Page 1

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 2r i = 10 A,

or, 0 =

B=

2.

3.

4.

2rB mN N = = 2  A m A A

d=1m

 i 10 7  4   10 –6 = 20 × 10 T = 2 T B= 0 = 2r 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 2r 2    8  10  4 i = 100 A, d = 8 m  i B= 0 2r 4  10 7  100 = 2.5 T 28 –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 2r

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 2r direction of traveling current.   (b)The minimum B  0  0i 2r i 2r 0 If r = B net = 0 2B r 0 B net = 0 r< 2B     r > 0 B net = B  0 2B 2r –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 2r 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 2r

 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 2r 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 4r

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 4d At P B1 due to 1 is 0  i B2 due to 2 is 0 4d At Q  i B1 due to 1 is 0 4d B2 due to 2 is 0 At R B1 due to 1 is 0  i B2 due to 2 is 0 4d At S  i B1 due to 1 is 0 4d B2 due to 2 is 0  i 14. B = 0 2 Sin  4d  0ix  i 2 x = 0 = 2 4d x x2 2  d2  4d 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= 4dx / 2 4d 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 (Sin1  Sin 2 ) 4r =

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 , 2d

B2 =

 0i  i (2  Sin) = 0 4d 4d

 i 1 B1 – B2 = B2  0  100 2d  0i

4d d2 

2

 4

4 d2 

2

 ℓ = 3.92 d +

4d d2 

2 4

2 d2 

2

 4

=

 0i 4d 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 2a / 4  =

 0  2i   2i a / 2 = 2 0i 2Cos = 0 2 4a 4a  5 a( 5 / 4)

Magnetic field due to DC  i 2Sin (90° – B) BDC = 0  4 23a / 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  = 6a  43a   2 2 0i 2 2 0i 2 0 i Now B = – =  3a 6a 3a

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  4r  i For AC B  B = 0 (Sin60  Sin60) 4r  i For BD B  B = 0 (Sin60) 4r  i For DC B  B = 0 (Sin60) 4r  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) 4x

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  i9 = 0 (Sin1  Sin 2 ) = 0  i  6 3  3 = 0 4r 4 2  9 0i 27 0i net B = 3 = 2 2  8 2 0 i  i8 (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  4r  i[Sin(90  ( / 2))  1]   0 4  Sin( / 2)

 0i(Cos( / 2)  1)  4   Sin( / 2)

i A

3   3 a = = 23 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 2x 35.6


Magnetic Field due to Current

 23. BAB  0i  2  iSin  2Sin = 0 4b b   0i = = BDC b  2  b 2  BBC  0i  2  iSin  2  2Sin = 0 4  =

(  / 2)

2

 Sin (ℓ +b) =

 Sin  =

2

2

 /4b /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 2r  = , 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 42r 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 2d 

2  10 5 10  10  2

d=

=

P

 0  20  20 2d

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 2r    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 = 2x 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. 2d  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 2x 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  i2r o 1  ir = = 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 2R 2R  r Now, the B due to both will balance each other  0 4i  0i 4r Hence = R= 2r 2R  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 2d 23r 2r 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 × 2b B= 0  2b 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  2a 2b 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  (2r  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     4a 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 2m e f 2m e 10 8  9.1 10 31 qB B B= n= = = = 1421 turns/m 2m 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

35. MAGNETIC FIELD DUE TO CURRENT  

r &lt; B 2 7.  0 = 4 × 10 –7 T-m/A,  = 30 A, B = 4.0 × 10 –4 T Parallel to current. 2. i = 10 A, d = 1 m 4. i = 100 A, d = 8 m N m   ne...

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