, B 2 and B 3 be magnetic fields at the
1 qx × 4pe0 ( R2 + x 2 )3/ 2 2 iA m B= 0× 2 4 p ( R + x 2 )3 / 2 q qv Hence, i = = T 2 pR E=
and A = p R2 \
given point due to the wires along x, y and z axis respectively, then
E 1 1 c2 = × = B m 0e0 v v
More than One Correct Options 1. B1 =
B2 =
m 0 N1I1 4p ´ 107 ´ 50 ´ 2 = 2 R1 2 ´ 5 ´ 102 m 0 N 2 I2 = 2 R2
= 4p ´ 104 T 4p ´ 107 ´ 100 ´ 2 2 ´ 10 ´ 10
® If F = 0 ® ® ® Either, E =  v ´ B , ® ® E ¹ 0, B = 0
2
= 4p ´ 104 T If current is in same sense, B = B1 + B2 = 8p ´ 104 T And if current is in opposite sense, B = B1  B2 = 0 ® ® ® ® ® ® 2. F = Fe + Fm = q ( E + v ´ B )
or
® ® ® E = 0, or v ´ B = 0
® ® Again, If v ´ B = 0 ® Either B = 0 ® ® or q = 0°, i.e., v  B .
é 1 ù êc = ú m 0e0 úû êë
106 3. The particle will describe a circle in xy plane with radius, 2 2 mv 1 ´ 8 + 6 r= = =5m qB 1 ´2 2pm T= = p s = 3.14 s qB
and 4.
® t = MB sin q U =  pE cos q q = 80° Hence, t = 0, U = pE = maximum. As PE (U) is maximum, equilibrium is unstable.
5. Fact. 6. Upward and downward components of force will cancel each other while leftward force is more than rightward force, hence net force is leftwards. ® ® ® ® 7. F = q E + q ( v ´ B ) ^
^
^
= q { E0 k + (v j) ´ ( B0 i )}
^
= q ( E0  vB0 ) k E0 , particle will deflect towards B0 positive zaxis. E If v > 0 , particle will deflect towards B0 negative zaxis. E If v = 0 , particle will move undeflected and B0 its KE will remain constant.
If v <
8. K = e V Þ K µ V will become double 2mK R= Þ R µ K will become 2 times. qB qB is independent of kinetic energy. w= 2pm 9. Use right hand thumb rule. 10. For cd to be in equilibrium, force on it must be repulsive while for ab to be in equilibrium, force on it must be attractive. Equilibrium of cd will be stable while that of ab will be unstable.
Match the Columns 1. ( a ® r), (b ® q), (c ® p), (d ® r) ® ® ® ® ® Fm = q ( v ´ B ) =  e ( v ´ B ) ® ® ® and Fm = q E =  e E 2. (a ® r), (b ® s), (c ® q), (d ® p) ® ® ® As Fm = q ( v ´ B ) By Fleming’s left hand rule, positively charged particles deflects towards left and negatively charged particles deflects towards right. 2 mK mv Again, r = = qB qB m Þ rµ q 3. (a ® p, s), (b ® p, q), (c ® p, r), (d ® p, s) Whenever a closed current carrying loop is placed in uniform magnetic field, net force experienced by it is zero. Also t = PE sin a
is maximum if a = 90°, i.e., in case (b) only. And U =  PE cos a U is positive if a is obtuse, i.e., in cases (a) and (d). and U is minimum if a = 0, i.e., in case (c). 4. (a ® q), (b ® r), (c ® s), (d ® s) Use right hand thumb rule. 5. (a ® q), (b ® r), (c ® q), (d ® r) 1
2
® F12 ® F14
3
® F24 ® ® F13 F23
® F21
4
® F34 ® F31
® F24 ® F32
® F43 ® F41
Direction of different forces on different wires is shown in figure.
107 ® I ® ® I ® ® ^ F = ( l1 ´ B ) + ( l 2 ´ B ) =  B0I l k 2 2
6. (a ® q, s), (b ® p, r), (c ® p, r), (d ® q, s)
® t = 0, because lines of action of force on the two wires are equal and opposite. ® ^ If B = B0 j
i
When the current is increased or the loop is moved towards the wire, magnetic flux linked with the loop increases. As a result of this, induced current will produce in the loop to decrease the magnetic field. Because initial magnetic flux linked with the loop is inward, induced magnetic flux will be outward and induced current will be anticlockwise and viceversa.
® ^ F = B0I l k Again, lines of action of force on the two wires are equal and opposite. t =0 ® ^ ^ If B = B0 ( i + j) ® F =0
7. (a ® r, s), (b ® r, s), (c ® q, r), (d ® p, r)
® t =0
y I/2 I/2 I
If I/2 x
I/2
Effective lengths of two conductors, ^
^
l1 = l2 = l i + l j If
® ^ B = B0 i
® ^ B = B0 k ® ^ ^ F = B0I l ( i  j)
® Þ F  = 2B0I l t =0
24
Electromagnetic Induction Introductory Exercise 24.1
1. Magnetic field inside the loop due to current carrying conductor is inwards. As the current in the conductor increases, magnetic flux linked with the loop increases as a result of which, induced current will produce in the loop to produce an outward magnetic field, i.e., induced current will be anticlockwise.
2. No. Emf is induced if the field is time varying. dfB 3. = induced emf dt é dfB ù 2 3 1 \ ê dt ú = [ V ] = [ML T I ] ë û
Introductory Exercise 24.2 1. If the outward magnetic flux increases, induced current will be in such a way that it produces inwards magnetic flux, i.e., it will be clockwise. 2. Magnetic flux linked with the coil will not change, hence induced current will be zero. 3. If the current in coil 1 (clockwise) increases, outward magnetic flux linked with the coil 2 increases and the coil 2 will produce induced current in clockwise direction to oppose the change in magnetic flux linked with it.
i
1
2 i'
i increasing i
1
2 i'
i decreasing
Hence, if the current in coil 1 increases, induced current will be in same sense and viceversa.
Introductory Exercise 24.3 1. fB = BS = B0 S e  at df e =  B = a B0 S e at dt 2. No. As,
Fm = i lB = 0
Because, i = 0 as the circuit is not closed. As net force acting on the bar is zero, no external force is required to move the bar with constant velocity.
109 f2  f1 t But, f1 = NB1 A cos q, f2 = NB2 A cos q NA cos q ( B2  B1 ) \ e = t et ÞA= N( B2  B1 )cos q
3. e =
=
B=
Induced emf in this portion, m 2vi de = B dxv = 0 × dx 4p d + x
80.0 ´ 103 ´ 0.4 50 ´ (600 ´ 106  200 ´ 106 ) ´
3 2
= 1.85 m 2 Side of square, a = A = 1.36 m Total length of wire = 50 ´ 4a = 50 ´ 4 ´ 1.36 = 272 m
5. (a) EMF induced in the bar ab, l dx m e = ò de = 0 × 2 viò 0d + x 4p m = 0 2vi [ln ( d + x )]0l 4p m 0vi d + l = ln 2p d m 0vi æ lö = ln ç1 + ÷ 2p d è ø
4. (a) Consider an elementary portion of length dx of the bar at a distance x from end a. Magnetic field at this point,
(b) Magnetic field in the region ab is inwards, hence by Fleming’s left hand rule, positive charge will move up and a will be at higher potential.
i
Or d
a dx
m0 2i × 4p a + x
x v l
b
Use Fleming’s right hand rule. (c) No. As flux linked with the square loop will remain same.
Introductory Exercise 24.4 1. Potential difference across an inductor, di d V =L =L (3 t sin t ) dt dt = 3 L [sin t + t cos t ]
Introductory Exercise 24.5 1. (a) Total number of turns on the solenoid, 40 ´ 102 l N= = d 0.10 ´ 102 = 400 L= =
m 0N2A l 4p ´ 107 ´ (400)2 ´ 0.90 ´ 104 40 ´ 102
(b)
= 4.5 ´ 105 H di e = L dt =  4.5 ´ 105 ´ = 4.5 ´ 103 V = 4.5 mV
0  10 0.10
110
Introductory Exercise 24.6 1. Consider a current i is flowing in the outer loop. i
r
R
Magnetic field at the centre of the loop. m i B= 0 2R
As R >> r, magnetic field inside smaller loop may assumed to be constant. Hence, magnetic flux linked with the smaller loop, m pr 2i fm = B ´ pr 2 = 0 2R fm pm 0r 2 M= = 2R i
Introductory Exercise 24.7 1. (a) V0 = i0 R = 36 ´ 103 ´ 175 = 6.3 V (b) i = i0 (1  e  t / t ) L where, t = R Now, at t = 58 ms i = 4.9 mA \ 4.9 = 36 (1  e 58 / t ) 31.1 e 58 / t = Þ 36 Þ t = 397 ms L = 397 ms R L = 175 ´ 397 ´ 106 Þ = 69 mH (c) t = 397 ms [ e] [ V ][ t ] 2. = [L] = [ i] é di ù êë dt úû [V ] and [ R] = [ i] é L ù [L] \ êë R úû = [ R] = [T ] 3. (a) Initially E=L Þ
(b)
di dt
di E = dt L 12.0 = = 4 A/s 3.00 E = VL + VR
di + iR dt di 1 Þ = [ E  iR] dt L 1 = ´ [12  1 ´ 7] 3.00 di 5 Þ = = 1.67 A/s dt 3 L 3 (c) t = = R 7 i = i0 (1  e  t / t ) E 12 = (1  e t / t ) = (1  e 1.4/ 3 ) R 7 Þ i = 0.639 A E 12 (d) i0 = = = 1.71 A R 7 E2 4. (a) P = Ei = (1  e  t / t ) R (12)2 = (1  e 7t / 3 ) = 20.6 (1  e 2.33t ) W 7 (b) Rate of dissipation of energy, PR = i2 R = i02 R (1  e 7t / R )2 = 20.6 (1  e 2.33t )2 W (c) Rate of increase of magnetic energy di PL = ei = L i dt = 20.6 ( e 2.33t  e 4.67t ) W (d) Clearly, P = PR + PL Þ
E=L
5. No. E = VL + VR and VR cannot be negative in RL circuit.
111 R1 R2 = 2W R1 + R2 E 5 i0 = = = 0.5 A R + 8 10 L 1 t= = R + 8 10
6. Consider the system as a combination of two batteries (E1 = 10 V and E2 = 0) as shown A
R1 = 4W
8W
i
R=
B
i2
i1
R2 = 4W E1 = 10V
1H
E2 = 0 D
C
ß
R
8W
1H
E
E=
E1 R2 + E2 R1 =5 V R1 + R2
i = i0 (1  e  t / t ) i = 0.5 (1  e 10t ) A \Current through inductor i = 2.5 (1  e 10t ) A In loop ABCDA di i1 R1 + 8i + L  E1 = 0 dt i1 ´ 4 + 8 ´ 0.5 (1  e 10t ) + 1 (5e 10t )  10 = 0 i1 = (1.5  0.25 e 10t ) A
Introductory Exercise 24.8 1.
[ q] [ i][ T ] ] [C ] = = [V ] [V ] [V ] [T ] [ e] = [L] = [ i] é di ù êë dt úû Þ [ LC ] = [ L C ] = [T ]
2. In LC oscillations, magnetic energy is equivalent to kinetic energy in spring block system. dq dx i= Þv= dt dt Also L is equivalent to inertia (m) in electricity, hence 1 Magnetic energy = Li2 is equivalent to 2 1 kinetic energy = mv2. 2
q = 18 ´ 106 ´ 0.75 ´ 3.40 = 46.5 ´ 106 C = 46.5 mC di æ 1 ö (b) e =  L =  L çq÷ dt è LC ø 4 q 4.8 ´ 10 = = = 23.3 V C 18 ´ 106
Þ
4. i0 = wq0
1 LC
where, w = Þ
V0 =
q0 i = 0 C wC
V0 = i0
20 ´ 103 L = 0.1 ´ C 0.5 ´ 106
= 20 V
3. In LC oscillations, di 1 di (a) =q Þ q =  LC dt LC dt
Introductory Exercise 24.9 dfm di =  m 0 nNA dt dt 25 =  4p ´ 107 ´ ´ 10 ´ 5.0 ´ 104 ´ (  0.2) 0.01 e=
1. (a) B = m 0 ni fm = NBA = m 0n NAi
112 2
6
= 3.14 ´ 10 V = 3.14 mV 3.14 ´ 106 e (b) E = = 2pR 2 ´ 3.14 ´ 25 ´ 102 ´ 10 = 2 ´ 107 V/m 2. B = (2.00t 3  4.00t 2 + 0.8) T dB = (6.00t 2  8.00t ) T/s dt dB is negative, dt hence B is decreasing in that interval. dB For t > 1.33 s, is positive, hence B is dt increasing for t > 1.33 s. (a) For point P2, dfm2 dB induced emf, V2 = =  pR2 dt dt Induced electric field at P2, V R2 dB E= 2 =× 2pr2 2r2 dt
From, t = 0 to t = 1.33 s,
=
R (6.00t 2  8.00t ) 2r2
F =  eE =
R2 (6.00t 2  8.00t ) 2r2
= 8.0 ´ 1021 N As magnetic field is increasing in this region, induced electric field will be anticlockwise and hence, electron will experience force in clockwise sense, i.e., downward at P2. (b) For point P1, dfm1 dB Induced emf, V1 = =  pr12 dt dt Induced electric field at P1, V 1 dB E =  1 =  r1 2pr1 2 dt 1 =  r1(6.00t 2  8.00t ) = 0.36 V/m 2 At, t = 2.00 s magnetic field is increasing, hence, induced electric field will be anticlockwise, i.e., upward at P1 and perpendicular to r1.
AIEEE Corner Subjective Questions (Level 1) 1. < e > = 
f2  f1 B ( A2  A1 ) =t t A1 = pr 2 = 3.14 ´ (0.1)2 = 3.14 ´ 102 = 0.0314 2 æ 2pr ö A2 = a 2 = ç ÷ è 4 ø 2
\
æ 2 ´ 3.14 ´ 0.1 ö =ç ÷ = 0.025 4 è ø 100 (0.025  0.0314) <e>=0.1 = 6.4 V
2. f1 = NBA = 500 ´ 0.2 ´ 4 ´ 104 = 0.04 Wb f2 =  NBA =  0.04 Wb Average induced emf, (f  f ) <e>= 2 1 t Average induced current, (f  f ) <e> <i>= = 2 1 R Rt
Charge flowing through the coil q=<i>t ( f2  f1 ) (  0.04  0.04) =Þq=R 50 0.08 = = 1.6 ´ 103 C 50 = 1.6 mC = 1600 mC 3. f1 = NBS, f2 =  NBS Induced emf, ( f  f ) 2 NBS <e>= 2 1 = t t Induced current < e > 2 NBS <i>= = R Rt Charge flowing through the coil, 2 NBS q=<i>t= R 4.5 ´ 106 ´ 40 qR Þ B= = 2 NS 2 ´ 60 ´ 3 ´ 106
113 Hence, equal force in direction of motion of coil is required to move the block with uniform speed.
= 0.5 T ® ^ ^ 4. B = (4.0 i  1.8 k ) ´ 103 T, ® ^ S = (5.0 ´ 104 k ) m 2 ®® f = B × S =  9.0 ´ 107 Wb 5. e = Blv = 1.1 ´ 0.8 ´ 5 = 4.4 V By Fleming’s right hand rule, north end of the wire will be positive. 6. A = pr 2 = 3.14 ´ (12 ´ 102 )2 = 0.045 m 2 (a) For t = 0 to t = 2.0 s 0.5  0 dB = slope = = 0.25 T/s dt 2.0  1 df dB e= m =A dt dt =  0.045 ´ 0.25 =  0.011 V e = 0.011 V (b) For, t = 2.0 s to t = 4.0 s dB = slope = 0 e = 0 dt (c) For, t = 4.0 s to t = 6.0 s 0  0.5 dB = slope = =  0.25 dt 6.0  4.0 df dB e= m =A = 0.11 V dt dt 7. (a) When magnetic flux linked with the coil changes, induced current is produced in it, in such a way that, it opposes the change. Magnetic flux linked with the coil will change only when coil is entering in (from 3L L L to x =  ) or moving (from x = x=2 2 2 3L to x = ) of the magnetic field. 2 Because, of induced current, an opposing force act on the coil, which is given by BLv B2L2v F = ilB = BL = R R F F0 =
B2L2v R
(b) When the coil is entering into the magnetic field, magnetic flux linked with the coil increases and the induced current will produce magnetic flux in opposite direction and will be counterclockwise and viceversa. i i0 = i0 x –i0
8. Consider an elementary section of length dl of the frame as shown in figure. Magnetic flux linked with this section, l dl i
x
dfm = BdB =
a
m 0 2i × adl 4p x + l
Total magnetic flux linked with the frame, m ai a dl fm = ò dfm = 0 ò 2p 0 x + l m ai = 0 [ln ( x + a )  ln x ] 2p Induced emf  dfm m ai é 1 1 ù dx e= =  0 ×ê  ú dt 2p ë x + a x û dt =
m 0a 2i m 0a 2iv v= 2px ( x + a ) 2px ( x + a )
9. As solved in Qusetion 4. Introductory Exercise 24.3. i d v l
F0 –3L –L O L 2 2 2
3L 2
x
BLv R
114 m iv æ lö e = 0 ln ç1 + ÷ dø 2p è
Effective emf E r  E1r2 E= 21 r1 + r2 0.008 ´ 15.0  0.004 ´ 10.0 = 15.0 + 10.0
Here,
i = 10 A v = 10 ms 1 l = 10.0 cm  1.0 cm = 9.0 cm d = 1.0 cm
= 0.0032 V 15 ´ 10 rr r= 12 = =6W r1 + r2 25 E 0.0032 i= = = 0.003 A = 0.3 mA R+ r 5+6
4p ´ 107 ´ 10 ´ 10 æ 9.0 ö e= ln ç1 + ÷ 2p 1.0 ø è e = (2 ´ 10V ) ln (10) V 10. Induced current e Blv = R R Force needed to move the rod with constant speed = Magnetic force acting on the rod Blv ie., F = i lB = lB R 2 2 2 B2l2v (0.15) ´ (50 ´ 10 ) ´ 2 = = R 3 F = 0.00375 Þ i=
11. Suppose the magnetic field is acting into the plane of paper. Rods 1 and 2 can be treated as cells of emf E1 ( = Blv1 ) and E2 ( = Blv2 ) respectively. 2
1 B Ä R
v2
v1
r2
12. (a) e =  L
(b) Current flowing from b to a is decreasing, hence, a must be at higher potential. 13. (a) i = 5 + 16t, e = 10mV = 10 ´ 103 V di d e = L (5 + 16t ) Þ 10 ´ 103 = L dt dt 3 10 ´ 10 L= = 0.625 mH 16 (b) at t = 1 s i = 5 + 16 (1) = 21 A Energy stored in the inductor, 1 1 U = Li2 = ´ 0.625 ´ 103 ´ (21)2 2 2 = 0.138 J dU di P= = Li = 0.625 ´ 103 ´ 21 ´ 16 dt dt = 0.21 W 14. From t = 0 to t = 2.0 ms V 0 5.0  0 = =0 t  0 2.0 ´ 103
r1
ß
Þ
R r2
E1 r1
E r
V = 2500 t di = 2500 t dt 2500 ò di = L ò tdt i 2500 t ò 0 di = L ò 0 tdt 1250 2 i= t L t = 2.0 ms 1250 i= ´ (2.0 ´ 103 )2 150 ´ 103 L
ß E2
di =  0.54 ´ (  0.030) dt = 1.62 ´ 102 V
R i
Now, E1 = Blv1 = 0.010 ´ 10.0 ´ 102 ´ 4.00 = 0.004 V E2 = Blv2 = 0.010 ´ 10 ´ 0 ´ 102 ´ 8.00 = 0.008 V
Þ
at
= 3.33 ´ 102 A
115 From t = 2.0 ms to t = 4.0 ms V  5.0 0  0.50 = t  2.0 ´ 103 (4.0  2.0) ´ 103 V =  2500 ( t  2.0 ´ 103 ) + 5.0 =  2500 t + 10.0 di L =  2500 t + 10.0 dt 1 di = (  2500 t + 10.0) dt L 1 i = [  1250 t 2 + 10.0 t ] L at t = 4 s 1 i= [  1250 ´ (4.0 ´ 103 )2 150 ´ 103 + 10.0 (4.0 ´ 103 )] 2
= 3.33 ´ 10 A e di 0.0160 15. (a)e = L Þ L = = dt di / dt 0.0640
18. (a)e = M
di is constant, induced emf is dt constant. (b) Coefficient of mutual induction remains same whether current flows in first coil or second. di Hence,  e = M1 = 0.27 V dt As,
19. (a) Magnetic flux linked with the secondary coil, f2 = Mi1 f2 0.0320 ´ 400 M= = i1 6.52 = 1.96 H (b) f1 = Mi2 = 1.96 ´ 2.54 = 4.9784 Wb Flux per turn through primary coil f 4.9784 = 1 = N1 700
= 0.250 H (b) Flux per turn Li 0.250 ´ 0.720 f= = N 400 = 4.5 ´ 104 Wb i i di 16. e = M =M 2 1 dt t 12  4 3 Þ 50 ´ 10 = M × 0.5 50 ´ 103 ´ 0 . 5 M= = 3.125 ´ 103 H 8 = 3.125 mH If current changes from 3 A to 9 A in 0.02 s. i i di e = M =M 2 1 dt t 9 3 = 3.125 ´ 103 ´ 0.02 = 0.9375 V 17. (a) Magnetic flux linked with secondary coil, fm2 = M i1 6.0 ´ 103 ´ 1000 f M= 2 = =2H 3 i1 dfm2 di (b) e==M 1 dt dt 0 3 = 2 ´ = 30 V 0.2 fm 600 ´ 5 ´ 103 (c) L = 1 = =1H 3 i1
di = 3.25 ´ 104 ´ 830 dt = 0.27 V
= 7.112 ´ 103 Wb/turn. 20. Same as Question 2. Introductory Exercise 24.4 21.
i = i0(1  e  t / t ) Rt E = (1  e L ) R Rt di E  L = e dt L Power supplied by battery, Rt E2 P = Ei = (1  e L ) R Rate of storage of magnetic energy Rt Rt di E2 P1 = Li = (1  e L ) e L dt R 10 ´ 0.1
Rt
P1 =e L =e P L 2 22. (a) t = = = 0.2 s R 10
= e 1 = 0.37
1
R
L
K E
116 i But i = 0 2
E 100 (b) i0 = = = 10 A R 10 (c) i = i0(1  e

t t
i = 10 (1  e

1 0.2 )
5
= 10 (1  e ) = 9.93 A 23. (a) Power delivered by the battery, Rt E2 P = Ei = (1  e L ) R 12.8 ´ 0.278
(3.24)2 (1  e 3.56 ) 12.8 = 0.82 (1  e 1 ) = 0.518 W = 518 mW (b) Rate of dissipation of energy as heat Rt E2 P2 = i2 R = (1  e L )2 R = 0.82 (1  e 1 )2 = 0.328 W = 328 mW (c) Rate of storage of magnetic energy P1 = P  P2 = 190 mW di 24. E = VL + VR = L + iR dt =
L
R
VL
VR
Rt L
æ i0 = i0 ç1  e ç 2 è
)
Rt L
ö ÷ ÷ ø
1 2 1.25 ´ 103 L t = ln 2 = ´ 0.693 R 50.0 = 17.3 ´ 106 = 17.3 ms 1 1 æ1 ö (b) U = Li2 = ç L i02 ÷ 2 2 è2 ø 1 i= i0 2 Rt ö æ i i0 ç1  e L ÷ = 0 ç ÷ 2 è ø Rt 2 1 e L = Þ 2 L 2 Þ t = ln R 2 1 e

=
= 30.7 ms . 26. Steady state current through the inductor L, r i0
K
R
E
(a) Initially, i = 0 di E 6.00 = = = 2.40 A/s dt L 2.50 (b) When, i = 0.500 A di E  iR 6.00  0.500 ´ 8.00 = = dt L 2.50 = 0.80 A/s Rt ö E æç (c) i = 1e L ÷ ÷ R çè ø 8.00 ´ 0.250 ö 6.00 æç ÷ 2.5 = 1e ÷ 8.00 ç è ø
\
0.8
= 0.750 (1  e ) = 0.413 A E 6.00 (d) i0 = = = 0.750 A R 8.00 Rt ö æ 25. (a) i = i0 ç1  e L ÷ ç ÷ è ø
S
E
E r When the switch S is open L t= R+ r i0 =
(a) i = i0 e t / t
æ (R + r) ö
E  ççè L ÷÷ø e r (b) Amount of heat generated in the solenoid i=
Þ
¥
¥
0
0
H = ò i2r dt = i02r ò e 2t / t dt E 2 ì t 2 t / t ¥ ü ]0 ý í [ e r î 2 þ ( R + r ) E2 = 2rL
=
117 When the switch S is open, current i2 flows in the circuit in clockwise direction and is given by i2 = i0e t / t E Here, i2 = R2 L t= R1 + R2
27. At any instant of time, 5 mH i1 10 mH i 20 V
i2 =
5W
L1 Þ
=
di1 di = L2 2 dt dt L1i1 = L2i2 i1 = 2i2
…(i)
In steady state, inductors offer zero resistance, hence 20 i= =4A 5 But i1 + i2 = i 4 8 i2 = A, i1 = A 3 3 28. When the switch is closed, i
i2 i1
E
L R1 R2
S
E (1  e  R2 t / L ) R2 di2 E  R2 t / L = e dt L Potential difference across L di V + L 2 = E e  R2 t / L = (12e 5t ) V dt i2 =
i2 i1 E
L R1 R2
S
i2
æ R1 + R 2 ö ÷÷ t L ø
E  ççè e R2
12 10t e = (6 e 10t ) A 2
29. For current through galvanometer to be zero, L1,R1
R3
P
i1
G
i1 L2,R2
R4 Q
i2
K
i2
E
VP = VQ di1 di L1 + i1 R1 = L2 2 + i2 R2 dt dt Also, i1 R3 = i2 R4 From Eqs.(i) and (ii), di di L1 1 + i1 R1 L2 2 + i2 R2 dt dt = i1 R3 i2 R4
…(i) …(ii)
…(iii)
In the steady state, di1 di2 = =0 dt dt R1 R2 R R \ = Þ 1 = 3 R3 R4 R2 R4 Again as current through galvanometer is always zero. i1 = constant i2 di1 / dt or = constant di2 / dt di1 dt = i1 or …(iv) di2 i2 dt From Eqs. (iii) and (iv),
118 L1 R3 R1 = = L2 R4 R2 30. (a) In LC circuit Maximum electrical energy = Maximum magnetic energy 1 1 CV02 = Li02 Þ 2 2 2
æ 1.50 ö æV ö ÷ L = C çç 0 ÷÷ = 4 ´ 106 ç ç 50 ´ 103 ÷ i è 0ø ø è
2
= 3.6 ´ 103 H L = 3.6 mH
Þ (b) f =
1 2p LC 1
=
2 ´ 3.14 3.6 ´ 10
3
´ 4 ´ 10
6
= 1.33 ´ 103 Hz = 1.33 kHz (c) Time taken to rise from zero to maximum value, T 1 1 t= = = 4 4f 4 ´ 1.33 ´ 103 = 3 ´ 103 s = 3 ms. 31. (a) w = 2pf = 2 ´ 3.14 ´ 103 = 6.28 rad/s 1 1 3 T = = 3 = 10 s = 1 ms f 10 (b) As initially charge is maximum, (i.e.., it is extreme position for charge). q = q0 cos w t q0 = CV0 = 1 ´ 106 ´ 100 = 104 4 q = [10 cos (6.28 ´ 103 ) t ] C. \ 1 (c) w = LC 1 1 ÞL= 2 = w C (6.28 ´ 103 )2 ´ 106 Þ
= 2.53 ´ 103 L = 2.53 mH
(d) In one quarter cycle, entire charge of the capacitor flows out. q 4CV = t T 6 4 ´ 10 ´ 100 = = 0.4 A 103 <i>=
32. (a) V0 =
q0 5.00 ´ 106 = C 4 ´ 104
= 1.25 ´ 102 V= 12.5 mV (b) Maximum magnetic energy = Maximum electric energy 1 2 q02 Li0 = 2 2C q Þ i0 = 0 LC 5.00 ´ 106 = 8.33 ´ 104 A Þ i0 = 0.090 ´ 4 ´ 104 (c) Maximum energy stored in inductor, 1 = L i02 2 1 = ´ 0.0900 ´ ( 8.33 ´ 104 )2 2 = 3.125 ´ 108 J (d) By conservation of energy, q2 1 2 1 2 + Li = Li0 2C 2 2 i0 But i = 2 q2 3 2 = Li0 2C 8 i 3 q = 0 3 LC = q0 2 2 1.732 = ´ 5.00 ´ 106 2 = 4.33 ´ 106 C 1 1 æ1 ö Um = Li2 = ç Li02 ÷ 2 4 è2 ø = 7.8 ´ 109 J 1 1 = 3 LC 2.0 ´ 10 ´ 5.0 ´ 106
33. (a) w =
= 104 rad/s ½ ½ di ½ ½ = w2Q ½ dt ½ = (104 )2 ´ 100 ´ 106 = 104 A/s (b) i = w
Q02
 Q2
= 104 (200 ´ 106 )2  (200 ´ 106 )2 = 0 (c) i0 = wQ0 = 104 ´ 200 ´ 106 = 2 A (d) i = w Q02  Q 2
119 Þ Þ
i0 = w Q02  Q 2 2 w Q0 = w Q02  Q 2 2 1.73 ´ 200 ´ 106 3 Q= Q0 = 2 2 = 173 mC
34. As initially charge is maximum q = q0 cos w t and i = i0 sin w t 1 1 where, w = = LC 3.3 ´ 840 ´ 106 » 19 rad./s i0 = wq0 = 19 ´ 105 ´ 106 » 2.0 ´ 103 A = 2.0 mA At t = 2.00 ms q2 q2 (a) Ue = = 0 (cos 2 wt ) 2C 2C (105 ´ 106 )2 [cos 2 (38 rad)] = 2 ´ 840 ´ 106 Ue = 6.55 ´ 106 J = 6.55 mJ 1 1 (b) Um = Li2 = Li02 (sin w t ) 2 2 1 = ´ 3.3 ´ (2 ´ 103 )2 sin 2(38 rad ) 2 = 0.009 ´ 106 J = 0.009 mj q2 1 (c) U = 0 = Li02 2C 2 = 6.56 ´ 106 J = 6.56 mJ 35. As the inward magnetic field is increasing, induced electric field will be anticlockwise. × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×E × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×x× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×
At a distance x from centre of the region, Magnetic flux linked with the imaginary loop of radius x fm = p x 2B  dfm dB e= =  p x2 dt dt Induced electric field, e 1 dB E= = x 2p x 2 dt At a, E=
1 dB , towards left. r 4 dt
At b , E=
1 dB , upwards. r 2 dt
At c, E =0 36. Inside the solenoid, B = m 0ni dB di = m 0n dt dt Inside the region of varying magnetic field 1 dB 1 di E= r = m 0nr 2 dt 2 dt (a) r = 0.5 cm = 5.0 ´ 103 m 1 di E = m 0rn 2 dt 1 = ´ 4p ´ 107 ´ 5.0 ´ 103 ´ 900 ´ 60 2 = 1.7 ´ 104 V/m (b) r = 1.0 cm = 1.0 ´ 102 m 1 di E= rn m0 dt 1 = ´ 4p ´ 103 ´ 5.0 ´ 103 ´ 900 ´ 60 2 = 3.4 ´ 104 V/m
120
AIEEE Corner Objective Questions (Level 1) di dt [ V ][ T ] [ ML2 T 3 A 1 ] [T ] [L] = = [ i] [A ]
1. V = L
8.
= [ ML2 T 2 A 2 ] 2. M µ n1n2 3. Both will tend to oppose the magnetic flux changing with them by increasing current in opposite direction. 4. Moving charged particle will produced magnetic field parallel to ring, Hence fm = 0 Velocity of particle increases continuously due to gravity. 5. Induced electric field can exist at a point where magnetic field is not present, i.e., outside the region occupying the magnetic field. 6. At, t = 1 s 2W a
i
4V
2H
2F – + a
b
q = 4t 2 = 4 C dq i= = 8t = 8 A dt di = 8 A/s dt di d 2q As, = = Positive dt dt Charge in capacitor is increasing, current i must be towards left. di q Vab =  2I + 4  L dt C 4 =  2 ´ 8 + 4  2 ´ 8 _ =  30 V 2 di d 7. e = M =M ( i0 sin w t ) dt dt = w Mi0 cos w t Maximum induced emf = w Mi0 = 100p ´ 0.005 ´ 10 = 5p
1 2 1 L 2 Li0 = CV02 Þ i0 =2 ´ 2 2 C 4 ´ 106
= 2 ´ 103 V 1 9. e = B l2w, is independent of t. 2 df Df 10. e = = dt t Df =et = iRt = 10 ´ 103 ´ 0.5 ´ 5 = 25 ´ 103 Wb = 25 mWb. 11. As inward magnetic field is increasing, induced electric field must be anticlockwise. Hence, direction of induced electric field at P will be towards and electron will experience force towards right (opposite to electric field). 12. f = at ( t  t ) = att  at 2 df e = = at  2at dt e at  2at i= = R R 2 t 2 t ( at  2 at ) H = ò i R dt = ò dt 0 0 R 3 t 1 é ( at  2 at ) ù = ê ú R êë 3 ´ (  2a ) úû 0 1 3 3 = [  a t  a 3t 3 ]  6 Ra =
13. E =  L
a 2t 3 3R
di dt
di + 15  iR dt =  5 ´ 103(  103 ) + 15  5 ´ 1 = 15 V
14. VBA =  L
15.
di = 10 A/s, at t = 0, i = 5A dt di = 10 A/s dt
121 di  E =0 dt = 5 ´ 3 + 1 ´ 10  10 = 15 V æ d 2q ö q ÷ = w2q0 = 0 = çç ÷ dt LC ø max è
VA  VB = iR + L
æ di ö 16. ç ÷ è dt ø max 17. V = L
di dt
E 12 = = 40 A R 0.3 1 1 U0 = Li02 = ´ 50 ´ 103 ´ (40)2 2 2 = 40 J t Rt ö æ  ö E æç 26. i = i0 ç1  e t ÷ = 1e L ÷ ç ÷ Rç ÷ è ø è ø 25. i0 =
Rt
18. fm = BA cos q Þ Þ Þ Þ Þ ® 19. A
df dq e =  m = BA sin q dt dt dq iR = BA sin q dt dq dq R = BA sin q dt dt BA dq = sin q dq R BA 3p/ 2 q= sin q dq = 0 R òp/ 2 ^ ^ ^ ^ ® = ab k, B = 20t i + 10t 2 j + 50 k ®® fm = B × A = 50 ab df e =  m =0 dt
20. E = Vb + iR Vb = E  iR = 200  20 ´ 1 . 5 = 170 V V N 1 21. s = s Þ Vs = ´ 290 = 10 V Vp N p 2 ip N = s is Np Þ
is =
Np Ns
ip = 2 ´ 4 = 8 A
22. Vr = 0, hence magnetic flux linked with the coil remain same.  df \e= =0 dt 1 23. s = at 2 2 Due to change in magnetic flux linked with the ring, magnet experiences an upward force, hence, a<g 1 2 s < gt Þ s < 5 m 2 di 24. VA  VB = L =at dt
di E  L = e dt L Rt di VL = L = Ee L dt at t = 0 VL = E = 20 V at t = 20 ms 
R ´ 20 ´ 10 3

R
L VL = Ee Þ 5 = 20e 50L R = ln 4 Þ R = (100 ln 4) W 50L e 1 df 1 dB 27. i = = × = NA R R dt R dt 10 ´ 10 ´ 104 = ´ 108 ´ 104 20 =5A
28. In the steady state, inductor behaves as short circuit, hence entire current flows through it. fm = AB cos q But, q = 90° fm = 0 \ e 1 dfm 30. i = =R R dt dq  nBA d = (cos q) Þ dt R dt nBA dq = sin q R dt nBA Þ dq = sin q dq R 2nBA nBA p sin q dq = Q1 = R ò0 R nBA 2p Q2 = sin q dq = 0 R ò0 Q2 \ =0 Q1 29.
31. According to Lenz’s law, induced current always opposes the cause producing it.
122 t e t
Rt e L
ö ö Eæ ÷ ÷ = ç1 ÷ Rç ÷ ø è ø 5 ´2 ö 15 æç = 1  e 10 ÷ = 3 (1  e 1 ) ÷ 5 ç è ø 1ö æ = 3 ç1  ÷ A eø è
æ 32. i = i0 ç1 ç è
33. Velocity of AB is parallel to its length. 34. Velocity of rod is parallel to its length. 35. Vc  Va = Vc  Vb = BRV and Va  Vb = 0
df 37. E = dt 38. Magnetic flux linked with the change, hence e  1 df i= = × =0 R R dt 1 39. e = Blv cos q = Bl2w cos q 2
coil does not
l ö æ çQ v = w÷ 2 ø è
Ascos qvaries from 0 to 1 1 e varies from 0 to B l2w. 2
36. Induced current always opposes the cause producing it.
JEE Corner Assertion and Reason 1. Magnetic flux linked with the coil is not changing with time, hence induced current is zero.
8. L = m rm 0n 2lA, for ferromagnetic substance, mr > g and L does not depends on i.
2. Both Assertion and Reason are correct but Reason does not explain Assertion.
9. As soon as key is opened a
3. Induced electric field is nonconservative but can exert force on charged particles. 4. i = 2t  8 di =2 dt
R1 R2
E
L
di = 2 ´2 = 4 V Va  Vb = L dt æ di ö 5. ç ÷ = ( imax ) w = 1 ´ 2 = 2 A/s è dt ø max 6. Va  Vb = Vc  Va Vc > Va > Vb
i
b
i = i0 =
E R1
10. Inductors oppose change in current while resistor does not.
7. Fact.
Objective Questions (Level 2) 1. By conservation of energy 1 1 L i02 = mv02 2 2 m i0 = v0 k
iR
R
2. Wire AB behaves as a cell of emf, E = Blv
iC
E
q
+ –
C
123 E Blv = R R ic = 0 1 1 Uc = CE2 = CB2l2v2 2 2
\ fm = B tan q x 2 df dx e =  m =  2B tan q x dt dt = 2B tan q vx R = r l = r (2x tan q) where, r = resistance per unit length of the conductor. e Bv \ i= = = constant. R r
iR =
3. Apply Fleming’s left hand rule. 4. For SHM, v = w A cos wt e = Blv = Blw A cos wt ì ï e0 cos wt e =í ï e0 cos wt î
for nT < t < (2n  1) for
T 2
(2n  1)T > t > nT 2
5. fm = BA At any instant when wires have moved through a distance x, A = ( a + 2 x )2 fm = B( a + 2x )2 df dx e = m = 4B( a + 2x ) dt dt = 4B( a + 2x ) v0 e 4B ( a + 2x ) v0 Bv0 i = = = R l ´ 4 ( a + 2x ) l 6. A = l2 dA dl = 2l =  2 la dt dt
at
dl ö æ ça =  ÷ dt è ø
fm = BA df dA e= m =B = 2Bla dt dt l=a e = 2a aB
7. At this instant, direction of motion of wire PQ is perpendicular to its length. e = Blv \ 8. q = CV = CBlv = 20 ´ 106 ´ 0.5 ´ 0.1 ´ 0.2 = 0.2 mC Plate A is positive while plate B is negative. æ1 ö 9. fm = BA = B ç lx ÷ è2 ø l v q
But l = 2x tan q
x
10. fm = BA cos q = BA cos wt df e =  m = w BA sin wt dt 2 But A=b e = b2Bw sin wt \ 11. Induced emf dB = (1)2 ´ 2 ´ 103 dt = 2 ´ 103 V W = qe = 1 ´ 106 ´ 2 ´ 103 = 2 ´ 109 J e = a2
12. In the steady capacitor = 0.
state,
current
through
20 =4A 5 f1 = 0, f2 = iL L = 4 ´ 500 ´ 102 = 2 Wb Df = f2  f1 = 2 Wb. 1 1 æ1 ö 13. Li2 = ç Li02 ÷ 2 2 è2 ø i0 i= 2 t æ  ö i i0 ç1  e t ÷ = 0 ç ÷ 2 è ø 2 1 e t / t = 2 æ 2 ö÷ t = t ln ç ç 2  1÷ è ø æ L 2 ö÷ = ln ç ç R è 2  1 ÷ø m i 14. B = 0 2pa m iqv F = qvB = 0 2pa iL =
15. Consider an elementary section of loop of width dx at a distance x from wire AB
124 B
18. Induced current opposes change in magnetic flux. P
Q
19. VL = E  iR
i
20. The rod can be assumed as a cell of emf E = Blv The equivalent circuit is shown in figure,
C a S
R
dx
x
2W
m i dfm = BdA = 0 C dx 2px m iC b dx m 0iC b fm = 0 ò = ln 2p a x 2p a m C f b M = m = 0 ln i 2p a
R C
x E
dx
F
v
a B b
D
Q
fm =
m 0iy b ln 2p a
b m 0i ln dfm dy a e= = dt 2p dt m 0iv b e= ln a 2p e m 0iv b i= = ln R 2pR a Consider an elementary portion of length dx of the rod at a distance, x from the wire PQ. Force on this portion, dF = i dxB m 2i = i 0 × dx 4p x b dx m0 F=i × 2i 4p òa x b é m iv b ù m i = ê 0 ln ú 0 ln 2 p R a 2 p a ë û = 17. E =
1 dB r Þ E µr 2 dt
Þ 3W
i=
E Blv 0.50 ´ 0.25 ´ 4 = = 2+3 5 5
21. Outside the region of magnetic field, induced electric field, r 2 dB Br 2 E= = 2 R dt 2R F = qE 1 t = qER = qBr 2 2
P A
12W
E
4W
= 0.1 A
16. From previous question
i
i E
2W
b A
1 é m 0iv b ù ln ú vR êë 2p aû
2
22. VA  V0 = B (2 R)V Þ VA  V0 = 2 BRV L hL 23. L1 = , L2 = h+1 h+1 R1
L1
L2
R2 E
hR hR , R2 = R1 = h+1 h+1 1 1 1 = + Le L1 L2 h+1 h+1 = + hL L (h + 1) é 1 1 ù = + L êë h 1 úû hL Le = ( h + 1 )2 hR L L Similarly, Re = \ t= e = 2 R R ( h + 1) e
125 24. i = i0e t / t T / t
Bi0 = i0e T t= æ1ö ln ç ÷ è Bø L 25. Given, i02 R = P, =t R when, choke coil is short circuited, Total heat produced = Magnetic energy stored in the choke coil 1 2 1 æPö 1 = Li0 = ( Rt )ç ÷ = Pt 2 2 è Rø 2 26. i = i0
Rt e L

Rt L
=1
Rt = 0 = not possible. L 27. To final time constant, short the battery and find effective resistance in series with inductor R
R/2
L
Þ R
R Re = 2 L 2L t= = Re R 28. When switch is at position 1. In steady state, 2
Þ
31. Initially, inductor offers infinite resistance, hence, di i = 0 and = maximum dt \ E = VL + VC + VR But VC = VR = 0 VL = E Þ 32. Same as Q.12 objective Questions (Level 2). 33. Let V0 = Potential of metallic rod, VB  V0 = B (2 R)V = 2 BR2w V0  VC = B (2 R)V = 2 BR2w Adding Eqs. (i) and (ii), we get VB  VC = 4 BwR2
L
1
34. e = Blvc vc =
i2
\ R E
L L R
i1
E R i2 = 0
i1 =
t = t ln 2 L t = ln 2 2
30. At the moment when switch is thrown to position 2, current in capacitor = current in inductor just before throwing the switch to position 2, E ic = Þ R
For current to be constant i = i0 e
When switch is thrown to position 2. E E i1 = , i2 = R R 1 2 1 æ1 2ö 29. Li = ç L i0 ÷ 2 4 è2 ø i i= 0 2 t æ  ö i i0 ç1  e t ÷ = 0 ç ÷ 2 è ø
v1 + v2 2
1 Bl (v1 + v2 ) 2 or dA e=B dt 1 dA = l ( dx1 + dx2 ) 2 1 æ dx1 dx2 ö e = Bl ç + ÷ dt ø 2 è dt 1 = Bl (v1 + v2 ) 2
…(i) …(ii)
dx2
v2
e=
dx1
v1
126 35. Initially, capacitor offer zero resistance and inductor offers infinite resistance. Effective circuit is given by R R R E = 5V R R
e A dB 37. i = = R R dt B A B [(2b)2  pa 2 ] = 0 = 0 R R B0(4b2  pa 2 ) = R As inward magnetic field is increasing, net current must be anticlockwise. Hence current in inner circle will be clockwise. 38. From Q. 48 Subjective Questions (Level 1). m ai a fm = 0 ln æç1 + ö÷ 2p xø è
ß
R/3
5R/6=5 W
Case 1 x = b, a = a m ai a fm1 = 0 ln æç1 + ö÷ 2p bø è m 0 ai æ b + a ö = ln ç ÷ 2p è b ø
i Þ E =5V
E =5V R/2
E =1A R t R t  1 ö E æç E æç  R2 C ö÷ 36 . i1 = 1  e L ÷ , i2 = e ÷ ÷ R1 çè R2 ç ø è ø i=
4W
5W 5W
L = 10 mH K C = 0.1mF ß i1
i2
R1= 10W
R2= 10W E
L = 0.1mH
fm2
x=ba a=a m 0 ai æ a ö÷ = ln ç1 + ç b a ÷ø 2p è
m 0ai æ b ö ÷ ln ç çb  a÷ 2p è ø fm2  fm1 <e>=t fm  fm1 <e> <e>= = 2 R Rt fm2  fm1 q=<i>t=R m 0 ai é æ b + a ö æ b öù =÷  ç ln ÷ú ê ln ç 2pR ë è b ø è b  a ø û  m 0ai æ b ö÷ ln ç 2 = ç b  a2 ÷ 2pR è ø æ ö m ai b ÷ q = 0 ln ç 2 ç b  a2 ÷ 2pR è ø =
E = 20V
6W
Case 2
C = 0.1mF
i = i1 + i2 t R t  1 ö E æç E  R2 C 1e L ÷ + e = ÷ R2 R1 ç ø è at t = 103 ln 2 10 ´ 10 3 ln 2 ö 10 3 ln 2 æ ÷ 3 3 20 ç 20 10 ´ 10 i= e 10 ´ 0.1 ´ 10 ç1  e ÷+ 10 ç ÷ 10 è ø 1ö æ æ1ö = 2 ç1  ÷ + 2 ç ÷ = 2 A 2ø è è2ø
39. Magnetic flux linked with the coil. m n iA fm = nBA = 0 2r dfm e = dt dfm Þ iR = dt
127 1 dq df R = m Þ dq = dfm R dt dt m 0n iA m 0nA i q= di = 2rR ò0 2rR 40. Induced electric field inside the region of varying magnetic fields, 1 dB 1 E= r = r(6t 2 + 2x ) = 3 r ( t 2 + x ) V/m 2 dt 2 R At, t = 2.0 s and r = = 1.25 cm 2 = 1.25 ´ 102 m E = 3 ´ 1.25 ´ 102 ´ (4 + x ) = 0.3 V/m F = eE = 1.6 ´ 1019 ´ 0.3 = 48 ´ 1021 N 1 dB 41. E = r Þ E µr 2 dt 42. As inward magnetic field is increasing, induced electric field must be anticlockwise. df dB 43. e = m = pa 2 = pa 2B0 dt dt e 1 44. E = = aB0 2pa 2 45. t = qEa = ia qEa a= = ma 2 qB0 = 2m
1 aB0a 2 ma 2
q´
46. P = tw = t( at ) = ia 2 × t q2B02 = ma 2 ´ t m m2
dB = 2 T/s, A = 0.2 ´ 0.4 = 0.08 m 2 dt 0.08 [Q R = r ´ ( b + 2l)] \ i= ´ 2 = 16 A 1 ´ 1.0 As outward magnetic field is increasing, induced current must be clockwise. dA dB dB 48. e = B + A = Blv + A dt dt dt At t = 2 s, B = 4 T, A = 0.2 ´ (0.4  vt ) = 0.06 m 2 v = 5 cm/s = 0.05 m / s \ e =  4 ´ 0.2 ´ 0.05 + 0.06 ´ 2 =  0.04 + 0.12 = 0.08 V e 49. F = ilB = lB R 0.08 = ´ 0.2 ´ 4 1 ´ 0.8 = 0.008 N 50. When terminal velocity is attained, power delivered by gravity = power dissipated in two resistors mgv = 0.76 + 1.2 1.96 v= = 1 m/s 0.2 ´ 9.8 51. e = Blv = 0.6 ´ 1 ´ 1 = 0.6 V e2 P1 = R1 Þ 52. P2 =
At t = 1 s q2B02a 2 P= 4m 47. i =
e A dB = × R R dt
Þ
R1 =
e 2 (0.6)2 = = 0.47 W P1 0.76
R2 =
e 2 (0.6)2 = = 0.3 W P2 1.2
e2 R2
128
More than One Correct Options 1 æ1ö 1. e = B ç ÷v = BLv 2 è2ø By Fleming’s left hand rule, P must be positive w.r.t. Q. 2. dfm = BdA = Ba dx I dx
x
a
x
m 0a i dx 2px m ai fm = 0 ln 2 2pi fm m 0a M= = ln 2 i 2p If the loop is brought close to the wire, upward magnetic flux linked with the loop increases, hence induced current will be clockwise. =
3. f = Li = HenryAmpere. V dt Volt second V L= = = di / dt di Ampere 4. t =
e 80 = = 20 A R 4 q = it = 20 ´ 0.1 = 2 C Current is not given as a function of time, hence heat produced in the coil cannot be determined. i=
L =1s R E (1  e  t / t ) R = 4 (1  e t )
i = i0(1  e  t / t ) =
At t = ln 2, i = 2A Power supplied by battery, P = EI = 16 J/s. Rate of dissipation of heat in across resistor = i2 R = 8 J/s VR = iR = 4 V Va  Vb = E  VR = 4 V 5. In both the cases, magnetic flux linked with increases, so current i2 decreases in order to oppose the change. 6. f1 = BA = 4 ´ 2 = 8 Wb, f2 = 0 f f 8 e= 2 1 = = 80 V t 0.1
7. In LC oscillations, 1 w 1 ,f = w= = 2p 2p LC LC 1 T = = 2p LC f q i0 = w q0 = 0 LC q0 æ di ö = wq0 = ç ÷ LC è dt ø max q æ di ö ( VL )max = L ç ÷ = 0 dt è ø max C 8. If magnetic field increases, induced electric field will be anticlockwise and viceversa. 9. q = 2t 2 i=
dq = 4t dt
di = 4 A/s dt dq As = Positive dt Charge on the capacitor is increasing, hence current flows from a to b. 1H
a i
b
2F c
4W
d
+ –
t = 1 s, q = 2 C, i = 4 A di = 4 A/s dt di = 1 ´4 = 4 V Va  Vb = L dt q 2 Vb  Vc = = = 1 V c 2 Vc  Vd = iR = 4 ´ 4 = 16 V Va  Vd = 4 + 1 + 16 = 21 V 1 10. Va  Vb = Bl2w 2 1 Vc  Vb = Bl2w 2 Va  Vc = 0 [Direction of velocity of rod ac is parallel to length ac]
129
Match the Columns [ B] =
1.
[ ML T 2 ] [F] = [ i][ l] [ A ][L ]
4. i1 =
E 9 = = 1.6 A R1 6 i1
= [ ML0 T 2 A 1 ] [ V ][ dt ] [ ML2 T 3 ][T ] [L] = = [ di] [A ]
i1
2. i = i0 (1  e
L =1s R E =5A i0 = R VR = iR = E (1  e t ) VL = E  VR = Et  t
At t = 0, VL = E = 10 V, VR = 0 t =1s 1ö æ VL = E (1  e 1 ) = ç1  ÷ 10 V eø è 10 V VR = e
at
3. In LC oscillations, 1 w= = LC
1 1´
1 4
= 2 rad/s
q0 = 4 C i0 = wq0 = 8 A
i2 =
When, q = 2 C VL = VC =
E (1  e R2
At
R2 t
q = 8V C
æ di ö 1 æ di ö = 8 A/s. ç ÷= ç ÷ è dt ø 2 è dt ø max di =1 ´8 = 8V VC  VL = L dt
L
) = 3 (1  e t / 3 )
t = (ln 2) s q 21/ 3 = i2 R2 = q (1  e  t / 3 ) 1 ö æ = q ç1  1/ 3 ÷ 2 ø è
VL = E  i2 R2 = qe  t / 3 = VR 2
VR1 = i1 R1 = 9 V Vbc = VL + VR2 = 9 V (a ® s), (b ® s), (c ® p), (d ® p). 5. Induced emf f(Wb)
4
2
æ di ö = w2q0 = 16 A/s. ç ÷ è dt ø max
When,
R2
) t=
L R1
E
= [ ML2 T 2 A 2 ] [ LC ] = [T 2 ] [ fm ] = [ B][ S ] 0 = [ ML T  2 A 1 ] [L2 ] = [ ML 2 T  2 A 1 ] t / t
i2
t
e = slope of f  t graph 4 0 = =2V 2 0 e 2 i = = =1A R 2 q = it = 1 ´ 2 = 2 C As current i is constant H = i2 Rt = (1)2 ´ 2 ´ 2 = 4J
25
Alternating Current Introductory Exercise 25.1
VDC 100 = = 10 W I 10 V 150 Z = AC = = 15 W I 10 X L = Z 2  R2 = (15)2  (10)2 R=
1.
Þ Þ
=5 5 W XL XL 5 5 L= = = w 2 pf 2 ´ 3.14 ´ 50 \
» 7.7 H As \
» 0.036 H VL = IX L = 50 5 V = 111.8 V
2. For phase angle to be zero,
XL = XC 1 wL = wC 1 1 L= 2 = w C ( 2 pf )2 C 1 = (360)2 ´ 106 XL = XC Z =R V 120 I= = =6A Z 20
Introductory Exercise 25.2 1. Resonating frequency,
wr =
1 LC
=
6
0.03 ´ 2 ´ 10 =
fr =
2. Resistance of arc lamp,
1 104 6
wr 104 = 2 p 2 ´ 3.14 ´ 6
» 1105 Hz Phase angle at resonance is always 0°.
VDC I 40 = =4W 10 Impedance of series combination, V 200 Z = AC = = 20 W I 10 R 4 1 Power factor = cos f = = = Z 20 5 R=
131
AIEEE Corner Subjective Questions (Level1) 1. (a) X L = w L = 2 pfL
3. (a) Power factor at resonance is always 1,
= 2 ´ 3.14 ´ 50 ´ 2 = 628 W X (b) X L = w L Þ L = L w XL 2 = = 2 pf 2 ´ 3.14 ´ 50
as Z = R, Power factor = cos f = (b) P =
I 0 E0 cos f E20 = 2 2R (150)2 = = 75 W 3 ´ 150
(c) Because resonance is still maintained, average power consumed will remain same, i.e., 75 W.
= 6.37 mH 1 1 (c) X C = = wC 2 pfC 1 = 2 ´ 3.14 ´ 50 ´ 2 ´ 106
4. (a) As voltage is lag behind current,
inductor should be added to the circuit to raise the power factor. R (b) Power factor = cos f = Z R 60 250 Þ Z= = = W cos f 0.720 3
= 1592 W = 1.59 kW 1 1 (d) X C = ÞC= wC w XC 1 = = 1.59 mF 2 ´ 3.14 ´ 50 ´ 2 2
R = 1. Z
X C = Z 2  R2
2
2. (a) Z = R + ( X L  X C )
2
250 ö 2 = æç ÷  (60) è 3 ø
2
æ 1 ö ÷ = R2 + çç w L wC ÷ø è
æ 1 = (300)2 + çç 400 ´ 0.25 400 ´ 8 ´ 106 è
= 58 W 1 C= w XC 1 = 2pf X C
2
ö ÷ ÷ ø
= 367.6 W V0 120 = = 0.326 A Z 367.6 X  XC (b) f = tan 1 L R 1 212.5 = tan »  35.3 ° 300 As X C > X L voltage will lag behind current by 35.3°. (c) VR = I 0 R = 0.326 ´ 300 = 97.8 V, VL = I 0 X L = 32.6 V VC = I 0 X C = 0.326 ´ 312.5 = 101.875 V » 120 V I0 =
=
1 2 ´ 3.14 ´ 50 ´ 58
= 54 mF For resonance, 1 wr = LC 1 Þ L= 2 wr C 1 = (2pf )2 C Þ
L=
1 (2 ´ 3.14 ´ 50)2 ´ 54 ´ 106
= 0.185 H
132 (b) w = 1000 rad/s 60 \ I= = 1.2 ´ 102 A 1000 ´ 5
5. V ( t) = 170 sin (6280 t + p / 3) volt
i ( t) = 8.5 sin (6280t + p / 2) amp. V 170V O
0.25
0.50
0.75 1.00
1.25
t (ms)
(c) w = 10000 rad/s 60 \ I= = 1.2 ´ 103 A 10000 ´ 5 7. VR = (2.5 V ) cos [( 950 rad/s ) t ]
–170V
(a) I = i
VR R
(2.5 V ) cos [(950 rad/s) t ] 300 = ( 8 . 33 mA) cos [(950rad / s) t ] (b) X L = w L = 950 ´ 0.800 = 760 W (c) VL = I 0 X L cos ( w t + p / 2) Þ VL =  I 0 X L sin w t =  6.33 sin [(950 rad / s) t ] V 8. Given, L = 0.120 H, R = 240 W, C = 7.30 mF, I rms = 0.450 A , f = 400 Hz X L = w L = 2pfL = 2 ´ 3.14 ´ 400 ´ 0.120 = 301.44 W 1 1 XC = = wC 2pf C 1 = 2 ´ 3.14 ´ 400 ´ 7.3 ´ 106 =
O
(b) f =
1 — 12
1 — 3
7 — 12
5 — 6
13 — 12
4 — 3
t (ms)
w 6280 = = 1000 Hz 2p 2 ´ 3.14 = 1 kHz
p p p (c) f =  = 2 3 6 p 3 = 6 2 As phase of i is greater than V, current is leading voltage. (d) Clearly the circuit is capacitive in nature, we have R cos f = Z 3 R 2 = ÞZ= Þ R 2 Z 3 170 V Also, Z= 0 = = 20 W 8.5 i0 Þ
cos f = cos
3 Z = 10 3 W 2 Z = R2 + X 2C Þ X C = Z 2  R2 R=
Again,
= 400  300 = 10 W 1 1 1 XC = Þ = wC w X C 6280 ´ 10 = 1592 . mF V V 6. I = = XL wL (a) w = 100 rad/s \
I=
60 = 0.12 A 100 ´ 5
= 54.43 W R R (a) cos f = = 2 Z R + ( X L  X C )2 240 = 2 (240) + (301.44  54.43)2 = 0.697 f = cos 1(0697 . ) » 458 . ° (b) Z = R2 + ( X L  X C )2 = (240)2 + (301.44  54.43)2 = 344 W (c) Vrms = I rms Z = 0.450 ´ 344 = 154.8 V » 155 V (d) Pav = Vrms I rms cos f = 155 ´ 0.450 ´ 0.697 = 48.6 W (e) PR = I 2rms R = (0.450)2 ´ 240 = 48.6 W (f) and (g) Average power associated with inductor and capacitor is always zero.
133
Objective Questions (Level1) 1. In an AC circuit, cos f is called power
13. V = 10 cos 100p t
factor.
at t =
2. DC ammeter measures charge flowing in
the circuit per unit time, hence it measures average value of current, but average value of AC over a long time is zero. Z = R2 + ( X L  X C )2
3.
= R
V = 10 cos 100p = 10 cos
æ 1 ö ÷ + çç w L w C ÷ø è
Hence, for X L < X C , Z decreases with increase in frequency and for X L > X C , Z increases with increase in frequency. p 4. As voltage leads current and f < , hence 2 either circuit contains inductance and resistance or contains inductance, capacitance and resistance with X L > X C . 5. RMS value of sine wave AC is 0.707 I 0 ,
but can be different for different types of AC’s.
15. X C =
X 1 = Z 3 1 é 1 ù f = sin Þ êë 3 úû 3p I E 17. f = , P = 0 0 cos f = 0 2 2 VDC 18. R = = 100 W I DC V 100 Z = AC = = 200 W I AC 0.5 16.
sin f =
X L = Z 2  R2 = 100 3 W L=
2
7. Z = R + ( X L  X C )
V0 I 0 [V0 and I 0 are peak voltage and 2 current through resistor only] V 9. Vrms = 0 = 170 V 2 w 120 f = = » 19 Hz 2p 2 ´ 3.14
LC
=
19. I rms =
Vrms = wCVrms XC = 100 ´ 1 ´ 106 ´
200 2 2
I rms = 20 mA
10. Current is maximum at
1
X L X L 100 3 = = w 2pf 2p ´ 50
æ 3ö ÷÷ H = çç è p ø
8. P =
w = wr =
p 3 = 10 ´ =5 3 V 6 2
1 1 1 Þ X C µ or X C µ wC w f
6. P = I v Ev cos f = 0 2
1 600
14. For purely resistive circuit f = 0. 2
2
1 s, 600
1 0.5 ´ 8 ´ 106
= 500 rad/s. I 0 E0 cos f 11. P = 2 100 ´ 100 p = cos ´ 103 = 2.5 W 2 3 1 12. X C = = ¥ if w = 0, i.e., for DC wC
20. V = VR2 + VL2 = (20)2 + (15)2
= 25 V, V0 = 25 2 V I V cos f 21. P = 0 0 =0 2 Þ cos f = 0 Þ f = 90° 22. R is independent of frequency. 23. L is very high so that circuit consumes
less power.
134 24. tan f =
Þ
XL X Þ tan 45° = L 100 R X L = 100 W w L = 100 W 100 L= » 16 mH 2 ´ 3.14 ´ 103
1 1 1 = = = = 5 ms 4 f 4 ´ 50 200 26. f = 60°
1 4 ´ 220 ´ I 0 V0 cos f 2 P= = 2 2 = 220 W
25. The minimum time taken by it in reaching
from zero to peak value =
T 4
JEE Corner Assertion and Reasons 1. X C and X L can be greater than Z because 2
2
Z = R + ( X L  X C)
Hence, VC = IX C and VL = IX L can be greater than V = IZ . 2. At
resonance X L = X C , with further increase in frequency, X L increases but X C decreases hence voltage will lead current. 1 , if dielectric slab is inserted 3. fr = 2p LC between the plates of the capacitor, its capacitance will increase, hence, fr will decrease. 4. q = Area under graph
1 1 ´ 4 ´ (2 + 3) + ´ 4 ´ (2 + 4) 2 2 = 22 C q 22 Average current = = = 3.6 A t 6 =
5. On inserting ferromagnetic rod inside the
inductor, X L and hence VL increase. Due to this current will increase if it is lagging and viceversa. 6. VR = VL = VC Þ R = X L = X C
Hence, f = 0 and I is maximum. as Z = R2 + ( X L  X C )2 is minimum. 7. I = I L  I C = 0 8. P = I 2rms R = ( 2)2 ´ 10 = 20 W 9. Inductor coil resists varying current. 10. I 0 =
E0 2
2 2
, f = tan 1
R +wL
wL R
11. At resonance, current and voltage are in
same phase and I 0 =
V0 . Hence, I 0 R
depends on R.
Objective Questions (Level2) Single Correct Options 1. For parallel circuit 1 é1 /
XL ù 1 4 f = tan ê ú = tan 1 / R 3 ë û = 53 ° 2. Current will remain same in series circuit
given by
I = I 0 sin ( w t  f) X ö æ = I 0 sin ç w t  tan 1 L ÷ R ø è 3. R = R1 + RL = 10 W
X L = w L = 10 W, 1 XC = = 10 W wC
135 Reading of ammeter V 10 2 I rms = rms = R 10 = 2 A = 1.4 A Reading of voltmeter, V = I rms RL = 5.6 V 1 1 4. X C = = wC 2p ´ 5 ´ 103 ´ 1 ´ 106 p = 100 W V 200 IR = = = 2 A, R 100 V 200 IC = = =2 A X C 100
8. For parallel RLC circuit,
I = I 2R + ( I C  I L )2 2
Þ
= V0 9.
7. As X C > X L voltage will lag with current.
Again V = VR2 + ( VL  VC )2 = 10 V V < VC R V 4 and cos f = = R = Z V 5 Hence, a, b and c are wrong.
\
VL 2 = tan 1æç ö÷ VR è7 ø
10. Clearly P is capacitor and Q is resistor,
as, VP = V Q , X C = R . \When connected in series, Z = X 2C + R2 = 2 R p and f = , leading. 4 1 p A, leading in phase by . \I = 4 2 4 11. I = I 2R + ( I C  I L )2
Here, I C < I or I L > I 12. I = I L  I C = 0.2 A 13. For a pure inductor voltage leads with
p current by . 2 14. VR = IR = 220V Hence it is condition of resonance, i.e., VL = VC = 200 V H1 I 2DC R I2 15. = 2 = =2 H2 I rms R ( I / 2)2 16. H = I 2rms R =
6. As voltage is leading with current, circuit
p is inductive, and as f = ,X L = R 4 R R or L= = w 100
1 æ 1 ö ÷ + çç wC 2 w L ÷ø R è
V = VL2 + VR2 = 72.8 V f = tan 1
5. Let i = i1 + i2
Another method p é ù i = 5 ê1 + cos æç + 100 w t ö÷ú 2 è øû ë é ù ö 2æ p = 5 ê2 cos ç + 50 w t ÷ú è4 øû ë p = 10 cos2 é + 50 wt ù êë 4 úû p 1 Average value of cos2 æç + 50 w t ö÷ = è4 ø 2 10 \ average value of i = = 5 A. 2
2
ö ÷÷ ø
2
[Question is wrong. It should be choose the correct statement]. where, i1 = 5 A, i2 = 5 sin 100 w t A Average value of i1 = 5 A Average value of i2 = 0 \Average value of i = 5 A
æV V æV ö I = ç 0 ÷ + çç 0  0 è R ø è XC XL
17.
V02 R I 20 R = 2 2 2( R + w2 L2 )
VL = IX L = Iw L I VC = IX C = wC
If w is very small, VL » = 0,VC » V0 . V 18. Resistance of coil, R = =4W I When connected to battery V 12 I= = = 1.5 A R+r 4+4
136 2
19. VR = V 
VC2
1 XC = wC
=6V
f = tan 1
VC 4 = tan 1 VR 3
=
20. VC = V 2  VR2 = 16 V 21.
22.
1 =8 W 50 ´ 2500 ´ 106
Z = R2 + ( X L  X C )2 = 5 W
p I = I 0 sin æç t + p ö÷ è2 ø p 3p I = I 0 at + p = 2 2 t =1s V I0 = 0 2R
2 Vrms R Z2 (12)2 ´ 3 = = 17.28 W (5)2
Average power = I 2rms R =
25. Already X C > X L , with increase in w, X C
further decrease in w, X C increases and X L decreases, hence, I will decrease.
3 = 3R wC V I Þ I0 = 0 = 0 2R 2 VDC 12 23. R = = =3 W I DC 4 X C¢ =
26. For maximum current
w = wr =
1 LC
1
=
6
1 ´ 10 =
´ 4.9 ´ 103
105 rad/s. 7
27. In resonance,
24. X L = Z12  R2 = (5)2  (3)2
Z = R2P + X 2C » 77 W
=4W
28. In resonance, cos f = 1.
More than One Correct Answers 1. VR2 + VL2 = 10000
…(i) …(ii) …(iii)
VL  VC = 120 VR2 + ( VL  VC )2 = (130)2 = 16900 On solving Vr = 50 V, VL = 86.6 V, VC = 206.6 V V 50 5 and cos f = R = = V 130 15 As VC > VL , circuit is capacitive in nature. 2.
i = 3 sin w t + 4 cos w t = R sin ( w t + f) 4 R = 5 and f = tan 1 3 2i 2R 10 im = 0 = = p p p If V = Vm sin w t current will lead with the voltage. If V = Vm cos w t current will lag with voltage.
P V = 1 A, R = = 60 W V I For AC, 100 Z= = 100 W 1 X C or X L = Z 2  R2 = 80 W
3. I =
XL 80 4 = = H w 2p ´ 50 5p 1 1 125 or C= = = mF w X C 2p ´ 50 ´ 80 p V or R¢ + R = I Þ R¢ = 100  60 = 40 W R ì 1 if R = Z 4. cos f = =í Z î 0 if R = 0 L=
5. As X L > X C , voltage will lead with the
current. Z = R2 + ( X L  X C )2 = 10 2 W
137 XL  XC p = = 45° R 4 R 1 cos f = = Z 2
Vrms = VR2 + ( VL  VC )2 = 100 V ,
f = tan 1
6. As X L > X C , w > wr
with increase in w, X L and hence, Z will increase while with decrease in w, Z will first decrease and then increase. V 7. X c = C = 50 W I VR = IR = 80 V VL = IX L = 40 V
V0 = 100 2 V V
8. I =
2
æ 1 ö ÷ R2 + çç wL C ÷ø w è
with change in L or C I may decrease or 1 . increase depending on effect on wL wC
Match the Columns 1. (a) ® (p, r), (b) ® (q, r), (c ® s), (d) ® (p)
4. (a ® q),
Concept based insertion.
R=
2. (a) ® (p, s) current and voltage are in
same phase so either X C = 0, X L = 0 or X C = X L ¹ 0. (b) ® (q) I =  I 0 cos w t p = I 0 sin æç w t  ö÷ 2ø è f = 90° Þ R = 0 (c) ® (r, s) current is leading with voltage p by , either X L = 0 or X C > X L 6 but X C and R are nonzero. p (d) ® (s) current lags with voltage by , R 6 and X L are both nonzero. 3. (a) ® (q, s), (b) ® (r, s), (c) ® (r, s),
(d) ® (r, s). V V2r and P = 2 Z Z with increase in L, C or f , Z may increase or decrease, hence power and current. I=
VR 40 = = 20 W I 2
(b ® p) VC = IX C = 2 ´ 30 = 60 V (c ® r) VL = IX C = 2 ´ 15 = 30 V (d ® s ) V = VR2 + ( VL  VC )2 = 50 V 5. (a ® s) R is independent of f .
1 f 1 (c ® r ) X L µ f
(b ® p) X C µ
(d ® q) 2
æ 1 ö ÷ Z = R2 + çç w L w C ÷ø è
i.e., first decreases then increases.