DC Pandey
DC Pandey
DC Pandey
of
Electricity & Magnetism
By DC Pandey
20 Current Electricity
Introductory Exercise 20.1
q 2 pr 2.0
1. i = , here q = e, t = =
t v 8.43 ´ 1028 ´ 1.6 ´ 10-19 ´ 3.14
ev
\ i= ´ (0.5 ´ 10-3 )2
2pr
1.6 ´ 10-19 ´ 2.2 ´ 106 = 1.88 ´ 10-6 ms -1
=
2 ´ 3.14 ´ 5 ´ 10-11 3. Yes.
= 1.12 ´ 10 -3
A As current always flows in the direction of
electric field.
= 1.12 mA
4. False.
2. No. of atoms in 63.45 g of Cu = 6.023 ´ 1023
In the absence of potential difference,
\No. of atoms in 1 cm 3 (8.89 g) of Cu
electrons passes random motion.
6.023 ´ 1023
= ´ 8.89 5. Current due to both positive and negative
63.54
ions is from left to right, hence, there is a net
= 8.43 ´ 1022 current from left to right.
As one conduction electron is present per dq
6. i = 10 + 4 t Þ = 10 + 4 t
atoms, dt
n = 8.43 ´ 1022 cm -3 or 8.43 ´ 1028 m -3 q 10
Þ ò0 dq = ò0 (10 + 4t ) dt
As i = neAvd
i Þ q = [10t + 2t 2 ]10
0 = 300 C
Þ vd =
neA
AIEEE Corner
Subjective Questions (Level 1)
q ne 337.5
1. i = = = = 2.11 ´ 1021
t t 1.6 ´ 10-19
Given, 2 pr 1 v
4. T = Þf = =
i = 0.7 , t = 1 s, e = 1.6 ´ 10-19 C v T 2 pr
it 0.7 ´ 1
\ n= = 2.2 ´ 106
e 1.6 ´ 10-19 =
2 ´ 3.14 ´ 5.3 ´ 10-11
= 4.375 ´ 108
= 6.6 ´ 1019 s -1
2. q = it = 3.6 ´ 3 ´ 3600 q
I= = ef
= 38880 C T
3. (a) q = it = 7.5 ´ 45 = 337.5 C = 1.6 ´ 10-19 ´ 6.6 ´ 1019
q = 10.56 A
(b) q = ne Þ n =
e
6
2 -8
5. (a) I = 55 - 0.65 t = 1.7 ´ 10 W - m
dq l = 24.0 m
I=
dt 2 2
æ dö æ 2.05 ö
A = p ç ÷ = 3.14 ´ ç ´ 10-3 ÷
Þ dq = Idt è2ø è 2 ø
Þ q = ò I dt = 3.29 ´ 10-6 m 2
8 8 2 24.0
\ q = ò Idt = ò (55 - 0.65 t ) dt R = 1.7 ´ 10-8 ´
0 0 3.29 ´ 10-6
8
é t2 ù
= 55 [ t ]80 - 0.65 ê ú = 0.12 W
ë 2 û0 L
10. R=r
= 440 - 20.8 = 419.2 C A
rL
(b) If current is constant A=
q 419.2 R
I= = = 52.4 A
t 8 If D is density, then
6. i µ vd D r L2
m = DV = DA L =
vd 2 i2 R
\ = 8.9 ´ 103 ´ 1.72 ´ 10-8 ´ (3.5)2
vd1 i1 =
i2 6 . 00 0.125
Þ vd 2 = vd1 = ´ 1.20 ´ 10-4 -2
= 1.5 ´ 10 kg = 15 g
i1 1 . 20
11. At 20°C,
= 6.00 ´ 10-4 ms -1
i R1 = 600 W, R2 = 300 W
7. vd =
neA At 50°C,
1 R1¢ = R1(1 + a 1Dt )
=
8.5 ´ 1028 ´ 1.6 ´ 10-19 ´ 1 ´ 10-4 = 600 (1 + 0.001 ´ 30) = 600 ´ 1.03
= 0.735 ´ 10-6 ms -1 = 618 W
= 0.735 mm/s R2¢ = R2(1 + a 2Dt )
l 103 = 300 (1 + 0.004 ´ 30) = 336
t= =
vd 0.735 ´ 10-6 \ R¢ = R1 ¢ + R2 ¢ = 618 + 336
= 1.36 ´ 109 s = 43 yr = 954 W
R¢ - R 954 - 900
8. Distance covered by one electron in 1 s a= =
R ´ Dt 900 ´ 30
= 1 ´ 0.05 = 0.05 cm
Number of electrons in 1 cm of wire R = 600 + 300 = 900 W
= 2 ´ 1021 = 0.002 ° C -1
\ Number of electrons crossing a given area 12. As both the wires are connected in parallel,
per second VAl = VCu
= Number of electrons in 0.05 cm of wire iAl RAl = iCu RCu
= 0.05 ´ 2 ´ 1021 = 1020 L L
q ne iAl r Al Al2 = iCur Cu Cu 2
i= = p dAl p dCu
t t
iCu r Cu LCu
1020 ´ 1.6 ´ 10-19 Þ dCu = dAl
= = 1.6 ´ 10 = 16 A iAl r Al LAl
1
L 2 ´ 0.017 ´ 6
9. R = r = 1 ´ 10-3
A 3 ´ 0.028 ´ 7.5
Given, = 0.569 ´ 10-3 m
r = 0.017 mW - m = 0.569 mm.
7
V 0.938 20
13. (a) E = = = 1.25 V/m Þ R2 = W = 1.82 W
L 75 ´ 10-2 11
E 1.25 and R1 = 20 - R2 = 20 - 1.82 W
(b) J = Þ r =
r 4.4 ´ 107 = 18.2 W
r = 2.84 ´ 10-8 W - m 17. The circuit can be redrawn as
E V
14. (a) J = =
r rL
Current density is maximum when L is 24 V 12W 8W
minimum, ie, L = d, potential difference
should be applied to faces with dimensions
2d ´ 3 d.
V 8 ´ 12
J min. = . Reff = = 4.8 W
rd 12 + 8
V VA V 24
(b) i = = I= = =5A
R rL Reff 4.8
Current is maximum when L is minimum 18. Here, A and C are at same potential and B
and A is maximum. and D are at same potential,
Hence, in this case also, V should be applied A
8W
B
to faces with dimensions 2d ´ 3 d
V (2d ´ 3 d ) 6Vd
and imax = = .
r ( d) r 24V 4W 6W
L
15. (a) R = r C
A D 12W
RA
r=
L Hence, the circuit can be redrawn as
d
[r = = 1.25 mm = 1.25 ´ 10-3 m] A,C
2
0.104 ´ 3.14 ´ (1.25 ´ 10-3 )2
=
12W
14.0
8W
24V 4W 6W
= 3.64 ´ 10-7 W - m
V EL 1.28 ´ 14
(b) i = = = = 172.3 A
R R 0.104 B,D
(c) i = neAvd 1 1 1 1 1
i \ = + + +
vd = R 4 8 12 6
neA 6+3 +2+4
172.3 =
= 24
8.5 ´ 1028 ´ 1.6 ´ 10-19 ´ 3.14 ´ (1.25 ´ 10-3 )2 15 5
= =
= 2.58 ´ 10-3 ms -1 24 8
16. For zero thermal coefficient of resistance, 8
R= W
DR = 0 5
RC a CDT + RFe a Fe DT = 0 = 1.6 W
R1 - a Fe - 5.0 ´ 10-3 V 24
= = = 10 i= =
R2 aC - 0.5 ´ 10-3 R 1.6
= 15 A
R1 = 10 R2
19. Given circuit is similar to that in previous
Also, R1 + R2 = 20
question but 4 W resistor is removed. So the
Þ 10 R2 + R2 = 20 effective circuit is given by
8
A,C V 12
i= =
R 36 / 13
13
= A
12W
24 V 8W 6W 3
12 + 6
21. (a) i = =3 A
1+2+3
B,D i = 3A
A
1 1 1 1 12 V 1W
= + +
R 8 12 6
B
1 3+2+4 9 3
= = = G 2W
R 24 24 8
8 C
R = W = 2.67 W
3 6V 3W
V 24
i= = =9 A
R 2.67 D
20. VG = 0
6W
A B VA = VG + 12 = 12 V
4W VA - VB = 3 V
12V 12W 6W VB = 12 - 3 = 9 V
3W VB - VC = 6 V
D C
2W VC = 9 - 6 = 3 V
ßA VG - VD = 6 V, VD = - 6 V
6W 4W (b) If 6 V battery is reversed
4W 12 - 6
12 B C 12W i= =1A
1+2+3
3W 2W
D i = 1A
A
i2
100W 200W R2 RV2 6000
i1 D R2¢ = = = 1200 W
R2 + RV2 5
10 V As R1 ¢ = R2 ¢
10 1 Hence,
\ i1 = i2 =
= A
300 30 reading of V1 = reading of V2
Hence, reading of voltmeter 1200
= ´ 200 = 100 V
1200 + 1200
= Potential difference between B and C
20 (b) Current distribution is shown in figure
= 200 ´ i2 = V
3 E
= 6.67 V
26. (a) (i) When S is open. i i2
i1
E V1 V2
i2 S i1
R1 R2
V1 V2
E
R1 S R2 i=
R1 ¢ + R2 ¢
200 1
R1 i= = A
3000 2400 12
V1 = E= ´ 200
RV1 + RV2 5000 RV1 3000 1
i1 = i= ´
= 120 V R1 + RV1 5000 12
RV2 2000 1
V2 = E= ´ 200 = A
RV1 + RV2 5000 20
R1 2000 1
= 80 V i2 = i= ´
R1 + RV1 5000 12
(ii) When S is closed, 1
E A =
30
\Current flowing through
1 1
V1 V2 S = i1 - i2 = -
20 30
R1 S R2 1
= A
60
Now, R1 and V1 are in parallel and their 27. Effective emf of 2 V and 6 V batteries
effective resistance connected in parallel
R1 RV1 E r + E2r1 2 ´ 1 - 6 ´ 1
R1 ¢ = =
6000
= 1200 W E¢ = 1 2 =
R1 + RV1 5 r1 + r2 1+1
= -2V
Similarly, rr 1
R2 and V2 are in parallel with their effective and r¢ = 1 2 = W
r1 + r2 2
resistance,
= 0.5 W
11
2V 1W 32. V1 = E - i1r Þ E - 1.5 r = 8.4 …(i)
4V 0.5W V2 = E + i2r Þ E + 3.5 r = 9.4 …(ii)
On solving, we get
6V 1W r = 0.2 W
ß E = 8.7 V
0.5W 0.5W
33. In case of charging
4V 2V V = E + i r = 2 + 5 ´ 0.1 = 2.5 V
Net emf, E = 4 - 2 = 2 V 34. Clearly current through each branch is zero.
2W
28. (a)
A B
2W 4W 8W 8W Þ 4W 4W 4W
– – 2V
E1 E2 2V 2V 2V
+ +
2V 2V
E
35. i1 =
As E1 > E2 R+G
Current will flow from B to A. R
G
(b) E1 is doing positive work
i1
(c) As current flows from B to A through
resistor, B is at higher potential.
E
29. i2 R = 2 W < 5 W
Clearly X is doing negative work. On shunting the galvanometer with
resistance S,
A i
S
R=2.0W E B R R'
G
P 0.5
(a) P = Vi Þ V = = = 5.0 V i2
i 1.0
(b) E = V - iR = 5 - 2 = 3 V E
300W 100 R
400W Re = 3 + 2 +
100 + R
V 100 R
=5 +
100 + R
Effective resistance of 300 W resistor and
voltmeter 3.4
i= = 0.04
300 ´ 1200 100 R
R¢ = = 240 W 5+
300 + 1200 100 + R
60 4R
i= Þ 0.2 + = 3.4
400 + 240 100 + R
60 Þ R = 400 W
= A
640 Reading of voltmeter,
3 100 R 100 ´ 400
= A V =i´ = 0.04 ´
32 100 + R 100 + 400
\Reading of voltmeter,
3 = 3.2 V
V = iR ¢ = ´ 240 If the voltmeter had been ideal,
32
= 22.5 V Reading of voltmeter
R¢ 100
38. V2 = V, = ´ 3.4 = 3.24 V
R1 + R2 ¢ 105
L R
rR2 120 42. 1 = 1
R2 ¢ = = L2 R2
r + R2 3
L1 8
= 40 W Þ = (L1 + L2 = 40 cm)
40 40 - L1 12
V2 = 120
60 + 40 Þ L1 = 16 cm from A.
= 48 V
13
ig Maximum power dissipated by the circuit
43. S = (G + R)
i - ig 2
P ¢max = Imax Re
i - ig 3
Þ R= S -G = 15 ´ ´ 2.4 = 54 W
ig 2
20 - 0.0224 47. Total power of the circuit, P = P1 + P2 + P3
= ´ 0.0250 - 9.36
0.0244 = 40 + 60 + 75
= 12.94 W = 175 W
E V2 V2
44. (a) i = As P = Þ R=
RV + r R P
Rv (120)2
V
= = 82.3 W
175
i 48. Thermal power generated in the battery
r
R
E
E i
V = iRV = RV r
RV + r
r 1 E
(b) = 1% =
RV + r 100 P1 = i2r = i ( E - V )
RV = 99r = 99 ´ 0.45 = 0.6 W
= 44.55 W Power development in the battery by electric
V RV forces
(c) =
E RV + r P2 = IE = 2.6 W
As RV decreases, V decreases, decreasing 49. The given circuit can be considered as the
accuracy of voltmeter. sum of the circuit as shown.
45. (a) When ammeter is connected 3
2W 2W A
E 21/6A 2W 16
IA = 35 5/16A
A 14/6A 2/16A
RA + R + r 16
3W + 3W
When ammeter is removed 7V 1V
E R + R+ r
I= = A IA ß
R+ r R+ r 2W 2W
I
(b) A = 99% 2A 1A
I 3W 1V
R+ r 99 7V
=
RA + R + r 100
1 1 \ P1 = 7 ´ 2 = 14 W,
Þ RA = ( R + r) = (3.8 + 0.45)
99 99 P2 = - 1 ´ 1 = 1 W
RA = 0.043 W E1 - E2 12 - 6
50. (a) i = = = 0.5 A
IA R+ r R1 + R2 4 + 8
(c) As = , as RA increases, I A
I RA + R + r (b) Power dissipated in R1 = I 2 R1 = 1 W
decreases, decreasing the accuracy of and power dissipated in R = I 2 R2 = 2 W
ammeter. (c) Power of battery E1 = E1I
r max 36
46. Imax = = = 15 A = 12 ´ 0.5 = 6 W (supplied)
R 2.4
Power of battery E2 = E2I
For the given circuit
= - 6 ´ 0.6 = - 3 W (absorbed)
1 3
Re = R + R = R
2 2
14
E 12 1 1 1 1 1
51. I = = =2A 54. (a) = + + +
R+ r 5+1 R 4 6 14 4
4W 4W
(a) P = EI = 12 ´ 2 = 24 W
4W
(b) P1 = I 2 R = 22 ´ 5 = 20 W 8W 4W
2W
6W ÞA 8W 6W B
(c) P2 = I 2r = 22 ´ 1 = 4 W
2W
A B
52. (a) 4W 4W
1.60W ß
I1 4W
I2 2.40W
6W
I3 4.80W A 14W B
I
4W
1 31
28.0V =
R 42
1 1 1 1 42
= + + R=
R R1 R2 R3 31
1 1 1 1 1 1 1 1
= + + (b) = + +
R 1.60 2.40 4.80 Re 2 R 2 R R
R = 0.80 W
V 28.0 R R R R
(b) I1 = = = 17.5 A R Þ
R1 1.60
R R
V 28.0 A B A B
I2 = = = 11.67 A R R
R2 2.40
R R
V 28.0
I3 = = = 5.83 A ß
R3 4.80 2R
(c) I = I1 + I2 + I3 = 35.0 A 2R
(d) As all the resistance connected in A B
parallel, voltage across each resistor is
28.0 V. R
V 2 (28)2
(e) P1 = = = 490 W Wheatstone bridge is balanced
R1 1.6 R
V 2 (28)2 Þ Re =
P2 = = = 326.7 W 2
R2 2.4 i2+ i3 4W
V 2 (28)2
P3 = = = 163.3 W (c) 2W i3 i3 2W
R3 4.8
i2 3W
V2 i1+ i2 3W i1 i + i
1 2
(f) As, P =
R A 1W i1 B
i1 1W
Resistor with least resistance will dissipate ß
maximum power. 4W
V2
53. (a) P = Þ V = PR 2W
R 3W 3W 2W
= 5 ´ 15 ´ 103 = 2.74 ´ 102
A B
= 274 V 1W 1W
2
V (120)2
(b) P = = = 1.6 W
R 9 ´ 103
15
ß 2W 8W C
2.4W 4W 2W 8W
(e) A 2W B Þ A
2W 6W 2W Þ 2W 6W 2W 2W B
A B A B 4W 10W
2W 2W 4W 10W
D
ß
6.4 8.5R C
1
1.52W W 8W
Þ A B 1W 2
11W ÞA B
A B A B O
2W 1W 10W
ß D
i3– i4 10W 5.8W
(d) i4 A B
5W i3
i3 5W
i4 By Star-Delta Method
10W 5W 10W R2 R3
RA =
5W
i1 + i2 + i3
i2 i2
i 1 + i 2 + i3 R1 + R2 + R3
A B R1 R3
i1 10W RB =
R1 + R2 + R3
10W ß 10W R2 R3
5W 5W RC =
R1 + R2 + R3
10W
10W 10W Þ 10W 10W R R R R
5W 10W
5W
R R R R
R
A B A B R R R R
10W 10W Þ
(f)
ß R R
R R R R R
5W R R
A B
25W A B
ß ß
10W 10W R 2R
A B
5W
ß A B R R 2R
Þ
5W R R
4.17W
A B A B
2R
A B
ß 2R
3R 6
W
Þ 5
A B
A B
2R
1W 1W R R R
E 2W F
C D
2W
R2 3 R
ß \ = = 0.6
R1 5 R
A
A Þ R2 = 0.6 R1
B A D
1W 2W 15W 15W
8W
56. F
2W Þ 5W 1W 5W
B 20W 6W
2W
2W 30W 40W
2W E
ß
1W 1W B 15W 15W
ß
0.71W A 2W 6W 40W 8W
A B B
Clearly C and D, E and F are at same
potential. C,E 20W 30W D,F
B D ß
F
30W
55. 2W 46W 8W
A B
A C E
50W
Let R be the resistance of each conductor,
and R1 be the effective resistance between A Here, C and E , D and F are at same
and F in first case then, potential.
\ R1 = 5 R Re = 23.3 W
If R2 be effective resistance between A and F r
2
in second case then, 2r r
r 3
57. (a) r r
r r r
a b b
r a b a
r r
ß
5 5
r r
8 3
A B
r
17
b ß
r
r
(b) r r 2r r r
Þ r
o a b a
r r a Þ
b a b
ß r r
4 2r
r r/ 3
3 r ß
Þ B r
b a
r r
(c) r r As Wheatstone bridge is balanced.
r
r
b
58. Re =
a Þ b a 2
r i1–i3
r r
r r (b)
i3
i3 i1
ß 2i1+i2 i1 i2
2r a i b
i1 2 i3 i1
r i3
b a Þb a
i1–i3
2r
ß
As Wheatstone bridge is balanced r
b r r
r r
r r r r r
r r
4 r 2r
r r Þ
(d) Þ r Þ a r b a 2r b
r r r r r
a r r r
ß
r 2r/3
8r/3
r r r r
r
(e) r a Þ a 2r b
r r Þ 2r b r r
a b
a b
r 2r/ 3
r r 8r/ 3
r ß 4/5r
a b
C
19
R V2
Pe = = n 2P
Re
A V
19. As bulb A is in series with entire circuit.
I2
E1 + E2 18
E 20. I = =
R + r1 + r2 R + 3
E
I2 = > I1 Vab = E2 - Ir2 = 0
RRV
RA + 18
R + RV 3- ´1 = 0
R+3
and V2 = E - I2 RA < V1
Þ R=3 W
17. Before connectivity resistance is parallel R
with ammeter 21. I2 = I
R + RV
A V
I I1 R
I1 A
I2
E V
E V R
I1 = , V1 = I1 RV = I
RA + RV RV R + RV
100 R
= E - I1 RA = 5
2500 R + 2500
After connecting resistance in parallel to the
ammeter. R + 2500 = 125 R
RA 2500
R= W » 100 W
24
A V
22.
R/10 R/10
I2
R/10 R/10
E R/10 R/10
E R/10 R/10
I2 = ,
RA
+ RV R/10 R/10
2
1
Reading of ammeter = I2
2 ß
E 1 R/5
= > I1
RA + 2 RV 2
R/5
2E
V = I2 RV = < 2V1 R/ 25
RA + 2 RV R/5
Þ
R R/5
18. Re =
n2
R/n R/5
R/n
R1 20 1
23. = = …(i)
R/n R2 80 4
R1 + 15 40 2
2
= =
V R2 60 3
P=
R
20
R1 15 2 VAB = - I (25 + 15)
+ =
R2 R2 3 1
= - ´ 40 » - 4 V
15 2 1 5 9
= - =
R2 3 4 12 30. (a) VAB = kL = 0.2 ´ 100 = 20 mV
Þ R2 = 36 W, RAB
VAB = E
R RAB + R
R1 = 2 = 9W
4 RAB
Þ 0.02 = ´2
V RAB + 490
24. (b) As V1 = , R1 = R2
2
Þ RAB + 490 = 100 RAB
RV ´ 100
= 50 490
100 + RV RAB = » 4.9 W
99
RV = 100 W 31. (c) When key is open,
E 2 2E
25. (d) = = 0.4 W I1 =
RPB + r 4+1 3R
VAB = IRAB = 1.6 W R 2R 3R
VPB 1.6
K = = = 0.016 V/cm
L 100
E 1.2 2R R Þ 3R
L= 1 = = 75 cm I1 I1
K 0.016
A A
26. (d) VAB = 3 ´ 2 + 3 + 1 ´ 4 - 2 + 6 ´ 1
E E
= 17 V 3/2R ß
1A 2A
3R
2W 2W I1
3A 2W 2W 1W 1W
A
B
A E
3V 2V
E1r2 + E2r1 When key is closed
27. (c) Ee = =2V R 2R
r1 + r2 2R/3 2R/3
rr
re = 1 2 = 0.5 W
r1 + r2
2R R Þ I2
For maximum power R = re I2
Ee2 (2)2 A A
and Pmax = = =2W E E
4re 4 ´ 0.5 ß
R 4R/3
28. (a) V = E
R+ r
æE ö æ 2.2 ö
r = ç - 1÷ R = ç - 1÷5 I2
èV ø è 1.8 ø
A
10
= W
9 3E
I2 =
E1 - E2 4R
29. (d) I =
R1 + R2 + r1 + r2 I1 8
\ =
10 - 5 1 I2 9
= = A
25 + 15 + 2.5 + 2.5 9
21
1
I Ig a
32. (b) S = G = 34 ´ 3663
I - Ig 33 R2 60° R1
I b
34
= 111 W
r Hence R1 and R2 are in parallel
2r R1 R2
r r r r Re =
r R1 + R2
r r
33. (b) A B Þ A
B
r = 2.5W
r r
r r 36. (c) Let RAD = RBC = x
r 2r
1W 1W 1W 1W 1W
5/3r ß
2r/3 1W 1W 1W 1W
r r
A B ÞA r
B
1W 1W
A B 1W 1W
r
5/3r 2r/3 Clearly x < 1 as 1 W resistor is in parallel
ß with some combination.
5/11r
Now RAB = x + 1 + x
= 2x + 1
5
Re = r As x < 1
11
11 1 < RAB < 3
r= ´ 1.5
5 R ( R + R0 )
37. (d) RAB = R + = R0
= 3.3 W 2 R + R0
r r r r Þ 2 R2 + RR0 + R2 + RR0 = 2 RR0 + R02
r Þ 3 R2 = R02
34. (b) r r r r R
A r r B ÞA r r B Þ R= 0
3
r
38.
r r r r
P P
ß R
R R R/2 R
2r R
2r Þ
r/2
A B ÞA 2r B R R/3
R
Q Q
2r
P ß
As the circuit is symmetrical about
perpendicular bisector of AB, all points 5 5
lying on it are at same potential. 6 R = 25W Ü 11 R R
L1
35. (c) R1 = R
L1 + L2 Q
R
Þ R1 = =3 W
6
l2
R2 = Þ R2 = 15 W
l1 + l2
22
39. Wheatstone bridge is balanced. 4
R1 = ´ pr = 2 = R2
A P 2pr
R 4 4
R R R R R3 = ´2 r =
R 2pr p
R R
R 1 1 1 1
2R/3 Ü R R
R
\ = + +
2R/3 R
R Re R1 R2 R3
B
R R 1 1 p 4+ p
2R/3 2R/3 = + + =
B 2 2 4 4
ßA 4
Re = W
R R R R 4+ p
7R/3 7R/3
Þ Þ 42. (d) Points C and D are shorted hence the
4R/3 4R/3 4R/3 4R/3 portion above line CD can be removed.
B
7
Þ Re =R
6
C D
L1 1
40. (d) R1 = R= R
L1 + L2 12
=3 W A B
ß
A C D
30°
B
O A B
ß
R/2 R/2
R/2 R
L2 11 A B Ü Ü
R2 = = R = 33 W A B
L1 + L2 12 R A B
R
R1 and R2 are in parallel,
R1 R2 3 ´ 33 43. (b) As AB is line of symmetry,
Re = = we can fold the network about AB.
R1 + R2 3 + 33
A R R
= 2.75 W
41. (a) Resistance per unit length of wire R R
R
4
= R R
2 pr R
A R R B
R/2 R/2 R
A R/2 R/2
R/2 R/2 A
Þ B
R1 R3 R2 R/2
R/2 R
ß
B 3R/2
A B
B
23
JEE Corner
Assertion and Reason
1. (d) V = IR, If V = 0 either I = 0 or R = 0 E2 R
P= is maximum at R = r.
2. (b) As all the resistors are in parallel ( R + r )2
potential difference is same, hence V V2
V2 8. (c) I = ,P= both I and P are inversly
P= is maximum if R is minimum. R R
R proportional to R hence both decrease with
I 2t r increase in R which increases with
3. (b) dH = I 2dRt = dH temperature.
A
I is same everywhere, hence portion having According to Ohm’s law V µ I not V = IR.
less area is more heated. As R can be variable also.
I 9. (d) Drift velocity is average velocity of all the
Again J =
A electrons but velocity of all electrons is not
\ J A > J B. constant.
Reason is also correct but does not explain L
10. (a) R = r
assertion. A
m
4. (b) Both assertion and reason are correct but r= 2
reason does not explain the cause of decrease ne t
in voltmeter reading. with increase in temperature, electron
5. (b) As RA < RV , more current passes through collide more frequently, i.e., t decreases,
increasing r and hence R.
ammeter when positions of ammeter and
E r + E2r1
voltmeter are interchanged and potential 11. (d) E = 1 2 \ E1 < E < E2
difference across voltmeter becomes less that r1 + r2
emf of cell. If E1 < E2
6. (c) During charging current inside the rr
r = 1 2 , r < r1, r < r2
battery flows from positive terminal to r1 + r2
negative terminal. Reason is false while R1 L1
assertion is true. 12. (d) =
R2 L2
E
7. (d) I = is maximum when R is zero Hence there is no effect of one while
R+ r
measuring using meter bridge.
hence reason is false.
100 ´ 100 A
Rnet = r + = r + 50
100 + 100 E, r
E
\ I0 = …(i)
r + 50
26
E 21. (a) Voltage sensitivity of voltmeter
I= =1A
R 1
+r µ
3 Resistance of voltmeter
R
V = I ´ =3 V Vs1 R +G
3 \ = 2
Ig Vs2 R1 + G
r
18. (c) S = G , G = r, S =
I - Ig 4 30 R2 + 50
=
1 1 20 2950 + 50
Ig = (I - Ig ) Þ Ig = I
4 5 30 ´ 3000
R2 + G = = 4500
= 0.006 A 20
10 5 Þ R2 = 4450 W
19. (d) I1 = = A
14 7 22. (b) For x = 0
8W B 6W
I1 VAB = E
P E
4W 3W k1 =
L
I I2 A EL1
E0 = k1L1 = …(i)
L
10 V
For x = x (say)
10 RAB
I2 = A VAB = E
7 RAB + x
VP - VA = I 2 ´ 4 RABE
40 k2 =
= A ( RAB + x ) L
7
40 RABEL2
VP - VB = I1 ´ 8 = A E0 = k2L2 = …(ii)
7 ( RAB + x ) L
VA - A B = 0 From Eqs. (i) and (ii),
Another method R + L2
R R L1 = AB
As, 1 = 3 , VB = VA ( RAB + x )
R2 R4
10 ´ 30
Þ 20 =
20. (b) For series connection 10 + x
V1 R
= 1 Þ x =5W
V2 R2
R1 3 23. (d) To obtain null point similar terminal of
\ = both the batteries should be connected.
R2 2
L rL 24. (c) Wheatstone bridge is balanced.
Now, R1 = r 1 = 12 , I1 20W 4W
A1 pr1 20W 4W
L rL 1.4A 1.4A
R2 = d 2 = 22 15W Þ
A2 pr2 I2
2
R1 L1 æ r2 ö 50W 10W 50W 10W
\ = ´ç ÷
R2 L2 çè r1 ÷ø R2
\ I1 = ´I
r1 R2L1 2 ´6 R1 + R2
= = =2 60
r2 R1L2 3 ´1 = ´ 1.4
r2 1 84
= =1A
r1 2
27
25. (b) L1 - L2
29. (c) r= R
E,B,H L2
R R L2r
Þ R=
A B R L1 - L2
R 490 ´ 10
D
C E Þ R = = 490 W
10V A,C,F 10
F H R More than One Correct Options
50W G 10W D G
10V 30. H = P1t1 = P2t2
ß H H
Þ t1 = , t2 =
P1 P2
I1 R R
I2
If connected in series
1 1 1
R = +
I3 R R P P1 P2
I Ü
Þ t = t1 + t2
If connected in parallel
10V 10V
P = P1 + P2
10 tt
I1 = = 2.5 A Þ t= 12
2 ´2 t1 + t2
26. (b) Effective resistance of voltmeter and E1r2 + E2r1 6 ´ 3 + 5 ´ 2
31. E = =
3 kW resistor, r1 + r2 2+3
3 ´6
R1 = = 2 kW = 5.6 V
2+6
As there is no load.
R1 2
V1 = E = ´ 10 = 5 V V = E = 5.6 V
R1 + R2 4
If E1 = E2, I = 0
27. (d) P1 = P2 = P3, Clearly R2 = R3 E - E2 6 - 5
R2 I= 1 = = 0.2 A
r1 + r2 2+3
R1
32. Let V = Potential difference between T1 and
i
T2.
R3
I1 A B
i T1
\ i2 = i3 = T2
2
2 I2
i 1 C
P1 = i2 R1, P2 = æç ö÷ ´ R2 = i2 R2
è2ø 4 V
1 2 I1 =
P3 = i R3 RA + RB
4
V
R2 = 4 R1, R3 = 4 R1 I2 =
RC
\ R1 : R2 : R3 = 1 : 4 : 4
E 2 Now,. I A = I B = I1
28. As E = kL1 Þ k = = = 250 IC = I2
L1 500
1 Also, V = IC RC = I1( RA + RB )
= V/cm
250 = I A RA + ID RB
1 I B I1 RC
V = kL2 = ´ 490 cm = =
250 IC I2 RA + RB
= 1.96 V
28
33. As R1 ¹ R2 E
Ie = >I
R
V1 V2 +r
2
A B If S2 is closed
E
V3 Ie = >I
r
R+
2
V1 ¹ V2
38. Vb - Va = - 10 + 2I = 2 V
Also, V3 = V1 + V2
2W I 10 V
34. As R1 = R2 a C b
I R1 R2
I =6A
V1 = V2 From b to a.
L Vc - Va = 2 ´ 6 = 12 V
R=r
A 39.
a a a
But
r r r r r r
L2 = 2L1
and R1 = R2 b d b d b d
\ A2 = 2 A1 r r r r r r
1 c c c
Also, vd µ (For constant current)
A ß ß ß
a
1 a
vd 2 = vd1 Þ vd1 = 2 vd 2 2r
2 r r r r
b r d
Again, vd µ E b d b d
\ E1 = 2 E2 r r 2r
2r
35. If E > 18 V current will flow from B to A and ß
c
vice-versa. ß ß r/ 2
36. V = kl a b d
a
If Jockey is shifted towards right, I and r r
hence k will decreases as k µ I.
2r 2r
Hence L will increase. b c
2r
If E1 is increased, k will increase, hence L 3
will decrease. ß c
r ß
If E2 is increased L will increase as V will
increase.
b a
If r is closed V will decrease hence L will 5r r
decrease. 3
E E ß
37. Ie = , Initially, I = 5r
Re + re R+ r
8
b a
If S1 is closed
29
d = [ML2T -3A -2 ]
[W ] [ML2T -2 ]
2 - Ve 4 - Ve 6 - Ve 4 - Ve [V] = =
+ + + =0 [ q] [AT]
1 2 1 2
Ve = 4 V, I1 = - 2 A, i2 = 0 , i3 = 2 A, i4 = 0 = [ML2T -3A -1 ]
2 -3 -2 2
[ R][ A ] [ML T A ][L ]
2. Current is same at every point and A1 < A2 [ s] = =
i [L] [L]
J = Þ J1 > J 2
A = [ML3T -3A -2 ]
i 1
vd = Þ vd1 > vd 2 [ s ]= = [M-1L-3T 3A 2 ]
neA [r ]
R r EA - EB
r= = Þ r1 > r2 5. I = =1A
L A R + rA + rA
V
k = Þ k1 > k2
L
4V 1W 1W
3. When switch S is closed 1V
V1 decreases, V2 increases,
\ Current through R1 decreases and through
R2 increases.
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 4. No. of electrons in 3 g mole of hydrogen atom
uncharged by inducing charge on it. = 3 ´ 6.022 ´ 1023
2. Yes. \ q = ne = 3 ´ 6.022 ´ 1023 ´ 1.6 ´ 10-19
3. On clearing, a phonograph record becomes = 2.9 ´ 105 C
charged by friction.
4. False. Direction of motion can be different Hence net field at O is same as produced by
from direction of force. A done but in opposite direction,02 i.e.,
s 1 q
5. E = Þ s = Ee0 = 3.0 ´ 8.85 ´ 10-12 E= ×
e0 4pe a 2
= 2.655 ´ 10-11 C/m 2 8. Net field at the centre (O) of wire is zero. If a
small length of the wire is cut-off, net field
6. q1 and q3 are positively charged as lines of
will be equal to the field
force are directed away from q1 and q3. q2 is
due to cut-off portion, i.e.,
negatively charged because electric field 1 dq
lines are towards q2. dE = × O
4pe0 R2
7. If a charge q is placed at A also net field at q
centre will be zero. dl R
1 2p R
= ×
4pe0 R2
q dl
=
8p 2e0 R3
32
® 1 q ®
9. E= × 3 r
4pe0 r
9 ´ 109 ´ 2 ´ 10-6 ^ ^ ^ ^
=- (3 i + 4 j) = - 144 (3 i + 4 j) N/C
(3 2 + 42 )3/ 2
A B
®
x 1m E = 20V/m
–2 O 2 4 1m
D C
(c) VB - VD = - 20 ´ 1 = - 20 V
Hence, E is uniform and negative (d) VC - VD = - 20 ´ 1 = - 20 V
From x = 0 to x = 2
34
P Q
R
q R —
2
O
B q
+
–(q +Q)
'+
3q
–2q C
–(q
q+Q
B
A +2q
A
–q R
q R 2R 3R
2R
1 æ q¢ + q + Q 2q - Q ö
VB = ×ç + ÷ =0
q B = - 2q 4pe0 è 2R 3R ø
Total charge inside a conducting sphere Þ 3 q¢ + q + f = 0 …(i)
appears on its outer surface, Again, VP = VC
\Charge on outer surface of A = 2q 1 éq + Q q 2q - Q ù 1 3 q + q¢
×ê + + ú= ×
and charge on outer surface of B 4pe0 ë R 2R 3 R û 4pe0 3R
= 2q - 2q = 0 6( q + Q ) + 3 q¢ + 2(2q - Q ) = 2(3 q + q¢ )
2. Let q¢ = charge on sphere B and charge f 4q + 4q + q¢ = 0
flows from sphere C to A.
On solving
36
5 24 3.
Q=- q, q¢ = - q 2q + q'
11 11
–(q') C
A B C B
q'
Charge on 0 - (q + Q ) - (q¢ + q + Q ) –2q A + 2q
–q
inner surface 6 18
=- q = q q A
11 11
Charge on q + Q = 6 q¢ + q + Q 3q + q = 9 q
outer surface 11 18 11
=- q
11
A B C
Charge on - q - 2q 4
+ q
inner surface 3
Charge on + 2q 4q 2
- + q
outer surface 3 3
AIEEE Corner
Subjective Questions (Level-1)
1 q (Q - q) 1 q1q2
1. F = × 4. F1 = × …(i)
4pe0 r2 4pe0 r 2
For maximum force 1 q2
F2 = × …(ii)
dF 1 Q - 2q 4pe0 r 2
= × =0
dq 4pe0 r2 [As both the spheres are identical, find
Q charge on both the spheres will be equal]
q= q - q2
2 q= 1
d 2F 1 -2 2
= × <0
dq2 4pe0 r 2 Þ q1 - q2 = 2q
Q From Eq. (ii),
Hence F is maximum at q = .
2 q2 = 4pe0 r 2F2
2. Minimum possible charge on a particle = e. (50 ´ 10-2 )2 ´ 0.036
9
1 e 2 9 ´ 10 ´ (1.6 ´ 10 )
-19 2 = = 10-12
\Fmin = × 2= 9 ´ 109
-2 2
4pe0 r (1 ´ 10 )
q = 10-6 C = 1 mC
= 2.3 ´ 10-24 N From Eq. (i),
1 q1q2
3. Fe = × …(i) (50 ´ 10-2 )2 ´ 0.108
4pe0 r 2 q1q2 = 4pe0r 2F1 =
9 ´ 109
Gm1m2
Fg = …(ii) = 3 ´ 10-12
r2
Fe q1q2 Also, q1 + q2 = 2q = 2 ´ 10-6
=
Fg 4pe0 Gm1m2 On solving
(3.2 ´ 10-19 )2 ´ 9 ´ 109 q1 = ± 3 mC
=
6.67 ´ 10-11 ´ (6.64 ´ 10-27 )2 and q2 = + 1 mC
= 3.1 ´ 1035
37
1 q1 Q 1 4q2
5. (a) F1 = × FAB = ×
4pe0 (3 a / 2)2 4pe0 L2
–a +a/2 +a 1 qQ
–a1 FAO = ×
O ® Q ® q2 4pe0 x 2
F2 F1
1 qq Q
1 4q1 Q = ×
F1 = × …(i) 4pe0 L2
4pe0 9a 2
1 qQ For net force on Q to be zero.
F2 = × 2
4pe0 ( a / 2)2 FAB = FAO
1 4q2 1 qq Q
1 4q2 Q × = ×
= × …(ii) 4pe0 L2 4pe0 L2
4pe0 a 2
9
For net force on Q to be zero Þ q= Q
4
F1 = F2 4
Þ Q= q
or q1 = 9 q2 9
1 4q1 Q -4
(b) F1 = × As Q is negative Þ q = q
4pe0 25a 2 9
+a (b) PE of the system
–a +Q 2
+ 1 é 4q2 qQ 4qQ ù
q1 O ® Q ® U= ×ê + + ú
F2 F1 q2 4pe0 êë L x L - x úû
1 4q2 Q 1 é 4q2 4qQ 8qQ ù
F1 = × = ×ê - - ú =0
4pe0 9a 2 4pe0 êë L 3L 3 L úû
For net force on Q to be zero. Hence, equilibrium is unstable.
F1 + F2 = 0 7. FBD of af placed at left can be given by
q1 25 N sin 60° O
Þ =
q2 9 N
Fe = T sin a l
mg = T cos a f
O dE cos f
Fe = mg tan a Q dE cos f
P
q2 = 4pe0r 2 tan a
Here, r = 2 l sin a dE
q = 16pe0 l2 sin 2 a tan a
2
dl dE sin f
q = 3.3 ´ 10-8 C.
9. Same as Q.7. Introductory Exercise 21.3. Charge on this portion
Q
10. See Q.7. Introductory Exercise 21.3. dq = l dl = dl
1 q ® L
11. E= ×1 r 1 dq
4pe0 r 3 \ dE = ×
9 -9 4pe0 ( a sec f)2
9 ´ 10 ´ ( - 8.0 ´ 10 ) ^ ^
= (1.2 i - 1.6 j ) 1 Q dl
((1.2)2 + (1.6)2 )3/ 2 = ×
4pe0 La 2 sec 2 f
^ ^
= - 18 2 (1.2 i - 1.6 j ) N/C. Now,
12. Consider an elementary portion on the ring l = a tan f
of length dl subtending angle df at centre ‘O’ Þ dl = a sec 2 f df
of the ring. 1 Q df
\ dE = ×
Charge on this portion, 4pe0 La
dE cos f Net Electric field at P.
dE cos f
dE E = ò dE cos f
dE
[ dE sin f components will cancel each other
df O as rod in symmetrical about P.]
dE sin f dE sin f
1 Q q
cos f df
4pe0 La ò- q
f = ×
R 1 2Q sin q
= ×
4pe0 La
dl dl L L
But sin q = =
2
L 4a + L2
2
2 a 2 + æç ö÷
dq = l dl = l Rdf è2ø
1 dq 1 l df 1 2Q
\ dE = × 2 = \ E= ×
4pe0 R 4pe0 R 4pe0 a 4a 2 + L2
39
14. (a) As shown in figure, direction of electric Clearly resultant field is along angle bisector
field at P will be along + ve y-axis. of field towards 9 and 10.
y 6E1
6E1
–Q 6E1
E
E1 E2 E
6E1
x
P
6E1
Q 6E1
E
(b) v = u + at
E1 5 ´ 106
E2 t= = 2.8 ´ 10-8 = 28 ns.
1.76 ´ 1014
–Q
(c) Dk = work done by electric field.
P = F × x = - eEx
= - 1.6 ´ 10-19 ´ 1 ´ 103 ´ 8 ´ 10-3
+Q = - 1.28 ´ 10-18 J
Loss of KE = 1.28 ´ 10-18 J
25
1 q 17. Here, ux = u cos 45° = ms -1
15. Let E1 = × 2
4pe0 R2 u E
12
11 1
E12 q
E11 E1
10 2
E10 E2 25
u y = u sin 45° = ms -1
9 E9 E3 3 2
E8 E4 a x = qE = 2 ´ 10-6 ´ 2 ´ 107
8 4 = 40 ms -1
E7 E E5
6
7 5 a y = - 10 ms -1
6 1 2
y = u yt + t
Resultant fields of two opposite charges can ay
be shown as given in figure. 25
y= t - 5t 2
2
40
at the end of motion, Case I.
t = T and y = 0 In between two charges : let potential is zero
5 at a distance x from q1 towards q2.
\ T= s
2 x 100–x
q1 q2
Also at the end of motion,
x=R 1 q1 1 q2
V= × + × =0
1 4pe0 x 4pe0 100 - x
\ x = ux t + a x t 2 -6 -6
2 1 2 ´ 10 1 3 ´ 10
2 = × - × =0
25 5 æ 5 ö 4pe0 x 4pe0 100 - x
R= ´ + 20 ´ ç ÷
2 2 è 2ø Þ 200 - 2x = 3 x
= 312.5 m x = 20 cm
m 2 sin 2q Case II.
18. (a) R=
qE Consider the potential is zero at a distance x
qER from charge q, on its left.
sin 2q =
mu 2 x 100 cm
1.6 ´ 10-19 ´ 720 ´ 1.27 ´ 10-3 q1 q2
=
1.67 ´ 10-27 ´ (9.55 ´ 103 )2 1 q1 1 q2
\ V= × + × =0
= 0.96 4pe0 x 4pe0 100 + x
-6 -6
2q = 88° or 92° 1 2 ´ 10 1 - 3 ´ 10
= × + × =0
q = 44° or 46° 4pe0 x 4pe0 100 + x
2mh sin q 200 + 2x = 3 x
T=
2E
x = 200 cm
1
2 ´ 9.55 ´ 103 ´ ´ 1.67 ´ 10-31 21. Let us first find the potential at a point on
= 2
the perpendicular bisector of a line charge.
1.6 ´ 10-19 ´ 720
Consider a line of carrying a line charge
= 1.95 ´ 10-11 s density l having length L.
® -19
® eE 1.6 ´ 10 ´ 120 ^
19. (a) a = - =- j
m 9.1 ´ 10-31 dl
^
= - 2.1 ´ 1013 i m/s
l
-2
Dx 2 ´ 10 4
(b) t = = 5
= ´ 10-7 s f
ux 1.5 ´ 10 3 L
q
vy = u y + a y t
4
= 3.0 ´ 106 ´ 2.1 ´ 1013 ´ ´ 10-7
3
= 0.2 ´ 106 m/s
® ^ ^ r
v = (1.5 ´ 105 ) i + (0.2 ´ 106 ) j
Consider an elementary portion of length dl
20. Absolute potential can be zero at two points on the rod.
on the x-axis. One in between the charges
and other on the left of charge a1 (smaller in Charge on this portion
magnitude). dq = l dl
1 l dl
\ dV = ×
4pe0 r sec f
R
Q
Q
But, q1 + q2 = Q Þ s =
4 p ( r + R2 ) 2
-Q
Hence, sin =
1 æ q1 q2 ö 4pa 2
VA = ×ç + ÷
4pe0 è r Rø Q
and sout =
2 2
1 æ s (4pr ) s (4p R ) ö 4pa 2
= ×ç + ÷
4pe0 çè r R ÷ (b) Entire charge inside the sphere appears
ø
on its outer surface, hence
1 Q ( r + R) Q+q
= × Q
4pe0 r 2 + R2 sin = - and sout =
4pa 2 4pa 2
1 æ q1 + q2 ö 1 Q (c) In case (a)
VB = ×ç ÷= ×
4pe0 è R ø 4pe0 R 1 Q
E= ×
2
43. q A = s (4pa ), q B = - s (4pb ) 2 4pe0 x 2
s +Q
–Q
–s
s
Q
a
x
A b
B
In case (b)
C c
E = E1 + E2
E1 = Field due to charge Q.
qC = s (4pc 2 )
E2 = Field due to charge on shell.
1 æ q A q B qC ö
VA = ç + + ÷ 1 Q
4pe0 è a b c ø E= ×
4pe0 x 2
s
= ( a + b + c) for x < a
e0
As field due to shell is zero for x < a.
45
1 Q+q (d) Let Q A be the charge on inner sphere.
and E= × , for x > a
4pe0 x 2 –Q
45. Let Q be the charge on the shell B,
b QA
—q –q
c
b –b R
—q —q
c c
C –q
q C 3R
A
B
a
A
B
b 1 é QA Q ù
VA = ×ê - =0
4pe0 ë R 3 R úû
Q
QA =
3
1 é q + Q - qù
VB = ê + ú =0 47. (a) At r = R
4pe0 ë b c û 1 é Q - 2Q 3Q ù
V= ×ê + + ú
æ b - cö 4pe0 ë R 2R 3 Rû
Q = qç ÷
è c ø 1 Q
= ×
Charge distribution on different surfaces is 4pe0 R
shown in figure.
At r = 3 R
46. (a) Let E1 and E2 be the electric field at P due 1 é Q - 2Q + 3Q ù
to inner shell and outer shell respectively. V= ×ê ú
4pe0 ë 3R û
1 2Q
+2Q = ×
P r 4pe9 3 R
30° mg mg cos q
A
mg sin 30° = Fe
–q +q +q –q 1 q2
mg sin 30° = ×
- q2 4pe0 r 2
= [4 - 2 ]
4pe0 a 1
r=q
1 é q 2
q q2
ù 2 4pe0 mg sin 30°
Uf = × ê- ´2 + ´2 - ´ 2ú
4pe0 êë a a 2 úû 91 ´ 109
= 2.0 ´ 10-6
2 q2 1
=- 0.1 ´ 10 ´
4pe0 a 2
» 20 cm
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 qB
O
C FCD
D F qC
FCO
FCA
FCB
q q
E D
1 (2 2 - 1) Q 2
2
FCB = × 1 q q
4pe0 a2 E= × 2
=
2 2 4pe0 (2a ) 16pe0 a 2
1 (2 2 - 1) Q
FCD = × 16. On touching two spheres, equal charge will
4pe0 a2
2 2 appear on both the spheres and for a given
1 (2 2 - 1) Q total charge, force between two spheres is
FCA = ×
4pe0 2a 2 maximum if charges on them are equal.
2
1 2 (2 2 - 1) Q 17. Charge distribution is shown in figure.
FCO = ×
4pe0 a2 +8Q
–2Q
Net force on C +2Q
F = FCA + FCO + FCB cos 45° + FCO cos 45° –4Q
+4Q
(2 2 - 1)Q é (2 2 - 1)Q (2 2 - 1)Q
= ê +
a2 êë 2 2
(2 2 - 1)Q 2q ù
+ + ú =0
2 1 úû
7Q
q=-
4
s 1 q
12. E = 18. V = ×
e0 4pe0 r
es If drops coalesce, total volume remains
F = eF =
e0 conserved,
4 4
Acceleration of proton pR3 = 1000 ´ pr 3
F se 3 3
a= =
m me0 R = 10r
1 1 1000q
s = ut + t 2 V¢ = × = 10V
a 4pe0 10q
u =0 1 éq Qù
19. VA = × + ú
25 me0 4pe0 êë r Rû
25 B q
t= =
a se
r
2 ´ 0.1 ´ 1.67 ´ 10-27 ´ 8.8 ´ 10-12 A
=
2.21 ´ 10-9 ´ 1.6 ´ 10-19
R
= 2 2 ms
49
1 é q + Qù
VB = ×ê ú
4pe0 ë R û
q é1 1 ù
VA - VB = × - O
4pe0 êë r R úû P
R 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
2Q × = × ×
4pe0 R + r 2 4pe0 2 R
–3Q
3Q 1 3°
=
–Q R + r 4R
Q
4R = 3 R + 3r
R
r=
3
25. Net charge on any dipole is zero.
26. For net force to be zero.
®® ^ ^ ^ T T cos q
21. f = E × S = (5 i + 2 j) × ( i ) = 5 V-m. q
1 Qq q
22. FDA = FDC = ×
4pe0 a 2 qE
q 7 sin q
A Q B
mg
Q
mg
T cos q = mg Þ T =
cos q
q Q qE
FDC or T sin q = qE Þ T =
D C sin q
1 q V1
FDB 27. E1 = × =
4pe0 a a
FDA 1 q V2
E2 = =
1 q2 4pe0 b2 b
FDB = × 2
4pe0 2a But E1 = E2
Net force on charge at D V1 V2
=
® a b
F0 = FDB + FDA cos 45° + FDB cos 45° = 0 V1 a
Þ =
1 q éq Q Qù V2 b
Þ 4pe0 × 2 ê2 + + ú =0
4 a ë 2 2û 28. Electric field on equatorial lines of dipole is
q = -2 2Q opposite to dipole moment.
23. As VB = 0, Total charge inside B must be zero 29. Potential difference between two concentric
and hence charge on its outer surface is zero spheres is independent of charge on outer
and on its inner surface is - q. sphere.
1 1 q
24. Vp = 30. E = × 2
V0 4pe0 r
50
1 q 1 q VR
V= × = Er E= × = 2
4pe0 r 4pe0 r 2 r
V 3000 36. When outer sphere is earthed field between
r =½ ½ = =6m
½ E ½ 500 the region of two spheres in non-zero and is
6 ´ ( - 3000) zero in all other regions.
q = 4pe0rV = = - 2 mC
9 ´ 109 ®®
37. W = F × s = qEs cos q
31. F1 = F2 W 4
1 q1q2 1 qq E= = = 20 N/C
× = × 1 2 qs cos q 0.2 ´ 2 ´ cos 60°
4pe0 r12 4p Ke0 r22
r 50 1 éê Q Q ù
ú
r2 = 1 = = 10 5 m 38. V1 = × -
K 5 4pe0 ê R d 2 + R2 úû
ë
» 22.3 m Q –Q
32. Electric field at a distance r from infinite line
charge R R
l
E=
2pe0r
dV = - E dr
V2 b r d
òV dV = - ò E dr 1 2
1 a
Þ V2 - V1 =
l
× ln
1 1 éê Q Q ù
ú
V2 = - +
2pe0 2 . ê R d 2
+ R 2 ú
4pe0 ë û
ql 1
W = q( V2 - V1 ) = ln é ù
2pe0 2 1 ê 2Q 2Q ú
V1 - V2 = - -
. ê R d 2
+ R 2ú
33. As negative charge is at less distance from 4pe0 ë û
the line charge, it is attracted towards the é ù
line charge. Q ê 1 1 ú
V1 - V2 = - +
. ê R d + R2 úû
2
34. r = (4 - 1)2 + (2 - 2)2 + (0 - 4)2 = 5 m 4pe ë
0
9 -8
1 q 9 ´ 10 ´ 2 ´ 10 39. Electric field inside a hollow sphere is
V= × = = 36 V
4pe0 r 5 always zero.
®® ®®
(b) and (c) are wrong. 40. W = F × r = q E × r
1 q
35. V = × ^ ^ ^ ^
= q ( E1 i + E2 j) - ( a i + b j)
4pe0 R
At a distance r from the centre, = q ( aE1 + bE2 )
JEE Corner
Assertion and Reason
1. Negative charge always moved towards If q1 and q2 have opposite sign, U decreases
increasing potential. with decrease in r.
On moving from A to B potential energy of dU
F=- Þ work done by conservative force
negative charge decreases hence its KE dr
increases. always decreases PE.
1 q1q2 dV
2. U = × 3. E = - = - (10) = 10 V/m along x-axis.
4pe0 r dr
1 q
4. V = ×
4pe0 R
51
Inside the solid sphere. ^ ^ ^
VA = - (4 i + 4 j) × (4 i ) = - 16 V
1 qr
E= × ^ ^ ^
4pe0 R3 VB = - (4 i + 4 j) × (4 i ) = - 16 V
R VA = VB
at r =
2 Hence, Assertion is false.
1 q V
E= × = 7. In the line going A and B, the energy of third
4pe0 2 R2 2 R
charge is minimum at centre.
Assertion is correct. 8. Dipole has both negative and positive
Reason is false as electric field inside the charges hence work done is not positive.
sphere is directly proportional to distance
9. Charge outside a closed surface can produce
from centre but not outside it.
electric field but cannot produce flux.
5. Gauss theorem is valid only for closed 1 qx a
surface but electric flux can be obtained for 10. E = × 2 2 3/ 2
is maximum at x =
4pe0 ( x + a ) 2
any surface.
1 q
6. Let V0 = Potential at origin, But V = × is maximum at x = 0.
4pe0 a 2 + x 2
®
P3 3a q
E4 E3
E2 P E1
Sort-cut Method
Entire charge resides on outer surface of
conductor and will be divided equally on two
outer surfaces.
61
4. Charge distribution is shown in figure. 5q
q=
q + q2 q 2
q1 = q4 = 1 =- e A
2 2 and C= 0
q - q2 5q d
q2 = 1 = q 5q d
2 2 \ V= =
q2 - q1 5q C 2e0 A
q3 = =-
2 2
\ Charge on capacitor = Charge on inner
side of positive plate.
C1 C2 C3 V1 V2
q1 = C1V = 1 ´ 10 = 10 mC
q2 = C2V = 2 ´ 10 = 20 mC V
q3 = C3V = 3 ´ 10 = 30 mC
q 800
2. Potential difference across the plates of (b) V1 = = = 800 V
C1 1
capacitor
q 800
V = 10 V V2 = = = 400 V
C2 2
q = CV = 4 ´ 10 = 40 mC
C1,V1
3. In the steady state capacitor behaves as
open circuit.
2W
4W C2,V2
A B
I
Now, if they are connected in parallel,
I
æ C V + C2V2 ö
Common potential, ç V = 1 1 ÷
ç C1 + C2 ÷ø
6W 30V è
1 ´ 800 ´ 2 ´ 400 1600
30 = = V
I= =3 A 1+2 3
6+4
1600 3200
Potential difference across the capacitor, q1 = C1V = mC, q2 = C2V = mC
3 3
VAB = 4 ´ I = 4 ´ 3 = 12 V 5. Common potential
\Charge on capacitor C V + C2V2
V= 1 2
q = CVAB = 2 ´ 12 = 24 mC C1 + C2
1 1 1 1 1
4. (a) = + = + But V = 20, V2 = 0, V1 = 100 V, C1 = 100 mC
Ce C1 C2 1 2
100 ´ 100 + C2 ´ 0
2 \ = 20
Ce = mC 400 + C2
3
Þ C2 = 400 mC
62
q –q C1
C2 R2
C2
At any instant, let charge on C2 be q, charge
on C1 at that instant = q0 - q
R1 R3
By Kirchhoff’s voltage law,
( q0 - q) q \Initial current
- IR - = 0 E
C C Ii =
dq q0 - 2q R1
Þ =
dt RC (b) After a long time, i.e., in steady state, both
q dq t dt
the capacitors behaves open circuit,
Þ ò 0 q0 - 2q = ò 0 RC E
If =
[ln ( q0 - 2q)]q0 1 R1 + R3
Þ = [ t ]t0
-2 RC 6. (a) Immediately after closing the switch,
capacitor behaves as short circuit,
63
A I2 (c) Potential difference across the capacitors in
I1 the steady state,
I V =E
E R1 R2 \Energy stored in the capacitor
1
U = CE2
2
C
(d) After the switch is open
S B
E E R2
\ I1 = and I2 = E R1
R1 R2
C
(b) In the steady state, capacitor behaves as
open circuit,
E Re = R1 + R2
\ I1 = , I2 = 0 t = R3C = ( R1 + R2 ) C
R1
AIEEE Corner
Subjective Questions (Level-1)
-3 A
e0 A Cd 1 ´ 1 ´ 10 k3e0
1. C = ÞA= =
d e0 8.85 ´ 10-12 C3 = 2 = k3e0 A
2d / 2 2d
= 1.13 ´ 108 m 2
e0 A1 e0 A2 Therefore, the effective capacitance,
2. C1 = and C2 = C2C3
d d C = C1 +
If connected in parallel C2 + C3
e A e A e0 A æ k1 k1k3 ö
C = C1 + C2 = 0 1 + 2 2 = ç + ÷
d d ç
2d è 2 k2 + k3 ÷ø
e0( A1 + A2 ) e0 A
= = 4. (a) Let the spheres A and B carry charges q
d d
and - q respectively,
where, A = A1 + A2 = effective area. q –q
Hence proved.
3. The arrangement can be considered as the a b
combination of three different capacitors as
shown in figure, where d
A B
C2 1 é q qù
\ VA = × -
4pe0 êë a d úû
C1
1 é q qù
VB = × - +
C3 4pe0 ëê b d ûú
Potential difference between the spheres,
A q é1 1 2ù
k1e0 V = VA - VB = + -
2 = k1e0 A 4pe0 êë a b d úû
C1 =
2d 4d q 4pe0
C= =
A V 1 + 1-2
k2e0
C2 = 2 = k2e0 A a b d
2d / 2 2d Hence proved.
64
(b) If d ® ¥ Let effective capacitance between A and B
4pe0 4pe0 × ab
= C= C AB = x
1 1 a+b
+ As the network is infinite,
a b
C PQ = C AB = x
If two isolated spheres of radii a and b are
connected in series, Equivalent circuit is shown in figure,
then, 2C
C1C2 A P
C¢ =
C1 + C2
C x
where, C1 = 4pe0a, C2 = 4pe0b
4pe0 × ab Q
\ C¢ = B
a+b
2Cx
RAB = C + =x
\ C¢ = C 2C + x
Hence proved.
Þ 2C 2 + Cx + 2Cx = 2Cx + x 2
5. (a) Þ x 2 - Cx - 2C 2 = 0
B
A On solving, x = 2C or - C
C
C But x cannot be negative,
C C C
Hence, x = 2C
A B
C 6. q = CV = 7.28 ´ 25 = 182 mC
-6
C q 0.148 ´ 10
7. (a) V = = -12
= 604 V
C 2C C 245 ´ 10
C
A B e A Cd
(b) C = 0 Þ A =
d e0
C
5C/3 2C/3 245 ´ 10-12 ´ 0.328 ´ 10-3
=
A B
A B
8.85 ´ 10-12
= 9.08 ´ 10-3 m 2
C
= 90.8 cm 2
(b) q 0.148 ´ 10
-6
C (c) s = = = 16.3 mC/m 2
A 9.08 ´ 10-3
C C C C
A A B 8. (a) E0 = 3.20 ´ 105 V/m
B
C C C E = 2.50 ´ 105 V/m
E 3.20 ´ 105
C k= 0 = = 1.28
E 2.50 ´ 105
4C/3 (b) Electric field between the plates of
A B A B
C/3 capacitor is given by
s
(c) E=
e0
2C 2C 2C 2C
A
P Þ s = e0E = 8.8 ´ 10-12 ´ 3.20 ´ 105
= 2.832 ´ 10-6 C/m 2
C C C C ¥
= 2.832 mC/m 2
B 9. (a) q1 = C1V = 4 ´ 660 = 2640 mC
Q
q2 = C2V = 6 ´ 660 = 3960 mC
65
As C1 and C2 are connected in parallel, For series combination,
V1 = V2 = V = 660 V US = 1.6 ´ 10-2 J = 0.016 J
1
C1 = 4.00mF US = C S V 2
2
2US 2 ´ 0.016
Þ CS = 2 = = 0.008 F
C2 = 6.00mF V (2)2
= 8 mF
Now, C P = C1 + C2 = 5 mF
or C2 = (5 - C1 ) mF
660 V 1 1 1 1
and = + =
(b) When unlike plates of capacitors are CS C1 C2 8
connected to each other, 1 1 1
Þ + =
Common potential C1 S - C1 8
C V - C1V1 6 ´ 660 - 4 ´ 660
V= 2 2 = On solving,
C1 + C2 6
C1 = 40 mF, C2 = 10 mF or vice-versa.
= 220 V 13. In the given circuit,
q1 = C1V = 4 ´ 220 = 880 mC q q
VA - VB = -E + =5
q2 = C2V = 6 ´ 220 = 1320 mC C1 C2
V 400 C1 E C2
10. E = = = 8 ´ 104 V/m +q –q +q –q
d 5 ´ 10-3 A B
Energy density, q q
1 1 Þ - 10 + =5
u = e0E2 = ´ 8.85 ´ 10-12 ´ ( 8 ´ 10-4 )2 10-6 2 ´ 10-6
2 2
Þ q = 10 ´ 10-6 C = 10 mC
= 2.03 ´ 10-2 J/m 3 q q
\ V1 = = 10 V, V2 = =5V
= 20.3 mJ/m 3 C1 C2
11. Dielectric strength = maximum possible 14. (a) In order to increase voltage range n
electric field times, n-capacitors must be connected in
V V series.
E= Þd=
d E Hence, to increase voltage range to 500V,
5500 5 capacitors must be connected in series.
= = 3.4 ´ 10-4 m Now, effective capacitance of series
1.6 ´ 107
combination,
ke A Cd 10
C= 0 ÞA= CS = Cn = = 2 pF
d ke0 5
1.25 ´ 10-9 ´ 3.4 ´ 10-4 Hence, no parallel grouping of such units is
=
3.6 ´ 8.85 ´ 10-12 required.
Hence, a series grouping of 5 such
= 1.3 ´ 10-2 m 2 capacitors will have effective capacitance
= 0.013 m 2 2 pF and can withstand 500 V.
12. Let C P and CS be the effective capacitance of (b) If n capacitors are connected in series and
parallel and series combination respectively. m such units are connected in parallel,
Ve = nV
For parallel combination,
mC
U P = 0.19 J Ce =
n
1
U P = C PV 2 Here, V = 100 V
2
2U 2 ´ 0.1 Ve = 300 V
Þ C P = 2P = = 0.05 F V
V (2)2 \ n = e =3
V
= 50 mF
66
C = 10 pF C1 + 2 90
Þ = =9
Ce = 20 pF C2 10
nCe 3 ´ 20 Þ C1 + 2 = 9C2
m= = =6
C 10 3
Þ C1 + 2 = 9 ´ C1
Hence, the required arrangement is shown 2
in figure. 25 4
Þ 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
15. Case I. C
C2 q¢ = E¢ V = V = 60 mC
V1 = V = 60 V 2
C1 + C2
e0 A pe0r 2
A B
(c) C = =
d d
If r is doubled, C will become four times, i.e.,
V1 V2
C ¢ = 4C
q¢ = C ¢ V = 480 mC
17. Heat produced = Energy stored in the
100 V
capacitor
1 1
C1 H = CV 2 = ´ 450 ´ 10-6 ´ (295)2
V2 = V = 40 V 2 2
C1 + C2
= 19.58 J
C1 2 e A 8.85 ´ 10-12 ´ 2
Þ = 18. (a) C = 0 =
C2 3 d 5 ´ 10-3
3
Þ C2 = C1 = 3.54 ´ 10-6 F
2
= 3.54 mF
Case II.
C2 (b) q = CV = 3.54 ´ 10-9 ´ 10000
V1 = = 10 V = 35.4 ´ 10-6 = 35.4 mC
C1 + C2 + 2
V 10000
(c) E = = = 2 ´ 106 V/m
2mF d 5 ´ 10-3
19. Given,
C1 C2 C3
A B
V1 V2
100 V
C1 + 2 V
V2 = = 90 V
C1 + C2 + C C1 = 8.4 mF, C2 = 8.2 mF
C3 = 4.2 mF, V = 36 V
67
(a) Effective capacitance, (Charge is shown in mC).
1 1 1 1 Hence, charge on 6 mF capacitor = 10 mC
= + +
Ce C1 C2 C3 40
and Charge on 4 mF capacitor = mC
1 1 1 3
= + + Þ Ce = 2.09 mF
8.4 8.2 4.2 21. (a)
C1=8.4mF C3=4.2mF
q = CeV = 2.09 ´ 36 = 75.2 mC
a
As combination is series, charge on each
capacitor is same, i.e., 75.2 mC. C2=4.2mF
1 1 b
(b) U = qV = ´ 75.2 ´ 36 ´ 10-6
2 2 C5=8.4mF C4=4.2mF
= 1.35 ´ 10-3 J = 1.35 mJ 8.4mF
a
(c) Common potential,
C V + C2V2 + C3V3 4.2mF 2.1mF
V= 1 1 = 10.85 V
C1 + C2 + C3 b
1 8.4mF
(d) U ¢ = (C1 + C2 + C3 )V 2
2 8.4mF
1 a
= ´ ( 8.4 + 8.2 + 4.2) ´ (10.85)2 ´ 10-6 2.52mF
2 a b 6.3mF
= 1.22 ´ 10-3 J = 1.22 mJ
b
20. The Given circuit can be considered as the 8.4mF
sum of three circuits as shown
(b) Charge supplied by the source of emf
3mF
+12 –12 q = CV = 2.52 ´ 10-6 ´ 220
= 554.4 mC
+6 +2 +4
5V 6mF q1 = q5 = q = 554.4 mC
2mF 4mF 4.2
–2
q2 = q
–6 –4 4.2 + 2.1
3mF 4.2
+12 –12 = ´ 554.4 mC = 369.6 mC
6.3
–24 +12 +24 2.1 2.1
and q3 = q4 = q= ´ 554.4 mC
+ 6mF 2mF 4.2 + 2.1 6.3
+24 –12 –24 4mF = 184.8 mC
q1 554.4
10V V1 = = = 66 V = V5
C1 8.4
+
3mF q 369.6
–4 +4
5V V2 = 2 = = 88 V
C2 4.2
4
–4
— q 184.8
+8+8/3 3 V3 = V4 = 3 = = 44 V
+ –8/3 4mF C3 4.2
–8 4
6mF +4
—
2mF 3 22. Let C1 and C2 be the capacitances of A and B
respectively.
ke A ke A
3mF \ C1 = 1 0 1 , C2 = 1 0 2
5V d1 d2
+20mC –20mC
+50 +40 C2
–10mC — mC — mC Now, V1 = V
3 3 C1 + C2
5V 6mF 4mF
2mF –50 C2
–40 130 13
+10mC — mC
3 — mC Þ = = …(i)
3 C1 + C2 230 23
10V
68
C1 C2=2mF
V2 = V
C1 + C2
4mF
C1 10
Þ = …(ii) 2mF
C1 + C2 23 C3=4mF
+
From Eqs. (i) and (ii), 20V
C1 10
= 3mF
C2 13
If dielectric slab of C1 is replaced by one for
which k = 5 then, C1=3mF
5e A 5
C1 ¢ = 0 1 = C1
d1 2
V2 ¢ C1 ¢ 5C1 50 2mF
\ = = =
V1 ¢ C2 2C2 26 2mF 2mF
50
V2 ¢ + V1 ¢ = 230 20V
26
Also, 3mF 3mF
V1 ¢ + V2 ¢ = 230
50
V1 ¢ = V1 ¢
26
V1 ¢ = 78.68 V
6mF
and V2 = 151.32 V
20V 20V 3mF
23. In this case 6mF
C1,V1
(b) q = CV = 3 ´ 20 = 60 mC
(c) Potential difference across C1
6
V1 = ´ 20 = 10 V
6+6
C2 C3
q1 = C1V1 = 3 ´ 10 = 30 mC
Common potential, (d) Potential difference across C2
C1V1 6
V= V2 = ´ 20 = 10 V
C2C3
C1 + 6+6
C2 + C3
1 ´ 110 q2 = C2V2 = 2 ´ 10 = 20 mC
V=
1 + 1.2
110 (e) Potential difference across C3
= 4
2.2 V3 = ´ V2 = 5 V
4+4
= 50 V
Charge flown through connecting wires, q3 = C3V3 = 4 ´ 2 = 20 mC
C2C3 25. (a) When switch S2 is open, C1 and C3 are in
= V
C2 + C3 series, C2 and C4 are in series their
effective capacitances are in parallel with
= 1.2 ´ 50 each other.
= 60 mC Hence,
24. (a) Hence, effective capacitance across the C1C3
q1 = q3 = V
battery is 3 mF. C1 + C3
69
1 ´3 C2C3Q
= ´ 12 = 9 mC q=
1+3 C1C2 + C2C3 + C3C1
C2C4 C1C2C3V
q2 = q4 = =
C2 + C4 C1C2 + C2C3 + C3C1
2 ´4 C12 (C2 + C3 ) V
= ´ 12 = 16 mC \ q1 = Q - q =
2+4 C1C2 + C2C3 + C3C1
(b) When S2 is closed, C1 is in parallel with C1C2C3V
q2 = q3 = q =
C2 and C3 is in parallel with C4. C1C2 + C2C3 + C3C1
C1 C3 e0 A e AV
27. C = , q = CV = 0
d d
S2 e0 A
(a) C ¢ = , q¢ = q
2d
C2 C4 (As battery is disconnected)
V1 V2 q¢
V¢ = =2V
C¢
B
1 2 e0 AV 2
(b) Vi = V =
Therefore, 2C 2d
C3 + C4 1 1 e A
V1 = V2 = V Uf = C ¢ V ¢2 = × 0 (2V )2
C1 + C2 + C3 + C4 2 2 2d
7 e AV 2
= ´ 12 = 8.4 V = 0
10 d
C1 + C2 e0 AV 2
V3 = V4 = V (c) W = Uf - Ui =
C1 + C2 + C3 + C4 2d
3 28. In the steady state, capacitor behaves as
= ´ 12 = 3.6 V
10 open circuit,
q1 = C1V1 = 1 ´ 8.4 = 8.4 mC I1 R1 P
q2 = C2q2 = 2 ´ 8.4 = 16.8 mC
I2 IC
q3 = C3V3 = 3 ´ 3.6 = 10.8 mC A S B
q4 = C4V4 = 4 ´ 3.6 = 14.4 mC C
26. Initial charge on C1 E1 E2 R2
Q = C1V0
Now, if switch S is thrown to right.
D
Let charge q flows from C1 to C2 and C3.
E1
By Kirchhoff’s voltage law, I1 = I2 = = 1 mA and IC = 0
R1 + R2
q
C2
E1 R2
Q–q VPD = I2 R2 =
–q R1 + R2
C1
–(Q–q) q When switch is shifted to B,
C3
–q At this instant,
E1 R2
q q Q-q VPD =
+ - =0 R1 + R2
C2 C3 C1 V E1
æ 1 I2 = PD = = 1 mA
1 1 ö Q R2 R1 + R2
q çç + + ÷=
÷
è C1 C2 C3 ø C1
70
E1 R2 q2 = C2V = 3 ´ 18 = 54 mC
E2 +
E + VPD R1 + R2 After closing the switch,
I1 = 2 =
R1 R1 q1 ¢ = C1V1 = 6 ´ 12 = 72 mC
( R1 + R2 ) E2 + E1 R2
= q2 ¢ = C2V2 = 3 ´ 6 = 18 mC
R1
Dq1 = 18 mC, Dq2 = - 36 mC
= 2 mA V 18
30. (a) I = = =2A
IC = I1 + I2 = 1 + 2 = 3 mA R1 + R2 9
29. (a) When switch S is open, no current pass V=18.0V
through the circuit,
I C1= 6mF
V=18.0V
+
R1 = 6W q
–
S
R1 C1 a b
+ C2= 3mF
a b R2 = 3W q
S –
C2 R2
C1C2
q= V = 2 ´ 18 = 36 mC
C1 + C2
Hence, Now, Va - 0 = IR2 Þ Va = 6 V
Vb - 0 = 0 q 36
and Vb - 0 = = = 12 V
Vb = 0 C2 3
18 - Va = 0 Þ Va = 18 V Va - Vb = - 6 V
Þ Va - Vb = 18 V (b) b is at higher potential.
(b) a is at higher potential. (c) When switch S is closed, in steady state,
(c) When switch S is closed, V =18V
V=18.0V
I
R1 q1 + C1
–
C1
R1 a S b
I
a b
S +
R2 q2 C2
R2 –
C2
I
Va - vb = 6 V
V
I= =2A q1 = C1V1 = 6 ´ 12 = 72 mC
R1 + R2
q2 = C2V2 = 3 ´ 6 = 18 mC
Vb - 0 = IR2 = 2 ´ 3 Charge flown through S
Vb = 6 V = q1 - q2 = 72 - 18 = 54 mC
(d) q1 = C1V = 6 ´ 18 = 108 mC
71
31. E1 R2 + E2 R1 V
Ee = =
I A R F R1 + R2 2
E
I–I1 R 3R
I1 Re = R + =
2 2
V R q C
q = q0 (1 - e - t / t )
CE
q0 =
C G 2
R B
(a) Consider the circuit as combination of two 3 RC
t=
cells of emf E and OV. 2
C CE
F
R
D
\ q= (1 - e -2t / 3 RC )
2
dq E -2t / 3 RC
I1 (b) I1 = = e
I2 R dt 3 R
A B q
I
In loop EDBA + I1 R - I2 R = 0
C
I q
I2 = + I1
E F RC
R E E -2t / 3 RC
V = (1 - e -2t / 3 RC ) + e
2R 3R
E
C 3R/2
= (3 - e -2t / 3 RC )
6R
V/2
3R = 5 ´ 10-3 J = 5 mJ
14. q = q0e - t / h
I = I0 e - t / h
P = I 2 R = I02e -2t / h R = P0e -2t / h
h
R R Þ h¢ =
2
C V + C2V2 E
15. Common potential = 1 1 =
C1 + C2 2
E
9
16. VA - VB = 6 + 3 ´ 2 - + 3 ´ 3 = 12 V
1
73
17. In the steady state, current through battery C C C
R
C/2
S
C
12V A B
I 2W
C
2mF
4W
A B C C C 3C/2 C
O
A B
I 6W
3C/14
12 3
I= = A
6+2 2
22.
Potential difference across the capacitor,
3 1mF 1mF
VAB = 6 ´ = 9 N
2
\ q = CVAB = 2 ´ 9 = 18 mC
x 1mF y
18. C2 and C3 are in parallel
Hence, V2 = V3
Again Kirchhoff’s junction rule 2mF
- q1 + q2 + q3 = 0
Þ q1 = q2 + q3 2mF 1mF
19. For the motion of electron
mu 2 sin 2q
R= =l …(i) x y
eE
mu 2 sin 2 q
and H= =d …(ii)
2eE 2mF
2
Dividing Eq. (ii) by Eq. (i), — mF
3
4d
tan q =
l
V 2Ve0 x y x y
20. V = Ed Þ d = =
E 6 2mF 8
— mF
2 ´ 5 ´ 8.85 ´ 10-12 3
=
10-7
k1e0 A k2e0 A
= 8.85 ´ 10-4 = 0.88 mm 23. C1 = +
2d 2d
21.
( k1 + k2 ) e0 A
P Q R S = (Parallel grouping)
A B 2d
1 d d
= + (Series grouping)
C2 2k1e0 A 2k2e0 A
2k1k2 e0 A
C2 =
P and Q are at same potential, hence k1 + k2 d
capacitor connected between them have no
C1 ( k1 + k2 )2 (2 + 3 )2 25
effect on equivalent capacitance. = = =
C2 4k1k2 4 ´ 2 ´ 3 24
74
24. 1
C d
A A 2
C C A
P C d
Q P Q C 3
C CC C
C
B C C d
B
4
A A 2d
C C
C C 5
— — C C
2 2 d
C C B
B B 6
A
Capacitance of all other capacitance is same,
2C e A
B i.e., C = 0 but that of formed by plates 4
d
25. Cases (a), (b) and (c) are balanced C
and 5 is as distance between these two
Wheatstone bridge. 2
plates is 2d.
26. The given arrangement can be considered as
the combination of three capacitors as shown The equivalent circuit is shown in figure.
in figure. —
C
C C
A 2 1 5 24 4 3
5 6
C2
C1 2 C
A B
3
C3 C
C
C —
B 3
k1e0 A
Hence, C1 =
2d
C
A A B
k2e0
C2 = 2 = k2e0 A C
d/2 d 4C 4C
— —
A 7 C 3
k3e0
C3 = 2 = k3e0 A
d/2 d A B A B
C C
Effective capacitance,
C2C3 e A ék k2k3 ù
C = C1 + = 0 ê 1 + ú 11C
C1 + C2 d ë2 k2 + k3 û —
7
27. Here, plate 1 is connected to plate 5 and
11 11e0 A 11
plate 3 is connected to plate 6. \Ceq = C= = ´ 7 mF = 11 mF
7 7d 7
75
JEE Corner
Assertion and Reason
q 6. As potential difference across both the
1. Capacitance = is constant for a given
V capacitors is same, charge will not flow
capacitor. through the switch.
2. Reason correctly explains the assertion. 7. C and R2 are shorted.
1
3. U = qV , W = qV 8. Time constant for the circuit,
2
t = RC
4. For discharging of capacitor
9. In series, charge remains same
q = q0e - t / t
q2 1
dq q and U= ÞU µ
= - 0 e -t / t 2C C
dt t
q 10. In series charge remains same
= - 0 e -t / t q q
RC \ V1 = , V2 =
C1 C2
Hence, more is the resistance, less will be
the slope. On inserting dielectric slab between the
5. Charge on two capacitors will be same only if plates of the capacitor, C2 increases and
both the capacitors are initially uncharged. hence, V2 decreases. So more charge flows to
C2.
Re = R Þ t = RC
C1 = 60mF
While discharging
Re = 2 R Þ t = 2 RC C2 = 20mF V0 C3 = 30mF
21. Common potential,
C V - C1V1 3 ´ 100 - 1 ´ 100 B C
V= 2 2 =
C1 + C2 1+3
q1 + q2 + q3 = 0
= 25 V C1(C A - V0 ) + C2( VB - V0 ) + C3( VC - V0 ) = 0
1 ´ 1.5 C V + C2VB + C3VC
22. q1 = ´ 30 = 18 mC V0 = 1 A
1 + 1.5 C1 + C2 + C3
C1 C2
60 ´ 6 + 2 ´ 20 + 3 ´ 30
1.0mF
a
1.5mF V0 =
60 + 20 + 30
P q1 q1 Q 49
q2 q2 = V
11
b 25. In the steady state, there will be no current
2.5mF 0.5mF
C3 C4 in the circuit.
30V
78
3mF C1C2
3mF 1mF B
1mF 29. H = ( V1 - V2 )2 is independent of
B 2 (C1 + C2 )
3mF
resistance.
3mF
2mF 3mF 1mF 30. Immediately after switch is closed, capacitor
1mF behaves like short circuit.
2W 10W 20W 10W
10V 10V V - t / RC V
31. i1 = e , i2 = e - t / RC
A A 2R R
5t
i1 1 6 RC
= e
C1=6mF C2=3mF i2 2
B
q1 q2 Increases with time.
C1=1mF rd d
32. R = =
q3 A sA
20W 10W ke0 A
C=
A 10V d
d ke A
C2 3 10 t = RC = ´ 0
\ V1 = V= 10 = V sA d
C1 + C2 3+6 3
e0 8.85 ´ 10-12
E1 18 = = =6s
26. I1 = = =3 A 6 7.4 ´ 10-12
R1 + r1 5 + 1
E2 15 33. i = i0e -t / t
I2 = = = 2.5 A i0
- ln 4
R2 + r2 4 + 2 Þ = i0e RC
2
3mF ln 4
A + – B Þ = ln 2
I1 I1 q RC
15V
18V 5W 4W
Þ ln 4 = ln 2 RC
2W
1W q I2 I2 Þ RC = 2
– + 2 2
D C Þ R= = =4W
2mF C 0.5
34. Potential difference across each capacitor is
In loop ABCD, equal, hence they are in parallel,
q q
- I2 R2 + - I1 R1 = 0 charge on each capacitor
3 2
5q q = CeV = 2 ´ 10 = 20 mC
Þ = 3 ´ 5 + 2.5 ´ 4 Þ q = 30 mC
6 As plate C contributed to two capacitors,
27. During discharging charge on plate,
q = q0 e - t / t C = 2q = + 40 mC
q0 = CE = 10 mC 35. Charge distribution on the plates of the
at t = 12 s, capacitor is shown in figure
q = 10e -12/ 6 = 10e -2 Q/2 CV +
Q Q/2
2
= (0.37)210 mC
C1C2
28. q = ( E1 - E2 ) –CV +
Q
C1 + C2 2
q C1
Vap = = ( E1 - E2 ) Q
C2 C1 + C2 CV +
Q¢ 2
\ V¢ = =
æ E - E2 ö C C
=ç 1 ÷C
çC + C ÷ 1 Q
è 1 2ø =V +
2C
79
t
36. Let q be the charge on C2 (or charge flown C2q0 æ - ö
q= ç1 - e t ÷
through the switches at any instant of time) C1 + C2 ç ÷
è ø
By Kirchhoff’s law C1 + C2
S1
t=
I C1C2 R
t
C æ - ö
or q= q0 ç1 - e RC ÷
C1 ç ÷
è ø
C1C2
+ where, C=
+ C1 + C2
q0 – q – q C2
– C1C2
37. H = ( V1 - V2 )2
q q -q 2 (C1 + C2 )
IR + - 0 =0 2
C2 C1 C1C2 æ q0 ö
= ç ÷
dq C2q0 - (C1 + C2 ) q 2 (C1 + C2 ) çè C1 ÷ø
=
dt C1C2 R C2q02 C q02
q dq t dt = =
ò0 C2q0 - (C1 + C2 ) q = ò0 C1C2 R 2C1 (C1 + C2 ) 2C12
1 q
38. Electric field in the gap will remain same.
=
C1 + C2
[ln|C2q0 - (C1 + C2 ) q]0 39. Electric field inside the dielectric slab
E V
1 E¢ = = .
= t k kd
C1C2 R
t
q 4 ´ 10-3
æ - ö 3. V1 = = = 40V
2. q1 = CE ç1 - e RC ÷ C 100 ´ 10-6
ç ÷
è ø
80
V2
Þ U µd
Q Qd
V= =
200W C e0 A
V1 900W Þ V µd
6. When switch S is open
100W C
A2 S
A1
C
I1 = 0 C
V1 40 2
I2 = = = A + –
900 900 45
2 80 E
V2 = I2 ´ 200 = ´ 200 = V
45 9 C ´ 2C 2
80 Ce = = C
E = V1 + V2 = 40 + C + 2C 3
9 2
440 q1 = CE
= V 3
9
When switch S is closed
E
4. Initially I1 = 0,I2 = I = Ce ¢ = 2C
R
C q2 = 2CE
Charge flown through the battery
4
I2 Dq = q2 - q1 = CE = positive
I1 B 3
7. Let charge q flows to C1 at it falls to the free
I S end of the wire.
+ – A
A C1 C1 –q
E
q
As the capacitor starts charging,
I2 decreases and I1 increases, q2 q3 q2–q q3–q
+ – + – + – + –
In the steady state 2mF 3mF C2 C3
E C2 C3
I1 = I = , I2 = 0
R By Kirchhoff’s voltage law,
At any instant q2 - q q3 - q q
+ - =0
P1 = I12 R, P2 = I22 R C2 C3 C1
Steady state potential difference across the q2 q
+ 3
capacitor, C2 C3
E q=
V= 1 1 q
2 + +
C1 C2 C3
1 CE2 V2 + V3
U = CV 2 = =
2 8 1 1 1
Q2 + +
5. F = independent of d. C1 C2 C3
2e0 A 150 + 120
Q q= = 180 mC
E= independent of d. 1 1 1
+ +
e0 A 2 3 1.5
Q 2 Q 2d q2 ¢ = q2 - q = 150 ´ 2 - 180
U= =
2C 2e0 A = 120 mC
81
e0 A - 6 ln 2
8. C = q0 12 - 1
d I= e 6 = ´
RC 6 2
Q 2 Q 2d 1
U= = ÞU µd =2 ´
=1 A
2C 2e0 A 2
Q Qd Potential difference across 1 W resistor
V= = Þ V µd
C e0 A Þ 1 ´1 = 1 V
e A 1 Potential difference across 2 W resistor
C = 0 ÞC µ
d d Þ 1 ´2 = 2 V
Q
E= Þ E is independent of d. \ By Kirchhoff’s voltage law, potential
e0 A
difference across capacitors = 1 + 2 = 3 V.
9. R = 1 + 2 = 3 W, C = 2 F 10. q = C1V1 = 1 ´ 10 = 10 mF
q0 = CV0 = 2 ´ 6 = 12C 6mF
At any instant
æ -t ö 1mF 4mF 1mF 4mF 9mF
q = q0 ç e RC ÷ 3mF
ç ÷
è ø
-t V1 V2 V3
dq q0 RC
I= = e
dt RC E E
at t = 0
q 12 q 10
I= 0 = =2 A V2 = = = 2.5 V
RC 3 ´ 2 C2 4
q 10
at t = 6 ln 2 V3 = = V
C3 9
C2
(a ® s), (b ® s), (c ® s).
a b 6. Charge distribution is shown in figure.
(a ® p), (b ® p, q), (c ® s), (d ® p, q, r).
C 4Q Q 2Q 7Q
3
2V V
V2 = , V1 = V2 =
3 3
2CV CV 7Q –3Q 3Q –2Q 2Q 0 0 7Q
q2 = , q1 = q3 =
3 3
4. Common potential
23 Magnetics
Introductory Exercise 23.1
1. [ Fe ] = [ Fm ] ®
Þ |Fm| = qvB sin q
[ qE] = [ qvB]
If Fm = 0, either B = 0 or sin q = 0,
Þ é E ù = [v] = [L T -1 ] i.e., q = 0
ê Bú
ë û ® ® ®
® ® ® 4. F = q ( v ´ B )
2. F = q ( v ´ B )
^ ^ ^
® ® ® ® = - 4 ´ 10-6 ´ 10-6 ´ 10-2 [(2 i - 3 j + k )
\ F ^ v and F ^ B ^ ^ ^
´ (2 i + 5 j - 3 k )]
Because cross product of any two vectors is
always perpendicular to both the vectors. ^ ^
= - 4 ´ 10-2(4 i + 8 j + 16 k )
^
® ® ®
3. No. As Fm = q ( v ´ B ) ^ ^ ^
= - 16 ( i + 2 j + 4 k ) ´ 10-2 N
a 0.2
r= = = 0.14
2 2
\Net magnetic field at P Consider one such cylinder of radius r and
B = ( BA + BD )2 + ( BB + BC )2 thickness dr.
m0 4 2 I Current passing through this hollow
= × cylinder,
4p r
di = jdA = j ( 2pr dr ) 2p br 2dr
10-7 ´ 4 2 ´ 5
= = 20 ´ 10-6 T (a) Total current inside the portion of radius r1,
0.2 / 2 r1
I1 = ò di = 2pbò r 2dr
= 20 mT 0
r1
Clearly resultant magnetic field is é r3 ù
= 2 pb ê 1 ú
downward. êë 3 úû 0
2. At point A 2
= p br13
I2 B1 3
××
I1
m
By ampere’s circuital law,
q
b
ò B × dl = m 0i1
2 pr1
2
B1 ´ 2 pr1 = m 0 æç p br13 ö÷
B2
è3 ø
m 0 I1
B1 = × m 0 br12
4p r1 Þ B1 =
3
B2 = 0 (Magnetic field inside a current (b) Total current inside the cylinder
carrying hollow cylinder is zero) R
m I i = 2 pbò r 2dr
\ Ba = B1 + B2 = 0 × 1 0
4p r1 2
-7
p bR3
=
10 ´1 3
= - 10-4 T
1 ´ 10-3 m 2 i m 0 bR3
B2 = 0 =
4p r2 3 r2
= 100 mT (upward)
88
AIEEE Corner
Subjective Questions (Level-1)
1. Positive. By Flemings left hand rule. ® ^ ^ ^
7. Let B = Bx i + By j + Bz k
2. Fm = evB sin q
Fe ® ® ®
Þ v= (a) F = q( v ´ B )
eB sin q ^ ^
4.6 ´ 10-15 7.6 ´ 10-3 i - 5.2 ´ 10-3 k
=
1.6 ´ 10-19 ´ 3.5 ´ 10-3 ´ sin 60° ^ ^
= - 7.8 ´ 10-6 ´ 3.8 ´ 10 3( Bz i - Bx k )
= 9.46 ´ 106 m / s Þ Bx = - 0.175 T, Bz = - 0.256 T
3. Fm = qvB sin q (b) Cannot be determined by this information.
= (2 ´ 1.6 ´ 10-19 ) ´ 105 ´ 0.8 ´ 1 ® ® ®
(c) As F = q ( v ´ B )
= 2.56 ´ 10-14 N
® ® ®
® ® F ^B
4. (a) Fm = e ( v ´ B )
®®
^ ^ Hence, B × F = 0
= - 1.6 ´ 10-19 [(2.0 ´ 106 ) i + (3.0 ´ 106 ) j]
® ^
^ ^ 8. B = B i
´ (0.03 i + 0.15 j)
^ ® ^
= - (6.24 ´ 10-4 N) k (a) v = v j
® ® ® ^
® ® ® ^
(b) = Fm e ( v ´ B ) = - (6.24 ´ 10-4 N) k F = q ( v ´ B ) = - qvB k
® ® ® ® ^
5. Fm = e ( v ´ B ) (b) v = v j
^ ^ ^ ® ® ® ^
(6.4 ´ 10-19 ) k = - 1.6 ´ 10-19[(2 i + 4 j) F = q ( v ´ B ) = qvB j
^ ^ ® ^
´ ( Bx i + 3 Bx j)] (c) v = - v i
^
6.4 ´ 10-19 k = - 1.6 ´ 10-19[2Bx k ]
^ ® ® ®
F = q( v ´ B ) = 0
6.4 ´ 10-19 ®
Bx = = - 2.0 T ^ ^
(d) v = v cos 45° i - v cos 45° k
- 3.2 ´ 10-19
® ® ® qvB ^
6. (a) As magnetic force always acts F = q( v ´ B ) = - j
perpendicular to magnetic field, magnetic 2
field must be along x-axis. ® ^ ^
(e) v = v cos 45° j - v cos 45° k
F1 = qv1B sin q1 ® ® ® qvB ^ ^
F1 5 2 ´ 10-3 F = q( v ´ B ) = (- j - k)
Þ B= = 2
qv1B sin q1 1 ´ 10-6 ´ 106 ´ 1 qvB ^ ^
2 =- ( j + k)
2
Þ B = 10-3 T
mv 2m k 2 m eV
® ^ 9. r = = =
or B = (10-3 T ) i qB eB eB
(b) F2 = qv2 B sin q2 2mV
Þ B= r
= 1 ´ 10-6 ´ 106 ´ 10-3 ´ sin 90° e
= 10-3 N 2 ´ 9.1 ´ 10-31 ´ 2 ´ 103
= ´ 0.180
F2 = 1 mN 1.6 ´ 10-19
89
= 0.36 ´ 10-4 T 13. The component of velocity along the
B = 3.6 ´ 10-4 T magnetic field (i.e., vx ) will remain
mv qBr unchanged and the proton will move in a
10. (a) r = Þv= helical path.
qB m
z
1.6 ´ 10-19 ´ 2.5 ´ 6.96 ´ 10-3 ®
B
= -27
3.34 ´ 10
vy cos wt
= 8.33 ´ 105 ms -1
T pm
(b) t = =
2 qB vy sinwt vy
-27
3.14 ´ 3.34 ´ 10
=
1.6 ´ 10-19 ´ 2.5
= 2.62 ´ 10-8 s At any instant,
1 Components of velocity of particle along
(c) k = eV = mv2
2 Y-axis and Z-axis
mv2 v¢y = vy cos q = vy cos wt
Þ V=
2e and v¢z = - vz sin q = vz sin wt
3.34 ´ 10-27 ´ ( 8.33 ´ 105 )2 qB
= where, w=
2 ´ 1.6 ´ 10-19 m
= 7.26 ´ 103 V ® ^ ^ ^
\ v = vx i + vy cos w t j - vz sin w t k
= 7.26 kV
14. For the electron to hit the target, distance
11. (a) - q. As initially particle is neutral, charge
G S must be multiple of pitch, i.e.,
on two particles must be equal and opposite.
(b) The will collide after completing half GS = np
rotation, i.e., For minimum distance, n = 1
T pm 2p mv cos q
t= = Þ GS = p =
2 qB qB
× × × ×
× +q × –q ×
×B 2p 2 mk cos 60°
×
× × × ×
×
× Þ p= (mv = 2 mk)
×
×
×
×
×
×
×
×
×
×
qB
× × × × ×
×
×
×
×
×
×
×
×
×
× 2p 2 mk cos 60°
×
×
×
×
×
×
×
×
×
×
Þ B=
× × × × × qp
× × × × ×
× × × q × ×
× × × × × 1
2 ´ 3.14 ´ 2 ´ 9.1 ´ 10-31 ´ 2 ´ 1.6 ´ 10-16 ´
10.0 2
12. Here, r = = 5.0 cm, =
2 1.6 ´ 10-19 ´ 0.1
mv mv
(a) r = ÞB= Þ B = 4.73 ´ 10-4 T
qB qr
15. (a) From Question 5 (c)
9.1 ´ 10-31 ´ 1.41 ´ 106
= Introductory Exercise 23.2
1.6 ´ 10-19 ´ 5 ´ 10-2 L
= sin q Þ L = R sin q
= 1.6 ´ 10-4 T R
By Fleming’s left hand rule, direction of R
R sin 60° =
magnetic field must be inward. 2
T pm mv mv0
(b) t = = Þ L= =
2 qB 2qB 2qB0
3.14 ´ 9.1 ´ 10-31 (b) Now, L ¢ = 2.1 L = 1.05 R
=
1.6 ´ 10-19 ´ 1.6 ´ 10-4 As L ¢ > R,
= 1.1 ´ 10-7 s
90
2
Particle will describe a semicircle and move mv
qvB0 =
out of the magnetic field moving in opposite R
direction, i.e., qB0 R
v=
v¢ = - v = - v0 i
^ m
qB0 R cos q
T pm vx = v cos q =
and t= = m
2 qB0 qB0Z
= (Q R cos q = Z)
® ^ ® ^
m
16. v = (50 ms -1 ) i, B = (2.0 mT ) j 2qE0Z q2B02Z 2
Now, vz = v2 - vx2 = -
As particle move with uniform velocity, m m2
® ® ® ® ® ^ ® ^
F = q(E + v ´ B ) = 0 18. Given, E = E j , B = B k,
® ® ® ^
® ^ ^
Þ E = B ´ v = - ( 0.1 N/C)k v = v cos q j + v sin q k
As protons are moving undeflected,
17. If v be the speed of particle at point (0, y, z )
® ® ® ®
then by work-energy theorem, F = 0Þ e (E + v ´ B ) = 0
z ® ^
B = –B j ^ ^
Þ e ( E j - vB cos q j) = 0
E = EK. E
vy v or v=
B cos q
q vx Now, if electric field is switched off
Z 2p mv sin q 2p mE tan q
p= =
O
x qB qB2
(Component of velocity along magnetic field
= vz = v sin q)
19. F = I l B sin q
1 F 0.13
mv2
W = DK = I= =
2 lB sin q 0.2 ´ 0.067 ´ sin 90°
But work done by magnetic force is zero, = 9.7 A
hence, network done = work done by electric Fm
20. For no tension in springs
force × ×I× ×I× ×
Fm = mg ××× ×××
= qEZ ××× ×××
××× ×××
1 Þ I lB = mg ××× ×××
××× ×××
\ qE0Z = mv2 mg 13.0 ´ 10-3 ´ 10 mg
2 I= =
2qE0Z lB 62.0 ´ 10-2 ´ 0.440
Þ v=
m = 0.48 A
As the magnetic field is along Y-axis, By Fleming left hand rule, for magnetic force
particle will move in XZ-plane. to act in upward direction, current in the
The path of particle will be a cycloid. In this wire must be towards right.
case, instantaneous centre of curvature of 21. (a) FBD of metal bar is shown in figure, for
the particle will move along X-axis. metal to be in equilibrium,
As magnetic force provides centripetal force Fm + N = mg Fm
to the particle, Þ Fm = mg - N N
z
Þ I lB = m - N mg
vz V
Þ lB = mg - N
v R
q
vx R
Þ V= ( mg - N )
q R lB
X
91
For largest voltage, ® ® ^ ^
l 3 = cd = - (40 ´ 10-2 ) i + (40 ´ 10-2 m ) j
N =0
® ® ®
R mg 25 ´ 750 ´ 10-3 ´ 9.8 ^
F3 = I ( l 3 ´ B ) = - (0.04 N) k
V= =
lB 50.0 ´ 10-2 ´ 0.450 ® ® ^ ^
l 4 = da = (40 ´ 10-2 m ) i - (40 ´ 10-2 m ) k
= 817.5 V
® ® ® ^ ^
(b) If I lB > mg F4 = I ( l 4 ´ B ) = (0.04 N) i + (0.04 N) k
I lB - mg = ma ® ^
I lB - mg V lB 24. M = IA M
a= = -g ^ ^
m Rm = 0.20 ´ p( 8.0 ´ 10-2 )2(0.60 i - 0.80 j)
817.5 ´ 50 ´ 10-2 ´ 0.45 ^ ^
= - 9.8 = (40 . 2 ´ 10-4 ) (0.60 i - 0.80 j)A-m 2
2 ´ 750 ´ 10-3
® ^ ^
= 112.8 m/s 2 B = (0.25 T) i + (0.30 T) k
^ ® ® ®
22. I = 3.50 A, l = - (1.00 cm ) i (a) t = M ´ B
^
Þ l = - (1.00 ´ 10-2 m ) i ^ ^
= (40.2 ´ 10-4 )( - 0.24 i - 0.18 j + 0.2 k )
^
® ^ ^ ^ ^
(a) B = - (0.65 T) j = ( - 9.6 i - 7.2 j + 8.0 k ) ´ 10-4 N-m.
® ® ® ^ ®®
Fm = I ( l ´ B ) = - (0.023 N) k (b) U = - M× B = - (40.2 ´ 10-4 )(0.15) J
® ^ » - 6.0 ´ 10-4 J
(b) B = + (0.56 T) k
® ® ® 25. Consider the wire is bent in the form of a
^
Fm = I ( l ´ B ) = (0.0196 N) j loop of N turns,
® L
^ Radius of loop, r=
(c) B = - (0.33 T) i 2pN
® ® ® Magnetic dipole moment associated with the
Fm = I ( l ´ B ) = 0
loop
® ^ ®
(d) B = (0.33 T) i - (0.28 T ) k i L2
M = NiA = Ni ´ pr 2 =
4pN 2
® ® ® ^ 2
Fm = I ( l ´ B ) = - (0.0098 N) j iL B
t = MB sin 90° =
® ^ ^
4pN
(e) B = + (0.74 T) j - (0.36 T ) k
Clearly t is maximum, when N = 1
® ® ® ^ ^
Fm = I ( l ´ B ) = - (0.0259 N) k + (0.0126 N) j and the maximum torque is given by
^ ^ i L2B
= (0.0126 N) j - (0.0259 N) K tm =
4p
® ^
23. B = (0.020 T) j 26. Consider the disc to be made up of large
® ® number of elementary rings. Consider on
^
l1 = ab = - (40.0 cm) j such ring of radius x and thickness dx.
^ Charge on this ring,
= - (40.0 ´ 10-2 m ) j
® ® ®
F1 = I ( l1 ´ B ) = 0
x dx
® ® ^
l 2 = bc = (40.0 cm) k
^
= - (400 ´ 10-2 m ) k
q 2q
® ® ® ^ dq = ´ 2px dx = 2 x dx
F2 = I ( l 2 ´ B ) = (0.04 N) i p R2 R
92
Current associated with this ring, y
dq w dq w q g
di = = = x dx f
T 2p pR2 b c
Magnetic moment of this ring
wq
dM = px 2di = 2 x 3dx h x
R e
a d
Magnetic moment of entire disc, z
wq R 1
M = ò dM = 2 ò x 3dx = w qR2 …(i) ®
R 0 4 ^
Mabcd = - i l2 k
Magnetic field at the centre of disc due to the ® ^
elementary ring under consideration Mefgh = i l2 k
m di m 0w q2 ®
dB = 0 = dx ^
Madeh = i l2 j
2x 2pR2
Net magnetic field at the centre of the disc, \Total magnetic moment of the closed path,
m wq R m wq ® ® ® ®
B = ò dB = 0 2 ò dx = 0 ^
M = Mabcd + Mefgh + Madeh = i l2 j
2pR 0 2pR
M pR3 31. Circuit is same as in Q.30
\ =
B 2m 0 ® ^ ^
M = i l2 j = j
27. (a) By principle of conservation of energy, ® ^
Gain in KE = Loss in PE B =2 j
KE = - PE cos q + ME ® ® ®
t = M ´B =0
K 0.80 ´ 10-3
f cos q = 1 - =1 - m0 I
ME 0.02 ´ 52 ´ 10-3 32. B1 = ×
4p r
10 m0 I
= × B2 =
13 4p r
10
q = cos -1 = 76.7° Here, B1 and B2 are perpendicular to each
13 other, hence,
10
(b) q = cos -1 = 76.7° 2
13
Entire KE will again get converted into PE l
28. DU = U2 - U1 = - MB - ( + MB)
= - 2 MB
= - 2 ´ 1.45 ´ 0.835 = - 2.42 J 1
l
-11
2pr 2 ´ 3.14 ´ 5.3 ´ 10
29. (a) T = = B = B12 + B22
v 2.2 ´ 106
m 0 2I 10-7 ´ 2 ´ 5
= 1.5 ´ 10-16 s = × =
e 1.6 ´ 10
-19 4p r 35 ´ 10-2
(b) i = = = 1.1 ´ 10-3 A
T 1.5 ´ 10-16 = 2.0 ´ 10-6 T
= 1.1 mA = 2.0 mT
(c) M = p r 2i 33. Clearly DBOC ~ DAOB
r2 AD
= 3.14 ´ (5.3 ´ 10-11 )2 ´ 1.1 ´ 10-3 \ =
r6 BC
= 9.3 ´ 10-24 A-m 2
Þ r2 = 2r
30. Suppose equal and opposite currents are
flowing in sides a d and e h, so that three = 100 mm
complete current carrying loops are formed,
93
A q
a a
p
q
B q
2
3
q
l O 2a I
1
C r a
D r2
4
and AD = 2BC = 200 mm m0 I
r = × (inwards)
q = cos -1 = 45° 4p a 2
BC m0 I
2 B3 = B4 = (sin 0 + sin 0)
4p 2a
m0 I 2I m I
BBC = [sin 45° + sin 45° ] = 0× (outwards)
4p r r 4p 2a 2
m0
= (outwards) Net magnetic field at P
4p
m I B = B1 + B2 - ( B3 + B4 )
BAD = 0 × (sin 45° + sin 45° ) m I
4p r2 = 0× (inwards)
4p 2a
m0 2 I
= × (inwards) m I m I q
4p r2 36. B = 2 ´ 0 × - 0 ´ =0
4p R 2 R 2p
Net magnetic field at O. Þ q = 2 rad.
2 m 0I é 1 1 ù
B = BBC - BAD = ê - ú
4p êë r1 r2 úû l
R
é 1 1 ù q
= 2 ´ 10-7 ´ 2 ê -3
- -3 ú l
êë 50 ´ 10 100 ´ 10 úû
-6
= 2 ´ 10 T = 2 mT (outwards)
37. (a) Consider a point P in between the two
34. Let us consider a point P ( x, y) where
conductors at a distance x from conductor
magnetic field is zero. Clearly the point must
carrying current I1 (= 25.0 A),
lie either in 1st quadrant or in 3rd quadrant.
l2
P(xy) I2 = 75.0 A
x P
l1 I1 = 25.0 A
r
Magnetic field at P
m 0 2I1 m 0 2I2 m I m I
B= × - × =0 B = 0 × 1 - 0 × 2 =0
4p y 4p x 4p x 4p r - x
Þ I1x = I2 y I1 I
Þ = 2
æI ö x r-x
Þ y=ç 1÷x r - x I2
çI ÷
è 2ø Þ =
x I1
35. q = 45°
m0 I I1 25.0
B1 = B2 = × (sin q + sin q) Þ x= r= ´ 40 = 10 cm
4p a I1 + I2 100.0
94
-2
(b) Consider a point Q lying on the left of the 2 ´ 0.0580 ´ 2.40 ´ 10
=
conductor carrying current I1 at a distance x 4p ´ 10-7 ´ 800
from it.
Þ I = 2.77 A
(b) On the axis of coil,
x m 2 NIA
B= 0× 2
4 p ( r + x 2 )3 / 2
I2 = 75.0 A
3/ 2
BC ( r 2 + x 2 )3/ 2 æ r2 + x 2 ö
I1 = 25.0 A = 3
Þç ÷ =2
B r ç r2 ÷
è ø
m 0 I1 m 0 I2 Þ x = 0.0184 m
B= × - × =0
4p x 4p r + x 41. Let the current I2 ( = I ) upwards
I1 I I1 I2
Þ = 2
x r+x I3
I1 25.0
Þ x= r= ´ 40
I2 - I1 50.0 I4
= 20 cm
m 2 N pr 2I B = - B1 + B2 - B3 + B4
38. B = 0 × 2
4 p ( r + x 2 )3 / 2 m 2
= 0 × [ - I1 + I2 - I3 + I4 ] = 0
4p r
But, x = R
m NI 4 2 Br I2 = I1 + I3 - I4
B= 0 ÞN= = 10 + 8 - 20
4 2r m 0I
= -2 A
4 2 ´ 6.39 ´ 10-4 ´ 6 ´ 10-2
= Negative sign indicates that current I is
4p ´ 10-7 ´ 2.5
directed downwards.
Þ N = 69 ® m I^
42. B KLM = - 0 i
39. For magnetic field at the centre of loop to be 4R
zero, magnetic field due to straight
conductor at centre of loop must be outward,
hence I1 must be rightwards. L M
I
At the centre of the loop
I2 I
N
I
K
R
® m I^
D B KNM = 0 j
4R
® ® ® m I ^ ^
I1 B = B KLM + B KNM = 0 ( - i + j)
4R
B = B1 - B2 ® ® ® m Iqv ^
(a) F = q ( v ´ B ) = - 0 k
m 0 2I1 m 0I2 4R
= × - =0
4p D 2R ® ® ^
pD (b) l1 = l 2 = - 2 R k
I1 = I2 ® ® ®
R F1 = I ( l1 ´ B ) = 2 IRB i
^
® ® ® ^
m 0 NI 2 BR F2 = I ( l 2 ´ B ) = 2 IRB i
40. (a) B = ÞI=
2R m 0N ® ® ® ^
F = F1 + F2 = 4 IRB i
95
43. (a) Length of each side ®® -6
òbB × dl = - m 0I1 = - 5.0 ´ 10 T-m
l
®® -6
òc B × dl = m 0( I2 - I1 ) = 2.5 ´ 10 T-m
a ®® -6
òdB × dl = m 0( I2 + I3 - I1 ) = 5.0 ´ 10 T-m
q
46. P1 y
x
2 pr r
l= a
n
qq
p P2
q=
n a
l pr r
a = cot q = cot q
2 n
m i Current density
B = n ´ 0 × (2 sin q) I 2I
4p a J= =
2
m 2n 2 sin q a pa 2
= 0× pa 2 - 2p æç ö÷
4p p r cot q è2ø
p Let us consider both the cavities are
m 0 i n 2 sin 2 carrying equal and opposite currents with
= n
p current density J.
2p 2r cos
n Let B1, B2 and B3 be magnetic fields due to
p complete cylinder, upper and lower cavity
m 0i n 2 sin 2
n respectively.
(b) lim B = lim
n ®¥ n ®¥ p (a) At point P1
2p 2r cos
n
m 0i ® m 2I ^ m 2J ´ pa 2 ^
Þ lim = B1 = - 0 × 1 i = - 0 × i
n ®0 2r 4p r 4p r
®® m I^
44. ò B × dl = 3.83 ´ 10-7 T-m =- 0 i
pr
2
(a) By Ampere’s circuital law 2 J ´ p æaö
® m 2 I2 ^ m 0 ç ÷
®® B2 = 0 × i= × è 2 ø ^i
ò B × dl = m 0I 4p r - a 4p r-
a
1 ®® 1 2 2
ÞI= B × dl = ´ 3.83 ´ 10-7
m0 ò
m 0I ^
4p ´ 10-7 =- i
æ aö
4p ç r - ÷
= 0.3A è 2ø
(b) If we integrate around the curve in the ® m 0 2 I3 ^ m0 ^
opposite direction, the value of line integral B3 = × i= i
4p r + a æ aö
4p ç r + ÷
will become negative, i.e., 2 è 2ø
- 3.83 ´ 10-7 T-m. ® ® ® ®
®® B = B1 + B 2 + B 3
45. ò B × dl = m 0I
é ù
As the path is taken counter-clockwise ê 4m 0I 1 1 ú^
=ê - + + i
®® 4p a aú
direction, ò B × dl will be positive if current is ê r r- r+ ú
êë 2 2 úû
outwards and will be negative if current is 2
® m 0I é 2r - a ù ^ 2
inwards. B = ê úi
®® 4pr êë 4r 2 - a 2 úû
ò B × dl = 0
a
96
2 2
® m I é 2r - a ù x w
\ (B ) = 0 ê 2 2ú
, towards left.
4pr êë 4r - a ûú
(b) At point P2
y
®
B1 y z
x A
B3 sin q l
q Fig.1
q P2
q
B 2 sin ® ®®
® B3 B × dl = m 0l l
B2
B2 cos q òWXYZ
B3 cos q ®® ®® ®® ®®
Þ ò B × dl + ò B × dl + ò B × dl + ò B × dl = m 0l l
® m 2I ^ m I ^ W®X X ®Y Y ®Z Z ®W
B1 = 0 × 1 j = 0 j
4p r pr B l + 0 + B l + 0 = m 0l l
® m0 2 I2 1
B2 = ×
^ ^
[ - sin q i - cos q j ] B = m0 l
4p 2 2
a
r2 + In Fig. 2.
4
- m 0I ^ ^
= [sin q i + cos q j ]
2p 4r 2 + a 2 B2 B1
P
® m 2 I3 ^ ^
B3 = 0 × [sin q i - cos q j ]
4p a2
r2 + Q B2
B1
4
Fig.2
m 0I ^ ^
= [sin q i - cos q j ]
2p 4r 2 + a 2 At point P,
1
® ® ® ® B1 = B2 = m0 l
B = B1 + B 2 + B 3 2
é2 ù B = B1 - B2 = 0,
m 0I
ê - 2 cos q ú ^j
= At point Q,
2p êr 2
4r + a úû 2
1
ë B1 = B2 =
m0 l
r 2r 2
but, cos q = =
2
a 4r + a 2
2 B = B1 + B2 = m 0 l
r2 +
4 ® ® ® ®
® m I dl ´ r m 0 q ( v ´ r )
® m 0I é 2 4r ù^ 48. B = 0 × = ×
\ B = ê - 2 j 4p r3 4p r3
2ú
2p êë r 4r + a úû y
m I é 2r 2 + a 2 ù ^ u
= 0 ê 2 új
4pr êë 4r + a 2 úû o x
® m I é 2r 2 + a 2 ù
(B ) = 0 ê 2 ú , upwards.
4pr êë 4r + a 2 úû
5 cm
19. Magnetic field at O due to P,
I
6 cm
P Q
q q
r P
3 cm
I I
5 cm
R/2 R/2
2 2 O
r = 5 - 3 = 4 cm
= 4 ´ 10-2 m R
3
sin q = m 0 2I m I
5 B1 = × = 0 (inwards)
3 4p R / 2 p R
-7
10 ´ 50 ´ 2 ´
\ BP = 5 Magnetic field at O due to Q,
4 ´ 10-2 m 2I m I
B2 = 0 × = 0 (inwards)
4p R / 2 p R
= 1.5 ´ 10-4 T
= 1.5 gauss. Net magnetic field at O,
2 m 0I
16. Magnetic field on the axis of current B = B1 + B2 =
carrying circular loop, pR
m 2M 20. As solved in Question 16,
B1 = 0 × 2 …(i)
4 p ( r + x 2 )3 / 2 3/ 2
B2 æ x 2 + R2 ö
=ç ÷
Magnetic field at the centre of current B1 çè R2 ÷ø
carrying circular loop, 3/ 2
m 2M æ x 2 + R2 ö
B2 = 0 × 3 …(ii) Þ ç ÷ =8
4p r ç R2 ÷
è ø
From Eqs. (i) and (ii), x 2 + R2
B2 ( r 2 + x 2 )3/ 2 Þ =4
= R2
B1 r3 Þ x= 3 R
(3 2 + 42 )3/ 2 21. Component of velocity of particle along
=
33 magnetic field, i.e.,
125 qE
= vy = t = aE t
27 m
125
Þ B2 = ´ 54 = 150 mT is not constant, hence pitch is variable.
27
mv 2 mK
® ® ® ® ® 22. r = =
17. F = I( l ´ B ) = I ( ba ´ B ) qB qB
® ® ® ® 2 mK
= - I ( ab ´ B ) = I ( B ´ ab) Now, R =
eB
18. Kinetic energy of electron,
2m (2K ) 2
1 R¢ = = R
K = mv2 = e V e (3 R) 3
2
100
23. Same as question 1. Introductory exercise y
23.6.
Note. Her diagram is wrong correct diagram 1 2
should be
4
A x C
I
O x
a 3
B x D
b
mv 2 mK 2 mqV ® - 1 æ m 0I ö ^ m I ^
24. r = = = [K = qV ] B2 = ç ÷ k = - 0 k,
qB qB qB 4 è 2b ø 8b
2 mV æ 1 ö ® 1 æm I ö ^ m I ^
Þ r= ç ÷ B3 = ç 0 ÷ k = 0 k
q è Bø 4 è 2a ø 8a
25. Magnetic field due to a conductor of finite ® ® ® ® ®
B = B1 + B 2 + B 3 + B 4
length.
m I m 0I é 1 1 ù ^
B = 0 × (sin a + sin b ) = - k
4p r 8 êë a b úû
Here, a = - q2, b = q1 and r = a 29. Current associated with electron,
m I q
\ B = 0 (sin q1 - sin q2 ) I= = ef
2a T
m I m ef
26. In case C, magnetic field of conductor 1-2 B= 0 = 0
and 2-3 at O is inward while those of 3-4 and 2R 2R
4-1 at O is outward, hence net magnetic field 30. Same as question 1(a). Introductory Exercise
at O in this case is zero. 23.5.
2 3 31. At point 1,
Magnetic field due to inner conductor is
I non-zero, but due to outer conductor is zero.
O
Hence, B1 ¹ 0
1 4 At point 2,
® ® ® Magnetic field due to both the conductors is
27. dF = I ( dl ´ B ) equal and opposite.
® ® Hence, B2 = 0
But B ||dl at every point,
32. Apply Fleming’s left hand rule or right hand
®
hence, dF = 0. thumb rule.
33. Magnetic field due to straight conductors at
28. B1 = B3 = 0 (Magnetic field on the axis of
O is zero because O lies on axis of both the
current carrying straight conductor is zero)
conductors.
f m 0I m 0If
Hence, B = × =
2p 2x 4px
34. Inside a solid cylinder having uniform
current density,
101
m 0Ir
B=
2 p R2
Here, r = R - x
m 0I ( R - x )
\ B=
2 p R2
r
35. Magnetic force is acting radially outward on
the loop.
JEE Corner
Assertion and Reason
1. For parabolic path, acceleration must be mv 2 meV
r= =
constant and should not be parallel or qB eB
antiparallel to velocity.
7. For equilibrium
2. By Fleming’s left hand rule.
® ®
3. Magnetic force on upper wire must be in Fe + Fm = 0
upward direction, hence current should be in ® ® ®
a direction opposite to that of wire 1. Þ q E = - q( v ´ B )
Reason is also correct but does not explain ® ® ® ® ®
Þ E =- v ´B = B ´ v
Assertion.
® ®
4. t = MB sin a 8. Pm = Fm × v
a = 90° ® ®
\t = MB ¹ 0 As Fm is always perpendicular to v ,
5. F2 = I lBO x1 ®
Pm = 0
y
® ®®
B Again, Pe = Fe× v , may or may not be zero.
1
2 4
9. Reason correctly explains Assertion.
10. Magnetic force cannot change speed of
F2 F4
particle as it is always perpendicular to the
speed of the particle.
3 v2
11. a =
R
x1 x2
x but R also depends on v.
F qvB
\ a= m =
F4 = I lB0x2 m m
\ F4 > F2 Þ a µv
Hence, net force is along X-axis.
6. Radii of both is different because mass of
both is different
102
T
q
R
y
O
mg
IAB0 = mgR
mgR mgR x
I= =
AB0 pR2B0
mg q
= dq = × (2p yx dx )
p RB0 4
pR3
3
2. As it is clear from diagram,
= 3 q cos 2 q sin q dq
l Current associated with this cylinder,
dq w dq 3 w q
di = = = cos 2 q sin q dq
T 2p 2p
I
Magnetic moment associated with this
cylinder,
(– 4,0) (2,0) 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
Effective length of wire, 3 0
M = ò dM = R2qò cos 2q sin 3 q dq
® ^ 2 p2
l = (4 m ) i
3 0
® ® ® = R2w qò cos 2q (1 - cos 2 q)sin q dq
F = I( l ´ B) 2 p/2
0
® 3 2 é cos 3 q cos 5 q ù
® F I ® ® = R wq ê - ú
a = = ( l ´ B) 2 êë 3 5 úû p/ 2
m m 1 2
2 ^ ^ ^ = R wq
= (4 i ´ ( - 0.02 k )) = 1.6 j m /s 2 5
0.1
5. As solved in question 5(c). Introductory
3. Impulse = Change in momentum
Exercise 23.2.
ò I lB dt = mv - 0 L
= sin q
lBò dq = mv R
mv m 2 gh mV
dq = = Here, L = d, R =
lB lB qB
qB d
4. Consider the sphere to be made up of large \ = sin q
number of hollow, coaxial cylinder of mV
q V sin q
different height and radius. Consider one or =
such cylinder of radius x, height y and m Bd
thickness. 6. Force on portion AC will more compared to
Now, y = 2 R cos q, x = R sin q, dx = R cos q dq that on portion CB.
103
mv
7. Consider an elementary portion of the wire i.e., b-a£
carrying current I1 of length dx at a distance qB
x from end B. qB ( b - a )
or v³
m
11. Consider an elementary portion of length dx
at a distance x from the pivoted end.
I1
B I2
dx dxe
a
2a
x
x
C a
B
b m 0I1
B1 =
2 R1
m I
x 2 - b2 B2 = 0 2
I¢ = I - I 2 R2
c 2 - b2
c2 - x 2 \ B = B12 + B22
= I
c 2 - b2 m0 æ I1 ö
2
æI ö
2
2 2 = ç ÷ +ç 2÷
m 0I ¢ m 0I ( c - x ) 2 çR ÷ çR ÷
I= = è 1ø è 2ø
2px 2px ( c 2 - b2 )
105
2 ® m 2I ^
æ ö 2 B1 = 0 × j
4p ´ 10 -7 ç ÷ æ 5 2 ö 4p a
ç 5 ÷
= ´ + ç ÷
2 ç 5 ´ 10-2 ÷ ç 5 ´ 10-2 ÷ ® m 2I ^
ç 2 ÷ è ø B2 = - 0 × i
è ø 4p a
4p ´ 10-7 ´ 2 ®
= = 4p ´ 10-5 T B3 = 0
2 ´ 10-2
® ® ® ® m 0i ^ ^
B = B1 + B 2 + B 3 = ( j - i)
23. Initially, net force on the particle is zero. 2pa
Hence,
26. Effective length, l = AC = 42 + 3 2
E
V=
B =5 m
Now, if electric field is switched off. C
mv E æ q = Sö
r= = ç ÷
qB SB2 èm ø
24. For equilibrium,
mg
f = [f = magnetic force per unit length on A B
l
the conductors]
m 0 2I1I2
Þ × =lg F = I lB = 2 ´ 5 ´ 2 = 20 N
4p r
27. At point P,
m 2I I
Þ r= 0× 1 2 1 qx
4p l g E= ×
4pe0 ( R2 + x 2 )3/ 2
10-7 ´ 2 ´ 100 ´ 50 m 2 iA
= B= 0× 2
0.01 ´ 10 4 p ( R + x 2 )3 / 2
= 0.01 m q qv
Hence, i = =
Clearly, equilibrium of conductor B is T 2 pR
unstable.
and A = p R2
® ® ®
25. If B1, B 2 and B 3 be magnetic fields at the E 1 1 c2 é 1 ù
\ = × = êc = ú
given point due to the wires along x, y and z B m 0e0 v v êë m 0e0 úû
axis respectively, then
AIEEE Corner
Subjective Questions (Level 1)
f2 - f1 B ( A2 - A1 )
1. < e > = - =-
t t Charge flowing through the coil
A1 = pr 2 = 3.14 ´ (0.1)2 q=<i>t
= 3.14 ´ 10-2 = 0.0314 ( f2 - f1 ) ( - 0.04 - 0.04)
2 Þq=- =-
æ 2pr ö R 50
A2 = a 2 = ç ÷
è 4 ø 0.08
= = 1.6 ´ 10-3 C
2 50
æ 2 ´ 3.14 ´ 0.1 ö
=ç ÷ = 0.025 = 1.6 mC = 1600 mC
è 4 ø
100 (0.025 - 0.0314) 3. f1 = NBS, f2 = - NBS
\ <e>=-
0.1 Induced emf,
( f - f ) 2 NBS
= 6.4 V <e>=- 2 1 =
t t
2. f1 = NBA = 500 ´ 0.2 ´ 4 ´ 10-4
Induced current
= 0.04 Wb < e > 2 NBS
f2 = - NBA = - 0.04 Wb <i>= =
R Rt
Average induced emf, Charge flowing through the coil,
(f - f ) 2 NBS
<e>=- 2 1 q=<i>t=
t R
Average induced current, qR 4.5 ´ 10-6 ´ 40
<e> (f - f ) Þ B= =
<i>= =- 2 1 2 NS 2 ´ 60 ´ 3 ´ 10-6
R Rt
113
= 0.5 T Hence, equal force in direction of motion of
coil is required to move the block with
® uniform speed.
^ ^
4. B = (4.0 i - 1.8 k ) ´ 10-3 T,
(b) When the coil is entering into the
® ^ magnetic field, magnetic flux linked with
S = (5.0 ´ 10-4 k ) m 2
the coil increases and the induced
®® current will produce magnetic flux in
f = B × S = - 9.0 ´ 10-7 Wb
opposite direction and will be
5. e = Blv = 1.1 ´ 0.8 ´ 5 = 4.4 V counter-clockwise and vice-versa.
By Fleming’s right hand rule, north end of i
BLv
the wire will be positive. i0 =
R
6. A = pr 2 = 3.14 ´ (12 ´ 10-2 )2 = 0.045 m 2
i0
(a) For t = 0 to t = 2.0 s
dB 0.5 - 0 x
= slope = = 0.25 T/s
dt 2.0 - 1 –i0
df dB
e=- m =-A
dt dt
= - 0.045 ´ 0.25 = - 0.011 V 8. Consider an elementary section of length dl
|e| = 0.011 V of the frame as shown in figure. Magnetic
flux linked with this section,
(b) For, t = 2.0 s to t = 4.0 s
dB
= slope = 0 e = 0 l
dt dl
i
(c) For, t = 4.0 s to t = 6.0 s
dB 0 - 0.5
= slope = = - 0.25
dt 6.0 - 4.0 x a
df dB m 0 2i
e=- m =-A = 0.11 V dfm = BdB = × adl
dt dt 4p x + l
7. (a) When magnetic flux linked with the coil Total magnetic flux linked with the frame,
changes, induced current is produced in it, in m ai a dl
such a way that, it opposes the change. fm = ò dfm = 0 ò
2p 0 x + l
Magnetic flux linked with the coil will m ai
change only when coil is entering in (from = 0 [ln ( x + a ) - ln x ]
3L L L 2p
x=- to x = - ) or moving (from x =
2 2 2 Induced emf
3L - dfm m ai é 1 1 ù dx
to x = ) of the magnetic field. e= = - 0 ×ê - ú
2 dt 2p ë x + a x û dt
Because, of induced current, an opposing
force act on the coil, which is given by m 0a 2i m 0a 2iv
= v=
BLv B2L2v 2px ( x + a ) 2px ( x + a )
F = ilB = BL =
R R 9. As solved in Qusetion 4. Introductory
Exercise 24.3.
F
B2L2v i
F0 =
R d
v
l
F0
x
–3L –L O L 3L
2 2 2 2
114
m iv æ lö Effective emf
e = 0 ln ç1 + ÷
2p è dø E r - E1r2
E= 21
Here, i = 10 A r1 + r2
v = 10 ms -1 0.008 ´ 15.0 - 0.004 ´ 10.0
=
l = 10.0 cm - 1.0 cm = 9.0 cm 15.0 + 10.0
d = 1.0 cm = 0.0032 V
4p ´ 10-7 ´ 10 ´ 10 æ 9.0 ö rr 15 ´ 10
e= ln ç1 + ÷ r= 12 = =6W
2p è 1.0 ø r1 + r2 25
E 0.0032
e = (2 ´ 10V ) ln (10) V i= = = 0.003 A = 0.3 mA
R+ r 5+6
10. Induced current
e Blv di
i=
= 12. (a) e = - L = - 0.54 ´ ( - 0.030)
R R dt
Force needed to move the rod with constant = 1.62 ´ 10-2 V
speed = Magnetic force acting on the rod (b) Current flowing from b to a is decreasing,
Blv hence, a must be at higher potential.
ie., F = i lB = lB
R 13. (a) i = 5 + 16t, |e| = 10mV = 10 ´ 10-3 V
2 -2 2
B2l2v (0.15) ´ (50 ´ 10 ) ´ 2 di d
= = |e| = L Þ 10 ´ 10-3 = L (5 + 16t )
R 3 dt dt
-3
Þ F = 0.00375 10 ´ 10
L= = 0.625 mH
11. Suppose the magnetic field is acting into the 16
plane of paper. (b) at t = 1 s
Rods 1 and 2 can be treated as cells of emf i = 5 + 16 (1) = 21 A
E1 ( = Blv1 ) and E2 ( = Blv2 ) respectively. Energy stored in the inductor,
2 1 1 1
U = Li2 = ´ 0.625 ´ 10-3 ´ (21)2
2 2
B
Ä = 0.138 J
dU di
P= = Li = 0.625 ´ 10-3 ´ 21 ´ 16
v2 R v1 dt dt
= 0.21 W
14. From t = 0 to t = 2.0 ms
V -0 5.0 - 0
= =0
r2 r1 t - 0 2.0 ´ 10-3
ß Þ V = 2500 t
di
L = 2500 t
dt
ß
R 2500
E2
R
E1 E ò di = L ò tdt
r1 i i 2500 t
r2 r Þ ò 0 di = L ò 0 tdt
1250 2
i= t
L
Now, E1 = Blv1 = 0.010 ´ 10.0 ´ 10-2 ´ 4.00 at t = 2.0 ms
= 0.004 V 1250
i= ´ (2.0 ´ 10-3 )2
E2 = Blv2 = 0.010 ´ 10 ´ 0 ´ 10-2 ´ 8.00 150 ´ 10-3
= 0.008 V = 3.33 ´ 10-2 A
115
From t = 2.0 ms to t = 4.0 ms di
18. (a)|e| = M = 3.25 ´ 10-4 ´ 830
V - 5.0 0 - 0.50 dt
=
t - 2.0 ´ 10-3 (4.0 - 2.0) ´ 10-3 = 0.27 V
V = - 2500 ( t - 2.0 ´ 10-3 ) + 5.0 di
As, is constant, induced emf is
= - 2500 t + 10.0 dt
di constant.
L = - 2500 t + 10.0
dt (b) Coefficient of mutual induction remains
1 same whether current flows in first coil
di = ( - 2500 t + 10.0) dt
L or second.
1 di
i = [ - 1250 t 2 + 10.0 t ] Hence, | e| = M1 = 0.27 V
L dt
at t = 4 s 19. (a) Magnetic flux linked with the secondary
1
i= [ - 1250 ´ (4.0 ´ 10-3 )2 coil,
150 ´ 10-3
f2 = Mi1
+ 10.0 (4.0 ´ 10-3 )] f2 0.0320 ´ 400
= 3.33 ´ 10 A -2 M= =
i1 6.52
di |e| 0.0160
15. (a)|e| = L Þ L = = = 1.96 H
dt di / dt 0.0640
(b) f1 = Mi2 = 1.96 ´ 2.54 = 4.9784 Wb
= 0.250 H
Flux per turn through primary coil
(b) Flux per turn f 4.9784
Li 0.250 ´ 0.720 = 1 =
f= = N1 700
N 400
= 7.112 ´ 10-3 Wb/turn.
= 4.5 ´ 10-4 Wb
di i -i 20. Same as Question 2. Introductory Exercise
16. |e| = M =M 2 1 24.4
dt t
-3 12 - 4 21. i = i0(1 - e - t / t )
Þ 50 ´ 10 = M × Rt
0.5 E -
= (1 - e L )
50 ´ 10-3 ´ 0 . 5 R
M= = 3.125 ´ 10-3 H Rt
8 di E - L
= e
= 3.125 mH dt L
If current changes from 3 A to 9 A in 0.02 s. Power supplied by battery,
di i -i E2
Rt
|e| = M =M 2 1 P = Ei = (1 - e L )
-
dt t R
9 -3
= 3.125 ´ 10-3 ´ Rate of storage of magnetic energy
0.02 Rt Rt
di E2 - -
= 0.9375 V P1 = Li = (1 - e L ) e L
dt R
17. (a) Magnetic flux linked with secondary coil, Rt 10 ´ 0.1
P1 - -
fm2 = M i1 =e L =e 1 = e -1 = 0.37
P
f 6.0 ´ 10-3 ´ 1000 L 2
M= 2 = =2H 22. (a) t = = = 0.2 s
i1 3 R 10
dfm2 di L R
(b) e=- =-M 1
dt dt
0 -3
= -2 ´ = 30 V K
0.2
fm 600 ´ 5 ´ 10-3
(c) L = 1 = =1H E
i1 3
116
E 100 i
(b) i0 = = = 10 A But i = 0
R 10 2
t Rt
- i0 æ - ö
(c) i = i0(1 - e t ) = i0 ç1 - e L ÷
2 ç ÷
-
1 è ø
i = 10 (1 - e 0.2 )
-
Rt
1
-5 e L =
= 10 (1 - e ) = 9.93 A 2
23. (a) Power delivered by the battery, L 1.25 ´ 10-3
Rt
t = ln 2 = ´ 0.693
E2 - R 50.0
P = Ei = (1 - e L )
R = 17.3 ´ 10-6 = 17.3 ms
(3.24)2
12.8 ´ 0.278 1 1 æ1 ö
= (1 - e 3.56 ) (b) U = Li2 = ç L i02 ÷
12.8 2 2 è2 ø
1
= 0.82 (1 - e -1 ) = 0.518 W i= i0
2
= 518 mW Rt ö
æ - i
(b) Rate of dissipation of energy as heat i0 ç1 - e L ÷ = 0
Rt ç ÷ 2
E2 - è ø
P2 = i2 R = (1 - e L )2 Rt
R - 2 -1
Þ e L =
= 0.82 (1 - e -1 )2 = 0.328 W 2
= 328 mW L 2
Þ t = ln
(c) Rate of storage of magnetic energy R 2 -1
P1 = P - P2 = 190 mW = 30.7 ms .
di
24. E = VL + VR = L + iR 26. Steady state current through the inductor
dt L, r
L R
i0
VL VR
K R
(a) Initially, i = 0 S
E
di E 6.00
\ = = = 2.40 A/s E
dt L 2.50 i0 =
r
(b) When, i = 0.500 A
di E - iR 6.00 - 0.500 ´ 8.00 When the switch S is open
= = L
dt L 2.50 t=
R+ r
= 0.80 A/s
E æç -
Rt ö (a) i = i0 e -t / t
(c) i = 1-e L ÷ æ (R + r) ö
R çè ÷
ø E - ççè L ÷÷ø
Þ e i=
8.00 ´ 0.250 ö r
6.00 æç -
÷
= 1-e 2.5
(b) Amount of heat generated in the solenoid
8.00 ç ÷ ¥ ¥
è ø H = ò i2r dt = i02r ò e -2t / t dt
-0.8 0 0
= 0.750 (1 - e ) = 0.413 A
E 6.00 E 2 ì t -2 t / t ¥ ü
(d) i0 = = = 0.750 A = í- [ e ]0 ý
R 8.00 r î 2 þ
æ -
Rt ö ( R + r ) E2
=
25. (a) i = i0 ç1 - e L ÷ 2rL
ç ÷
è ø
117
When the switch S is open, current i2 flows
in the circuit in clockwise direction and is
given by
27. At any instant of time,
5 mH i2 = i0e -t / t
E
i1 Here, i2 =
R2
10 mH L
t=
R1 + R2
i æ R1 + R 2 ö
E - ççè ÷÷ t
20 V i2 = e L ø
R2
5W
12 -10t
di1 di = e = (6 e -10t ) A
L1 = L2 2 2
dt dt
29. For current through galvanometer to be
Þ L1i1 = L2i2 zero,
i1 = 2i2 …(i) L1,R1 R3
P i1
In steady state,
i1 G
inductors offer zero resistance, hence
20 L2,R2 R4
i= =4A
5 i2 Q i2
But i1 + i2 = i
4 8
i2 = A, i1 = A
3 3 K
E
28. When the switch is closed,
i i2
VP = VQ
di1 di
L1 + i1 R1 = L2 2 + i2 R2 …(i)
dt dt
i1 L
Also, i1 R3 = i2 R4 …(ii)
E From Eqs.(i) and (ii),
R1
di di
L1 1 + i1 R1 L2 2 + i2 R2
R2 dt dt
S = …(iii)
i1 R3 i2 R4
In the steady state,
E di1 di2
(1 - e - R2 t / L )
i2 = = =0
R2 dt dt
di2 E - R2 t / L R1 R2 R R
= e \ = Þ 1 = 3
dt L R3 R4 R2 R4
Potential difference across L Again as current through galvanometer is
di always zero.
V + L 2 = E e - R2 t / L = (12e -5t ) V
dt i1
= constant
i2
i2
di1 / dt
or = constant
di2 / dt
i1 L
di1
E
R1 or dt = i1 …(iv)
di2 i2
R2 dt
S i2 From Eqs. (iii) and (iv),
118
L1 R3 R1
= =
L2 R4 R2 q0 5.00 ´ 10-6
32. (a) V0 = =
30. (a) In LC circuit C 4 ´ 10-4
Maximum electrical energy = Maximum = 1.25 ´ 10-2 V= 12.5 mV
magnetic energy
1 1 (b) Maximum magnetic energy = Maximum
Þ CV02 = Li02 electric energy
2 2
2 2 1 2 q02
æV ö æ 1.50 ö Li0 =
L = C çç 0 ÷÷ = 4 ´ 10-6 ç ÷ 2 2C
i ç 50 ´ 10-3 ÷
è 0ø è ø q
Þ i0 = 0
= 3.6 ´ 10-3 H LC
Þ L = 3.6 mH 5.00 ´ 10-6
Þ i0 = = 8.33 ´ 10-4 A
1 0.090 ´ 4 ´ 10-4
(b) f =
2p LC
1 (c) Maximum energy stored in inductor,
= 1
2 ´ 3.14 3.6 ´ 10 -3
´ 4 ´ 10 -6 = L i02
2
= 1.33 ´ 103 Hz 1
= ´ 0.0900 ´ ( 8.33 ´ 10-4 )2
= 1.33 kHz 2
(c) Time taken to rise from zero to maximum = 3.125 ´ 10-8 J
value, (d) By conservation of energy,
t=
T
=
1
=
1 q2 1 2 1 2
+ Li = Li0
4 4f 4 ´ 1.33 ´ 103 2C 2 2
i0
= 3 ´ 10-3 s = 3 ms. But i =
2
31. (a) w = 2pf = 2 ´ 3.14 ´ 103
q2 3 2
= 6.28 rad/s = Li0
2C 8
1 1 -3
T = = 3 = 10 s = 1 ms i 3
f 10 q = 0 3 LC = q0
2 2
(b) As initially charge is maximum, (i.e.., it is 1.732
extreme position for charge). = ´ 5.00 ´ 10-6
2
q = q0 cos w t = 4.33 ´ 10-6 C
q0 = CV0 = 1 ´ 10-6 ´ 100 1 1 æ1 ö
Um = Li2 = ç Li02 ÷
= 10-4 2 4 è2 ø
\ q = [10 cos (6.28 ´ 103 ) t ] C.
-4
= 7.8 ´ 10-9 J
1 1 1
(c) w = 33. (a) w = =
LC LC 2.0 ´ 10 ´ 5.0 ´ 10-6
-3
1 1
ÞL= 2 =
w C (6.28 ´ 103 )2 ´ 10-6 = 104 rad/s
½ di ½
½ ½ = w2Q
= 2.53 ´ 10-3
½ dt ½
Þ L = 2.53 mH
= (104 )2 ´ 100 ´ 10-6 = 104 A/s
(d) In one quarter cycle, entire charge of the
(b) i = w Q02 - Q2
capacitor flows out.
q 4CV = 104 (200 ´ 10-6 )2 - (200 ´ 10-6 )2 = 0
<i>= =
t T
-6 (c) i0 = wQ0 = 104 ´ 200 ´ 10-6 = 2 A
4 ´ 10 ´ 100
= = 0.4 A (d) i = w Q02 - Q 2
10-3
119
At a distance x from centre of the region,
i0
Þ = w Q02 - Q 2 Magnetic flux linked with the imaginary
2 loop of radius x
w Q0
Þ = w Q02 - Q 2 fm = p x 2B
2 - dfm dB
3 1.73 ´ 200 ´ 10-6 e= = - p x2
Q= Q0 = dt dt
2 2 Induced electric field,
= 173 mC e 1 dB
E= = x
34. As initially charge is maximum 2p x 2 dt
q = q0 cos w t At a,
and |i| = i0 sin w t 1 dB
E= r , towards left.
1 1 4 dt
where, w = =
LC 3.3 ´ 840 ´ 10-6 At b ,
1 dB
» 19 rad./s E= r , upwards.
2 dt
i0 = wq0 = 19 ´ 105 ´ 10-6
At c,
» 2.0 ´ 10-3 A = 2.0 mA
E =0
At t = 2.00 ms
36. Inside the solenoid,
q2 q2
(a) Ue = = 0 (cos 2 wt ) B = m 0ni
2C 2C
dB di
(105 ´ 10-6 )2 = m 0n
= [cos 2 (38 rad)] dt dt
2 ´ 840 ´ 10-6
Inside the region of varying magnetic field
Ue = 6.55 ´ 10-6 J = 6.55 mJ 1 dB 1 di
E= r = m 0nr
1 1 2 dt 2 dt
(b) Um = Li2 = Li02 (sin w t )
2 2 (a) r = 0.5 cm = 5.0 ´ 10-3 m
1 1 di
= ´ 3.3 ´ (2 ´ 10-3 )2 sin 2(38 rad ) E = m 0rn
2 2 dt
= 0.009 ´ 10-6 J = 0.009 mj 1
= ´ 4p ´ 10-7 ´ 5.0 ´ 10-3 ´ 900 ´ 60
q2 1 2
(c) U = 0 = Li02
2C 2 = 1.7 ´ 10-4 V/m
= 6.56 ´ 10-6 J = 6.56 mJ (b) r = 1.0 cm = 1.0 ´ 10-2 m
35. As the inward magnetic field is increasing, 1 di
E= rn
induced electric field will be anticlockwise. m0 dt
× × × × × × × × × × × ×
× × × × × × × × × × × × 1
× × × × × × × × × × × × = ´ 4p ´ 10-3 ´ 5.0 ´ 10-3 ´ 900 ´ 60
× × × × × × ×E × × × × × 2
× × × × × × × × × × × × = 3.4 ´ 10-4 V/m
× × × × × × × × × × × ×
× × × × × × × × × × × ×
× × × × × × × × × × × ×
× × × × × × ×x× × × × ×
× × × × × × × × × × × ×
× × × × × × × × × × × ×
× × × × × × × × × × × ×
× × × × × × × × × × × ×
× × × × × × × × × × × ×
× × × × × × × × × × × ×
120
AIEEE Corner
Objective Questions (Level 1)
di 1 2 1 L 2
1. V = L 8. Li0 = CV02 Þ i0 =2 ´
dt 2 2 C 4 ´ 10-6
[ V ][ T ] [ ML2 T -3 A -1 ] [T ]
[L] = = = 2 ´ 103 V
[ i] [A ]
1
9. e = B l2w, is independent of t.
= [ ML2 T -2 A -2 ] 2
2. M µ n1n2 df Df
10. |e| = =
3. Both will tend to oppose the magnetic flux dt t
changing with them by increasing current in Df =|e|t = iRt
opposite direction. = 10 ´ 10-3 ´ 0.5 ´ 5
4. Moving charged particle will produced = 25 ´ 10-3 Wb
magnetic field parallel to ring, Hence
= 25 mWb.
fm = 0
11. As inward magnetic field is increasing,
Velocity of particle increases continuously induced electric field must be anti-clockwise.
due to gravity.
Hence, direction of induced electric field at P
5. Induced electric field can exist at a point will be towards and electron will experience
where magnetic field is not present, i.e., force towards right (opposite to electric
outside the region occupying the magnetic field).
field. 12. f = at ( t - t ) = att - at 2
6. At, t = 1 s df
|e| = = at - 2at
2W 4V 2H 2F dt
a b |e| at - 2at
i – + i= =
a R R
2
t ( at - 2 at )
q = 4t 2 = 4 C t 2
H = ò i R dt = ò dt
dq 0 0 R
i= = 8t = 8 A
dt 3 t
1 é ( at - 2 at ) ù
di = ê ú
= 8 A/s R êë 3 ´ ( - 2a ) úû
dt 0
1
di d 2q = [ - a t - a 3t 3 ]
3 3
As, = = Positive - 6 Ra
dt dt
Charge in capacitor is increasing, current i a 2t 3
=
must be towards left. 3R
di q
Vab = - 2I + 4 - L -
dt C di
4 13. E = - L
= - 2 ´ 8 + 4 - 2 ´ 8 _ = - 30 V dt
2 di
di d 14. VBA = - L + 15 - iR
7. |e| = M =M ( i0 sin w t ) dt
dt dt = - 5 ´ 10-3( - 103 ) + 15 - 5 ´ 1
= w Mi0 cos w t = 15 V
Maximum induced emf = w Mi0 di
15. = 10 A/s, at t = 0, i = 5A
= 100p ´ 0.005 ´ 10 dt
= 5p di
= 10 A/s
dt
121
di E 12
VA - VB = iR + L - E =0 25. i0 = = = 40 A
dt R 0.3
= 5 ´ 3 + 1 ´ 10 - 10 = 15 V 1 1
U0 = Li02 = ´ 50 ´ 10-3 ´ (40)2
æ di ö æ d 2q ö q 2 2
16. ç ÷ = çç ÷ = w2q0 = 0
è dt ø max dt ÷ LC = 40 J
è ø max t Rt ö
di
æ - ö E æç -
17. V = L 26. i = i0 ç1 - e t ÷ = 1-e L ÷
ç ÷ Rç ÷
dt è ø è ø
Rt
18. fm = BA cos q di E - L
df dq = e
Þ e = - m = BA sin q dt L
dt dt di -
Rt
dq VL = L = Ee L
Þ iR = BA sin q dt
dt
dq dq at t = 0
Þ R = BA sin q
dt dt VL = E = 20 V
BA at t = 20 ms
Þ dq = sin q dq
R R ´ 20 ´ 10 -3 R
- -
BA 3p/ 2 VL = Ee L Þ 5 = 20e 50L
q= sin q dq = 0
R òp/ 2
Þ
R
= ln 4 Þ R = (100 ln 4) W
® ^ ® ^ ^ ^ 50L
19. A = ab k, B = 20t i + 10t 2 j + 50 k
|e| 1 df 1 dB
®® 27. |i| = = × = NA
fm = B × A = 50 ab R R dt R dt
df 10 ´ 10 ´ 10-4
e = - m =0 = ´ 108 ´ 10-4
dt 20
20. E = Vb + iR =5A
Vb = E - iR = 200 - 20 ´ 1 . 5 = 170 V 28. In the steady state, inductor behaves as
V N 1 short circuit, hence entire current flows
21. s = s Þ Vs = ´ 290 = 10 V through it.
Vp N p 2
ip 29. fm = AB cos q
N
= s But, q = 90°
is Np
\ fm = 0
Np
Þ is = ip = 2 ´ 4 = 8 A |e| 1 dfm
Ns 30. i = =-
R R dt
22. Vr = 0, hence magnetic flux linked with the dq - nBA d
Þ = (cos q)
coil remain same. dt R dt
- df nBA dq
\e= =0 = sin q
dt R dt
1 nBA
23. s = at 2 Þ dq = sin q dq
2 R
Due to change in magnetic flux linked with nBA p 2nBA
Q1 = sin q dq =
the ring, magnet experiences an upward R ò0 R
force, hence, nBA 2p
Q2 = sin q dq = 0
a<g R ò0
1 2 Q2
s < gt Þ s < 5 m \ =0
2 Q1
di 31. According to Lenz’s law, induced current
24. VA - VB = L =-at
dt always opposes the cause producing it.
122
æ -
t
ö Eæ -
Rt ö df
32. i = i0 ç1 - ÷ = ç1 -
e t e L ÷ 37. E = -
ç ÷ Rç ÷ dt
è ø è ø
5 ´2 ö 38. Magnetic flux linked with the coil does not
15 æç -
change, hence
= 1 - e 10 ÷ = 3 (1 - e -1 )
5 ç ÷ e - 1 df
è ø i= = × =0
æ 1ö R R dt
= 3 ç1 - ÷ A 1 æ l ö
è eø 39. e = Blv cos q = Bl2w cos q çQ v = w÷
2 è 2 ø
33. Velocity of AB is parallel to its length.
As|cos q|varies from 0 to 1
34. Velocity of rod is parallel to its length. 1
e varies from 0 to B l2w.
35. Vc - Va = Vc - Vb = BRV 2
and Va - Vb = 0
36. Induced current always opposes the cause
producing it.
JEE Corner
Assertion and Reason
1. Magnetic flux linked with the coil is not 8. L = m rm 0n 2lA, for ferromagnetic substance,
changing with time, hence induced current mr > g
is zero.
and L does not depends on i.
2. Both Assertion and Reason are correct but
9. As soon as key is opened
Reason does not explain Assertion.
a
3. Induced electric field is non-conservative but
can exert force on charged particles. R1
4. i = 2t - 8 R2
di E
=2 L
dt
di
Va - Vb = L = 2 ´2 = 4 V i
dt b
æ di ö E
5. ç ÷ = ( imax ) w = 1 ´ 2 = 2 A/s i = i0 =
è dt ø max R1
6. Va - Vb = Vc - Va 10. Inductors oppose change in current while
Vc > Va > Vb resistor does not.
7. Fact.
[ fm ] = [ B][ S ] R2 t
= [ ML T - 2 A -1 ] [L2 ] = [ ML 2 T - 2 A -1 ]
0 E -
i2 = (1 - e L ) = 3 (1 - e -t / 3 )
-t / t R2
2. i = i0 (1 - e )
L At t = (ln 2) s
t==1s q
R VL = E - i2 R2 = qe - t / 3 =
E 21/ 3
i0 = =5A
R VR 2 = i2 R2 = q (1 - e - t / 3 )
VR = iR = E (1 - e -t ) æ 1 ö
= q ç1 - 1/ 3 ÷
VL = E - VR = Et - t è 2 ø
At t = 0, VR1 = i1 R1 = 9 V
VL = E = 10 V, VR = 0 Vbc = VL + VR2 = 9 V
at t =1s (a ® s), (b ® s), (c ® p), (d ® p).
æ 1ö 5. Induced emf
VL = E (1 - e -1 ) = ç1 - ÷ 10 V
è eø f(Wb)
10
VR = V
e
3. In LC oscillations,
1 1
w= = = 2 rad/s 4
LC 1
1´
4
q0 = 4 C
2 t
i0 = wq0 = 8 A
æ di ö |e| = slope of f - t graph
ç ÷ = w2q0 = 16 A/s.
è dt ø max 4 -0
= =2V
When, q = 2 C 2 -0
q |e| 2
VL = VC = = 8V |i| = = =1A
C R 2
æ di ö 1 æ di ö
When, ç ÷= ç ÷ = 8 A/s. |q| = |i|t = 1 ´ 2 = 2 C
è dt ø 2 è dt ø max
As current i is constant
di
VC - VL = L =1 ´8 = 8V H = i2 Rt = (1)2 ´ 2 ´ 2 = 4J
dt
25 Alternating Current
Introductory Exercise 25.1
VDC 100 XL = XC
1. R= = = 10 W
I 10 1
Þ wL =
V 150 wC
Z = AC = = 15 W
I 10 1 1
Þ L= 2 =
X L = Z 2 - R2 = (15)2 - (10)2 w C ( 2 pf )2 C
=5 5 W 1
=
XL XL 5 5 (360)2 ´ 10-6
L= = =
w 2 pf 2 ´ 3.14 ´ 50 » 7.7 H
» 0.036 H As XL = XC
\ VL = IX L = 50 5 V \ Z =R
V 120
= 111.8 V I= = =6A
Z 20
2. For phase angle to be zero,
AIEEE Corner
Subjective Questions (Level-1)
1. (a) X L = w L = 2 pfL 3. (a) Power factor at resonance is always 1,
= 2 ´ 3.14 ´ 50 ´ 2 R
as Z = R, Power factor = cos f = = 1.
= 628 W Z
X I 0 E0 cos f E20
(b) X L = w L Þ L = L (b) P = =
w 2 2R
XL 2 (150)2
= = = = 75 W
2 pf 2 ´ 3.14 ´ 50 3 ´ 150
= 6.37 mH (c) Because resonance is still maintained,
1 1 average power consumed will remain
(c) X C = = same, i.e., 75 W.
wC 2 pfC
1 4. (a) As voltage is lag behind current,
= inductor should be added to the circuit
2 ´ 3.14 ´ 50 ´ 2 ´ 10-6
to raise the power factor.
= 1592 W = 1.59 kW R
1 1 (b) Power factor = cos f =
(d) X C = ÞC= Z
wC w XC R 60 250
Þ Z= = = W
1 cos f 0.720 3
= = 1.59 mF
2 ´ 3.14 ´ 50 ´ 2
X C = Z 2 - R2
2 2
2. (a) Z = R + ( X L - X C ) 2
250 ö
2 = æç ÷ - (60)
2
æ 1 ö è 3 ø
= R2 + çç w L - ÷
è wC ÷ø = 58 W
2 1
æ 1 ö C=
= (300)2 + çç 400 ´ 0.25 - ÷ w XC
400 ´ 8 ´ 10-6 ÷
è ø 1
=
= 367.6 W 2pf X C
V0 120 1
I0 = = = 0.326 A =
Z 367.6 2 ´ 3.14 ´ 50 ´ 58
X - XC
(b) f = tan -1 L = 54 mF
R
For resonance,
-1 212.5
= tan » - 35.3 ° 1
300 wr =
LC
As X C > X L voltage will lag behind 1
current by 35.3°. Þ L= 2
wr C
(c) VR = I 0 R = 0.326 ´ 300 = 97.8 V,
1
VL = I 0 X L = 32.6 V =
(2pf )2 C
VC = I 0 X C = 0.326 ´ 312.5 1
Þ L=
= 101.875 V » 120 V (2 ´ 3.14 ´ 50)2 ´ 54 ´ 10-6
= 0.185 H
132
5. V ( t) = 170 sin (6280 t + p / 3) volt (b) w = 1000 rad/s
60
i ( t) = 8.5 sin (6280t + p / 2) amp. \ I= = 1.2 ´ 10-2 A
1000 ´ 5
V
170V (c) w = 10000 rad/s
60
\ I= = 1.2 ´ 10-3 A
O
0.25 0.50 0.75 1.00 1.25 t (ms)
10000 ´ 5
JEE Corner
Assertion and Reasons
1. X C and X L can be greater than Z because 5. On inserting ferromagnetic rod inside the
2
Z = R + ( X L - X C) 2 inductor, X L and hence VL increase. Due
to this current will increase if it is lagging
Hence, VC = IX C and VL = IX L can be and vice-versa.
greater than V = IZ . 6. VR = VL = VC Þ R = X L = X C
2. At resonance X L = X C , with further Hence, f = 0 and I is maximum.
increase in frequency, X L increases but
as Z = R2 + ( X L - X C )2 is minimum.
X C decreases hence voltage will lead
current. 7. I = I L - I C = 0
1
3. fr = , if dielectric slab is inserted 8. P = I 2rms R = ( 2)2 ´ 10 = 20 W
2p LC
9. Inductor coil resists varying current.
between the plates of the capacitor, its
E0 wL
capacitance will increase, hence, fr will 10. I 0 = , f = tan -1
decrease.
2
R +wL 2 2 R
4. q = Area under graph 11. At resonance, current and voltage are in
1 1 V0
= ´ 4 ´ (2 + 3) + ´ 4 ´ (2 + 4) same phase and I 0 = . Hence, I 0
2 2 R
= 22 C depends on R.
q 22
Average current = = = 3.6 A
t 6
= 53 ° 3. R = R1 + RL = 10 W
2. Current will remain same in series circuit X L = w L = 10 W,
given by 1
XC = = 10 W
wC
135
Reading of ammeter 8. For parallel RLC circuit,
V 10 2 I = I 2R + ( I C - I L )2
I rms = rms =
R 10 2 2
= 2 A = 1.4 A æV ö æV V ö
Þ I = ç 0 ÷ + çç 0 - 0 ÷÷
Reading of voltmeter, è R ø è XC XL ø
2
V = I rms RL = 5.6 V 1 æ 1 ö
1 1 = V0 + çç wC - ÷
4. X C = = R2
è w L ÷ø
wC 2p ´ 5 ´ 103 ´ 1 ´ 10-6
p 9. V = VL2 + VR2 = 72.8 V
= 100 W VL 2
f = tan -1 = tan -1æç ö÷
V 200 VR è7 ø
IR = = = 2 A,
R 100 10. Clearly P is capacitor and Q is resistor,
V 200
IC = = =2 A as, VP = V Q , X C = R .
X C 100
\When connected in series,
[Question is wrong. It should be choose Z = X 2C + R2 = 2 R
the correct statement].
p
5. Let i = i1 + i2 and f = , leading.
4
where, i1 = 5 A, i2 = 5 sin 100 w t A 1 p
\I = A, leading in phase by .
Average value of i1 = 5 A 4 2 4
Average value of i2 = 0 11. I = I 2R + ( I C - I L )2
\Average value of i = 5 A
Here, I C < I or I L > I
Another method
12. I = I L - I C = 0.2 A
é p ù
i = 5 ê1 + cos æç + 100 w t ö÷ú 13. For a pure inductor voltage leads with
ë è 2 øû p
current by .
é 2æ p ö ù 2
= 5 ê2 cos ç + 50 w t ÷ú
ë è4 øû 14. VR = IR = 220V
p
= 10 cos2 é + 50 wt ù Hence it is condition of resonance, i.e.,
êë 4 úû
VL = VC = 200 V
p 1 H1 I 2DC R I2
Average value of cos2 æç + 50 w t ö÷ = 15. = 2 = =2
è4 ø 2 H2 I rms R ( I / 2)2
10
\ average value of i = = 5 A. I 20 R V02 R
2 16. H = I 2rms R = =
2 2( R + w2 L2 )
2
6. As voltage is leading with current, circuit
p 17. VL = IX L = Iw L
is inductive, and as f = ,X L = R
4 I
VC = IX C =
R R wC
or L= =
w 100
If w is very small,
7. As X C > X L voltage will lag with current.
VL » = 0,VC » V0 .
Again V = VR2 + ( VL - VC )2 = 10 V V
18. Resistance of coil, R = =4W
\ V < VC I
R V 4 When connected to battery
and cos f = = R = V 12
Z V 5 I= = = 1.5 A
R+r 4+4
Hence, a, b and c are wrong.
136
2 1
19. VR = V - VC2 =6V XC =
wC
VC 4
f = tan -1 = tan -1 1
VR 3 = =8 W
50 ´ 2500 ´ 10-6
20. VC = V 2 - VR2 = 16 V
Z = R2 + ( X L - X C )2 = 5 W
p
21. I = I 0 sin æç t + p ö÷ 2
Vrms R
è2 ø Average power = I 2rms R =
p 3p Z2
I = I 0 at + p = (12)2 ´ 3
2 2 = = 17.28 W
(5)2
t =1s
V 25. Already X C > X L , with increase in w, X C
22. I0 = 0
2R further decrease in w, X C increases and
X L decreases, hence, I will decrease.
3
X C¢ = = 3R 26. For maximum current
wC
1 1
V I w = wr = =
Þ I0 = 0 = 0 LC -6
1 ´ 10 ´ 4.9 ´ 10-3
2R 2
VDC 12 105
23. R = = =3 W = rad/s.
I DC 4 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.