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