CHAPTER – 35

MAGNETIC FIELD DUE TO CURRENT

1.

F F N N F = q B or, B = = = = q T A. sec . / sec . A m

0 2r i = 10 A,

or, 0 =

B=

2.

3.

4.

2rB mN N = = 2 A m A A

d=1m

i 10 7 4 10 –6 = 20 × 10 T = 2 T B= 0 = 2r 2 1 Along +ve Y direction. d = 1.6 mm So, r = 0.8 mm = 0.0008 m i = 20 A i 4 10 7 20 –3 B = 0 = = 5 × 10 T = 5 mT 2r 2 8 10 4 i = 100 A, d = 8 m i B= 0 2r 4 10 7 100 = 2.5 T 28 –7 0 = 4 × 10 T-m/A

Z axis X axis 1m

r

100 A

8m

= 5.

= 1 A,

r = 2 cm = 0.02 m,

–5 B = 1 × 10 T

We know: Magnetic field due to a long straight wire carrying current =

0 2r

P 2 cm i

7

6.

7.

4 10 1 –5 B at P = = 1 × 10 T upward 2 0.02 2 cm –7 net B = 2 × 1 × 10 T = 20 T Q –5 B at Q = 1 × 10 T downwards Hence net B = 0 (a) The maximum magnetic field is B 0 which are along the left keeping the sense along the 2r direction of traveling current. (b)The minimum B 0 0i 2r i 2r 0 If r = B net = 0 2B r 0 B net = 0 r< 2B r > 0 B net = B 0 2B 2r –7 –4 0 = 4 × 10 T-m/A, = 30 A, B = 4.0 × 10 T Parallel to current. B =–40 ×–10–4 T– – – B due to wore at a pt. 2 cm =

0 4 10 7 30 –4 = = 3 × 10 T 2r 2 0.02

net field =

3 10 4 10 4 2

4 2

–4

= 5 × 10

T 35.1

–

–

–

–

–

–

–

–

–

–

–

–

–

–

–

30 A

Magnetic Field due to Current 8.

i = 10 A. ( Kˆ ) B = 2 × 10

–3

T South to North ( Jˆ )

To cancel the magnetic field the point should be choosen so that the net magnetic field is along - Jˆ direction. The point is along - ˆi direction or along west of the wire. B=

0 2r

2 × 10

=

4 10 7 10 2 r

2 10 7

–3 = 10 m = 1 mm. 2 10 3 Let the tow wires be positioned at O & P

r=

9.

–3

R = OA, =

8 10 4 = 2.828 × 10

–2

(0.02)2 (0.02)2 =

m

7

4 10 10 –4 = 1 × 10 T (r towards up the line) (a) B due to Q, at A1 = 2 0.02 4 10 7 10 –4 = 0.33 × 10 T (r towards down the line) B due to P, at A1 = 2 0.06 O –4 –4 –4 net B = 1 × 10 – 0.33 × 10 = 0.67 × 10 T A1 2 10 7 10 –4 (b) B due to O at A2 = = 2 × 10 T r down the line 0.01 2 10 7 10 –4 = 0.67 × 10 T r down the line B due to P at A2 = 0.03 –4 –4 –4 net B at A2 = 2 × 10 + 0.67 × 10 = 2.67 × 10 T –4 r towards down the line (c) B at A3 due to O = 1 × 10 T –4 B at A3 due to P = 1 × 10 T r towards down the line –4 Net B at A3 = 2 × 10 T 2 10 7 10 –4 (d) B at A4 due to O = = 0.7 × 10 T towards SE 2.828 10 2 –4 B at A4 due to P = 0.7 × 10 T towards SW 2 2 Net B = 0.7 10 - 4 0.7 10 - 4 = 0.989 ×10–4 ≈ 1 × 10–4 T

10. Cos = ½ ,

A4

2 cm A2

= 60° & AOB = 60° 7

0 10 2 10 –4 = = 10 T 2r 2 10 2 –4 2 –4 2 –8 1/2 So net is [(10 ) + (10 ) + 2(10 ) Cos 60°]

O

B=

–4

A3

= 10 [1 + 1 + 2 × ½ ] 11. (a) B for X = B for Y

1/2

= 10

-4

×

2 cm

3 T = 1.732 × 10–4 T

Both are oppositely directed hence net B = 0 (b) B due to X = B due to X both directed along Z–axis 2 10 7 2 5 –6 Net B = = 2 × 10 T = 2 T 1 (c) B due to X = B due to Y both directed opposite to each other. Hence Net B = 0 –6 (d) B due to X = B due to Y = 1 × 10 T both directed along (–) ve Z–axis –6 Hence Net B = 2 × 1.0 × 10 = 2 T 35.2

A

(–1, 1)

(–1, –1)

2 cm

B

2 cm

(1, 1)

(1, –1)

Magnetic Field due to Current 12. (a) For each of the wire Magnitude of magnetic field i 0 5 2 = 0 ( Sin45 Sin45) = 4 5 / 2 2 4r

due to (1) B = due to (2) B = due to (3) B = due to (4) B =

0i 2 (15 / 2) 10

2 5 10 2

0i 2 (5 5 / 2) 10 0i 2 2.5 10

=

2

=

2

Bnet = [4 + 4 + (4/3) + (4/3)] × 10

2

2 15 10 2

2 15 10 2

2 5 10 2 =

5 2/2

Q3 –5

–5

–5

= (4/3) × 10

= 4 × 10

–5

32 –5 –5 –4 × 10 = 10.6 × 10 ≈ 1.1 × 10 T 3

At point Q2 due to (1) due to (2) due to (3) due to (4)

oi

2 (2.5) 10 2 oi

2 (15 / 2) 10 2 oi

2 (2.5) 10 2 oi

2 (15 / 2) 10 2

Bnet = 0 At point Q3 due to (1) due to (2) due to (3) due to (4)

4 10 7 5 2 (15 / 2) 10

2

4 10 7 5 2 (5 / 2) 10

2

4 10 7 5 2 (5 / 2) 10

2

4 10 7 5 2 (15 / 2) 10

2

= 4/3 × 10

= 4 × 10

–5

= 4 × 10

–5

= 4/3 × 10

Bnet = [4 + 4 + (4/3) + (4/3)] × 10 For Q4 –5 due to (1) 4/3 × 10 –5 due to (2) 4 × 10 –5 due to (3) 4/3 × 10 –5 due to (4) 4 × 10 Bnet = 0

–5

–5

=

–5

32 –5 –5 –4 × 10 = 10.6 × 10 ≈ 1.1 × 10 T 3

35.3

5 cm

P

= (4/3) × 10

4 5 2 10 7

4 5 2 10 7 –5

= 4 × 10

4 5 2 10 7

=

A

5 2

4 5 2 10 7

=

2 2.5 10 2

Q2 D

For AB for BC For CD and for DA . The two and 2 fields cancel each other. Thus Bnet = 0 (b) At point Q1

0i

3

4 Q1

C

B

Q4

Magnetic Field due to Current 13. Since all the points lie along a circle with radius = ‘d’ Hence ‘R’ & ‘Q’ both at a distance ‘d’ from the wire. So, magnetic field B due to are same in magnitude. As the wires can be treated as semi infinite straight current carrying i conductors. Hence magnetic field B = 0 4d At P B1 due to 1 is 0 i B2 due to 2 is 0 4d At Q i B1 due to 1 is 0 4d B2 due to 2 is 0 At R B1 due to 1 is 0 i B2 due to 2 is 0 4d At S i B1 due to 1 is 0 4d B2 due to 2 is 0 i 14. B = 0 2 Sin 4d 0ix i 2 x = 0 = 2 4d x x2 2 d2 4d d2 4 4

R

2 i

S

d

Q

1

P i

x/2

d

i

x

(a) When d >> x Neglecting x w.r.t. d 0ix ix = 0 2 B= 2 d d d B

1

d2 (b) When x >> d, neglecting d w.r.t. x 0ix 2 0i = B= 4dx / 2 4d 1 B d 15. = 10 A, a = 10 cm = 0.1 m r = OP =

O 10 A

A

3 0 .1 m 2

10 cm

B = 0 (Sin1 Sin 2 ) 4r =

10 7 10 1 3 0 .1 2

=

B

30

2 10 5 –5 = 1.154 × 10 T = 11.54 T 1.732

35.4

Q1

30

P P

Q2

Magnetic Field due to Current 16. B1 =

0i , 2d

B2 =

0i i (2 Sin) = 0 4d 4d

i 1 B1 – B2 = B2 0 100 2d 0i

4d d2

2

4

4 d2

2

ℓ = 3.92 d +

4d d2

2 4

2 d2

2

4

=

0i 4d d2

2

2

d2

2 4

156816 99 4 = = 3.92 = 40000 200

3.92 2 4 d2 2

=

0.02 d = = 3.92

0.02 = 0.07 3.92 B

2 3i

i/3 a/2 A 2i/3a

a2 a2 = 16 4 2

a 3a 2 4

=

3a/4 a A

O

a/2

a 13 13a 2 = 16 4

Magnetic field due to AB i (Sin (90 – i) + Sin (90 – )) BAB = 0 4 2a / 4 =

0 2i 2i a / 2 = 2 0i 2Cos = 0 2 4a 4a 5 a( 5 / 4)

Magnetic field due to DC i 2Sin (90° – B) BDC = 0 4 23a / 4 =

0i 4 2 0i a / 2 = 2 0i Cos = 4 3a 3a ( 13a / 4) a3 13

The magnetic field due to AD & BC are equal and appropriate hence cancle each other. Hence, net magnetic field is

2 0i 5

–

2 0i a3 13

i

D

i

9a 2 a 2 = 16 4

2

C

a 5 5a 2 = 4 16 2

i

C

a

D

i 2 2 2 2 0i B due to ABC = 2 0 = 6a 43a 2 2 0i 2 2 0i 2 0 i Now B = – = 3a 6a 3a

D0 =

i l

0i = 200 d

1 3.92 2 2 2 2 = 3.92 d 0.02 ℓ = 3.92 d 4 17. As resistances vary as r & 2r i & along ADC = Hence Current along ABC = 3 Now, i 2 2 2 2 2 0i B due to ADC = 2 0 = 4 3 a 3 a

18. A0 =

d

2 4

0i 1 1 d 2 200

99 200

=

2 4

2

=

0i

2

=

2 0i 1 1 a 5 3 13

35.5

a/2

a/4 B

Magnetic Field due to Current 19. B due t BC & B due to AD at Pt ‘P’ are equal ore Opposite Hence net B = 0 Similarly, due to AB & CD at P = 0 The net B at the Centre of the square loop = zero. i B = 0 (Sin60 Sin60) 20. For AB B is along 4r i For AC B B = 0 (Sin60 Sin60) 4r i For BD B B = 0 (Sin60) 4r i For DC B B = 0 (Sin60) 4r Net B = 0 21. (a) ABC is Equilateral AB = BC = CA = ℓ/3 Current = i AO =

i

i 0i2Cos 4 ( / 4) 0 Cot( / 4) 4 2Sin( / 4)Cos( / 4) 4x

C 2i A 30°

i

60°

i

i

B

C

i

A

Q

1 = 2 = 60° as AM : MO = 2 : 1 So, MO = 6 3 B due to BC at <. i i i9 = 0 (Sin1 Sin 2 ) = 0 i 6 3 3 = 0 4r 4 2 9 0i 27 0i net B = 3 = 2 2 8 2 0 i i8 (b) B due to AD = 0 2 = 4 4 8 2 0 i 8 2 0 i Net B = ×4= 4 r 22. Sin (/2) = x r = x Sin (/2) Magnetic field B due to AR 0i Sin(180 (90 ( / 2))) 1 4r i[Sin(90 ( / 2)) 1] 0 4 Sin( / 2)

0i(Cos( / 2) 1) 4 Sin( / 2)

i A

3 3 a = = 23 2 2 3

D

B

60° B

P

M 60°

C

O

C

B

45° 45° ℓ/8

D

A

C

r A

x

B

The magnetic field due to both the wire. 2 0i i Cot( / 4 ) 0 Cot( / 4 ) 4 x 2x 35.6

Magnetic Field due to Current

23. BAB 0i 2 iSin 2Sin = 0 4b b 0i = = BDC b 2 b 2 BBC 0i 2 iSin 2 2Sin = 0 4 =

( / 2)

2

Sin (ℓ +b) =

Sin =

2

2

/4b /4 (b / 2)

2 / 4 b2 / 4

=

=

2

b2

A

B

l

b 2 b2

= BAD

0ib 2 b 2

Net B =

C

D

2 0i b 2 b 2

2 0ib

+

2 b 2

=

2 0i( 2 b 2 ) b 2 b 2

=

2 0i 2 b 2 b

2 2r = , ℓ= n n n Tan = x= 2x 2Tan r 2 n i2Tan 2Sin 0i BAB = (Sin Sin) = 0 4( x ) 4

24. 2 =

=

A

B

l

0i2Tan( / n)2Sin( / n)n inTan( / n)Sin( / n) = 0 42r 2 2r

For n sides, Bnet =

0inTan( / n)Sin( / n) 2 2r

25. Net current in circuit = 0 Hence the magnetic field at point P = 0 [Owing to wheat stone bridge principle] –5 26. Force acting on 10 cm of wire is 2 ×10 N ii dF = 012 dl 2d

2 10 5 10 10 2

d=

=

P

0 20 20 2d

4 10 7 20 20 10 10 2 2 2 10 5

d -3

= 400 × 10 = 0.4 m = 40 cm

27. i = 10 A Magnetic force due to two parallel Current Carrying wires. F= 0 12 2r So, F or 1 = F by 2 + F by 3 10 10 10 10 = 0 0 2 5 10 2 2 10 10 2 = =

4 10 7 10 10 2 5 10 2

4 10 7 10 10

1 5 cm

2 3

2 10 10 2

2 10 3 10 3 3 10 3 –4 = = 6 ×10 N towards middle wire 5 5 5 35.7

Magnetic Field due to Current 28.

0 10i 0i40 = 2x 2(10 x )

i

10 A 10 40 1 4 = = x 10 x x 10 x 10 – x = 4x 5x = 10 x = 2 cm The third wire should be placed 2 cm from the 10 A wire and 8 cm from 40 A wire. 29. FAB = FCD + FEF A 10 10 0 10 10 = 0 C 2 1 10 2 2 2 10 2 A –3 –3 –3 = 2× 10 + 10 = 3 × 10 downward. E FCD = FAB + FEF As FAB & FEF are equal and oppositely directed hence F = 0 0i1i2 = mg (For a portion of wire of length 1m) 30. 2d 0 50 i2 –4 = 1 × 10 × 9.8 3 2 5 10

4 10 7 5 i 2

10

FRQ = =

2

0 i1i2 2 3 10

2

4 10 7 6 10

2 10 2

10

D

10

F

F

R

S I1 I2

A 10

P dx Q

P 1 cm

N (Towards right)

–4

+ 36 × 10

x

2 2 10 2

4 10 7 6 6

= 4 × 10

1 cm

50

–4

= 8.4 × 10

B

0 i1i2

2 3 10 2 2 10 2 Net force towards down = (8.4 + 7.6) × 10–4 = 16 × 10–4 N 32. B = 0.2 mT, i = 5 A, n 0 i B= 2r r=

2

2 10 7 6 6

40 A

mg

2 6 10 10 7

10

–4

= 9.8 × 10 2 5 10 3 –3 –3 –1 2 × i2 × 10 = 9.3 × 10 × 10 9 .8 i2 = 10 1 = 0.49 A 2 31. 2 = 6 A 1 = 10 A FPQ i i dx 30 dx ii ‘F’ on dx = 0 1 2 dx = 0 1 2 = 0 2x 2 x x 0 30 dx 2 –7 FPQ = = 30 × 4 × 10 × [logx]1 1 x x –7 = 120 × 10 [log 3 – log 1] –7 Similarly force of FRS = 120 × 10 [log 3 – log 1] So, FPQ = FRS 0 i1i2 0 i1i 2 FPS = 2 2 1 10 2 2 10 2 =

(10–x) x

n = 1,

–5

= 7.6 × 10

–4

N

r=?

n 0i 1 4 10 7 5 –3 –3 –1 = = 3.14 × 5 × 10 m = 15.7× 10 m = 15.7 × 10 cm = 1.57 cm 2B 2 0.2 10 3 35.8

Magnetic Field due to Current

n 0 i 2r n = 100, r = 5 cm = 0.05 m –5 B = 6 × 10 T

33. B =

2 0.05 6 10 5 2rB 3 –1 = = × 10 = 0.0477 ≈ 48 mA 7 n 0 6.28 100 4 10 5 34. 3 × 10 revolutions in 1 sec. 1 sec 1 revolutions in 3 10 5 i=

q 1.6 10 19 = A t 1 3 10 5

i=

0i 4 10 7.16 10 19 3 10 5 2 1.6 3 –10 –10 = 10 11 = 6.028 × 10 ≈ 6 × 10 T 10 0 . 5 2r 2 0.5 10 35. l = i/2 in each semicircle 1 (i / 2) i/2 ABC = B = 0 downwards i 2 2a A 1 0 (i / 2) upwards ADC = B = i/2 2 2a Net B = 0 r2 = 10 cm 36. r1 = 5 cm n2 = 100 n1 = 50 i=2A n i n i (a) B = 1 0 2 0 2r1 2r2 B=

=

50 4 10 7 2

B C

D

100 4 10 7 2

2 5 10 2 2 10 10 2 –4 –4 –4 = 4 × 10 + 4 × 10 = 8 × 10 n i n i (b) B = 1 0 2 0 = 0 2r1 2r2 37. Outer Circle n = 100, r = 100m = 0.1 m i=2A n 0 i 100 4 10 7 2 –4 B = = = 4 × 10 2a 2 0 .1 Inner Circle r = 5 cm = 0.05 m, n = 50, i = 2 A n 0 i 4 10 7 2 50 –4 B = = = 4 × 10 2r 2 0.05 Net B =

4 10 4 10 4 2

4 2

38. r = 20 cm, i = 10 A, F = e( V B) = eVB Sin = 1.6 × 10 =

–19

6

× 2 × 10 ×

=

downwards

32 2 10 8 = 17.7 × 10 6

0i Sin 30° 2r 2

= 16 × 10

–19

35.9

–4

N

–4

≈ 18 × 10

= 30°

V = 2 × 10 m/s,

1.6 10 19 2 10 6 4 10 7 10 2 2 20 10

horizontally towards West.

–3

= 1.8 × 10 = 1.8 mT

i

Magnetic Field due to Current

39. B Large loop = 0 2R ‘i’ due to larger loop on the smaller loop 2 = i(A × B) = i AB Sin 90° = i × r × 0 2r 40. The force acting on the smaller loop F = ilB Sin i2r o 1 ir = = 0 2R 2 2R 41. i = 5 Ampere, r = 10 cm = 0.1 m As the semicircular wire forms half of a circular wire, 1 4 10 7 5 1 0i So, B = = 2 2r 2 2 0 .1 –6 –6 –5 = 15.7 × 10 T ≈ 16 × 10 T = 1.6 × 10 T i i 2 42. B = 0 = 0 2R 2 3 2 2R

R

i

r

R i

r

10 cm

4 10 7 6

–6 120° = 4 × 10 2 6 10 t10 –6 –6 –5 = 4 × 3.14 × 10 = 12.56 × 10 = 1.26 × 10 T i 43. B due to loop 0 2r i Let the straight current carrying wire be kept at a distance R from centre. Given = 4i 4i B due to wire = 0 = 0 2R 2R r Now, the B due to both will balance each other 0 4i 0i 4r Hence = R= 2r 2R Hence the straight wire should be kept at a distance 4/r from centre in such a way that the direction of current in it is opposite to that in the nearest part of circular wire. As a result the direction will B will be oppose. –2 44. n = 200, i = 2 A, r = 10 cm = 10 × 10 n

=

n 0 i 200 4 10 7 2 –4 = = 2 × 4 × 10 2r 2 10 10 2 –4 –4 = 2 × 4 × 3.14 × 10 = 25.12 × 10 T = 2.512 mT

(a) B =

(b) B =

n 0ia 2 2

2 3/2

2(a d )

1 a2 = 2a 2(a 2 d2 )3 / 2 2 2 1/3 2 a + d = (2 a) –2 2 2/3 –2 10 + d = 2 10 –2 2 10 (1.5874 – 1) = d

n 0 i n 0ia 2 = 4a 2(a 2 d2 )3 / 2 2

2 3/2

2a

2

2/3 2

(a +d ) 2

3

2

a + d =2 a –2 2/3 2 (10 )(2 – 1) = d 2 –2 d = 10 × 0.5874 –1

= =

0ia 2 2(a 2 d2 )3 / 2 4 10 7 5 0.0016

2((0.0025 )3 / 2 –6 –5 = 40 × 10 = 4 × 10 T

3 2/3

–1 2

2

2/3

–1 2

(10 ) + d = 2 (10 ) –2 1/3 2 (10 ) (4 – 1) = d

10 2 0.5874 = 10 × 0.766 m = 7.66 × 10 45. At O P the B must be directed downwards We Know B at the axial line at O & P d=

2

a + d = (2a )

–2

= 7.66 cm.

a = 4 cm = 0.04 m

O 4 cm M

3 cm = 0.03 m

P

d = 3 cm = 0.03 m downwards in both the cases 35.10

3 cm

Magnetic Field due to Current 46. q = 3.14 × 10

–6

C,

r = 20 cm = 0.2 m,

w = 60 rad/sec.,

xQ

4 0 x a 2 Electric field = Magnetic field 0ia 2

2

2 a2 x 2 = =

3.14 10 6 60 q –5 = = 1.5 × 10 t 2 0 .2

i=

3/2

=

3/2

xQ

4 0 x 2 a 2

3/2

2 x 2 a2

3/2

0ia 2

9 10 9 0.05 3.14 10 6 2 4 10 7 15 10 5 (0.2)2

9 5 2 10 3 4 13 4 10

12

=

3 8

47. (a) For inside the tube B =0 As, B inside the conducting tube = o (b) For B outside the tube 3r d= 2 i i 2 i B = 0 = 0 = 0 2d 23r 2r 48. (a) At a point just inside the tube the current enclosed in the closed surface = 0. o Thus B = 0 = 0 A (b) Taking a cylindrical surface just out side the tube, from ampere’s law. i 0 i = B × 2b B= 0 2b 49. i is uniformly distributed throughout. i

So, ‘i’ for the part of radius a =

2

a 2 =

b Now according to Ampere’s circuital law B× dℓ = B × 2 × × a = 0

ia 2

ia 2

P

r/2 O

r

a b

=

b2

a

b

ia 1 = 0 2 2a 2b b –2 50. (a) r = 10 cm = 10 × 10 m –2 x = 2 × 10 m, i=5A i in the region of radius 2 cm 5 (2 10 2 )2 = 0.2 A (10 10 2 )2 –2 2 B × (2 × 10 ) = 0(0-2) B = 0

B=

2

4 10 7 0.2 4

=

0.2 10 7

4 10 10 (b) 10 cm radius –2 2 B × (10 × 10 ) = 0 × 5

4

= 2 × 10

–4

4 10 7 5

–5 = 20 × 10 10 2 (c) x = 20 cm –2 2 B× × (20 × 10 ) = 0 × 5

B=

B=

0 5 (20 10

2 2

)

=

4 10 7 5 400 10

4

–5

= 5 × 10

B x

35.11

Magnetic Field due to Current 51. We know,

B dl = i. Theoritically B = 0 a t A 0

Q

P

If, a current is passed through the loop PQRS, then ℓ 0i B will exist in its vicinity. B= S R 2( b) b Now, As the B at A is zero. So there’ll be no interaction However practically this is not true. As a current carrying loop, irrespective of its near about position is always affected by an existing magnetic field. P 52. (a) At point P, i = 0, Thus B = 0 (b) At point R, i = 0, B = 0 (c) At point , Applying ampere’s rule to the above rectangle l

B × 2l = 0K0

o

B ×2l = 0kl B =

B

A

dl

Bb

0k 2

B

Ba

B

l

B × 2l = 0K0

l

dl o

k B ×2l = 0kl B = 0 Bd 2 Since the B due to the 2 stripes are along the same direction, thus. BC k k Bnet = 0 0 = 0k C 2 2 D 53. Charge = q, mass = m We know radius described by a charged particle in a magnetic field B m r= qB Bit B = 0K [according to Ampere’s circuital law, where K is a constant] rq 0k m r= = q 0 k m –2

54. i = 25 A, B = 3.14 × 10 T, B = 0ni –2 –7 3.14 × 10 = 4 × × 10 n × 5

n=?

10 2

1 4 = 10 4 = 0.5 × 10 = 5000 turns/m 2 20 10 7 55. r = 0.5 mm, i = 5 A, B = 0ni (for a solenoid) –3 Width of each turn = 1 mm = 10 m 1 3 No. of turns ‘n’ = = 10 10 3 –7 3 –3 So, B = 4 × 10 × 10 × 5 = 2 × 10 T R 56. = 0.01 in 1 m, r = 1.0 cm Total turns = 400, ℓ = 20 cm, l 400 –2 B = 1× 10 T, n= turns/m 20 10 2 E E E i= = = R0 R 0 / l (2r 400 ) 0.01 2 0.01 400 n=

B = 0ni 35.12

B

l

Magnetic Field due to Current 2

10 = 4 × 10 E=

–7

×

400 20 10

2

E 400 2 0.01 10 2

10 2 20 10 2 400 2 10 2 0.01 4 10 7 400

57. Current at ‘0’ due to the circular loop = dB =

for the whole solenoid B = =

=

0 a 2indx 3/2 4 2 l 2 a x 2

B

dB 0

0 a 2nidx

0

=1V

2 4a 2 x 2

0ni 4

ni dx ℓ/2–x

2

a dx

0

3/2

2

2x a 3 1 2a 8

58. i = 2 a, f = 10 rev/sec, qe = 1.6 × 10

–19

c,

3/2

=

0ni 4 a

0

dx 2

2x 1 2a

n= ?,

3/2

me = 9.1 × 10 B B = 0ni n = 0i

–31

2x = 1 2a

2

ℓ/2

kg,

f 2m e f 2m e 10 8 9.1 10 31 qB B B= n= = = = 1421 turns/m 2m e qe 0i qe 0i 1.6 10 19 2 10 7 2 A 59. No. of turns per unit length = n, radius of circle = r/2, current in the solenoid = i, B = 0ni Charge of Particle = q, mass of particle = m f=

niqr q 0nir mV 2 qBr = qVB V = = = 0 r m 2m 2m 60. No. of turns per unit length = ℓ (a) As the net magnetic field = zero Bplate B Solenoid Bplate 2 = 0kdℓ = 0kℓ k ...(1) B Solenoid = 0ni …(2) Bplate 0 2 k Equating both i = 0 2 (b) Ba × ℓ = kℓ Ba = 0k BC = 0k Again

B=

Ba 2 Bc 2 =

2 0k 2 =

2 0k

Ba

0ni

2k n –3 61. C = 100 f, Q = CV = 2 × 10 C, t = 2 sec, –3 V = 20 V, V = 18 V, Q = CV = 1.8 × 10 C, 2 0k = 0ni

Bc

i=

2 10 4 Q Q –4 = = 10 A n = 4000 turns/m. t 2 –7 –4 –7 B = 0ni = 4 × 10 × 4000 × 10 = 16 × 10 T i=

35.13

A

C