Solution

PHYSICS

Class 12 - Physics
Section A
1.
(b) Increases
Explanation:
A change in the magnetic field induces an emf. When there is an emf, there has to be current. Hence, when the magnet is
moved inside a coil, the current in it increases.

2.
(b) 226 V
Explanation:
−4 2π
ε0 = N BAω = 30 × 1 × 400 × 10 × (1800 × )
60

= 226 V

3.
(d) 3.07 × 10-8 H
Explanation:
Frequency,
8
c 3×10 1 7
f = = = × 10 Hz
λ 360 12

Required inductance,
1 1
L= =
2 2 2
4π f C 2 1 7 −6
4π ( × 10 ) ×1.2× 10
12

= 3.07 × 10-8 H

4.
(c) 0 W, 0 W
Explanation:
P = V I cos ϕ

Average power consumed by inductor is zero as actual voltage leads the current by π

2
and (cos π

2
= 0) .
Average power consumed by capacitor is zero as actual voltage lags the current by π

2
and (cos π

2
= 0) .

5.
(d) Current
Explanation:
Current increases in a step-down transformer.

6. The SI unit of displacement current is ampere (A).

7.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
In case of a linearly polarised plane electromagnetic wave the average values of electric field and magnetic field are equal and
average values of electric energy and magnetic energy are also equal.

8. (a) 6 × 10-12 m
Explanation:
8
v 3×10 −12
λ = = = 6 × 10 m
c 19
5×10

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 Section B
9. Magnitude of Induced emf is directly proportional to the rate of area moving out of the field for a constant magnetic field.
dϕ dA
ε= − = −B
dt dt

For the rectangular coil, the rate of area moving out of the field remain same while it is not so for the circular coil. Therefore, the
induced emf for the rectangular coil remains constant.
10. Let a current I2 flows through the outer circular coil of radius R. The magnetic field at the centre of the coil is

As r << R, hence field B2 may be considered to be constant over the entire cross-sectional area of inner coil of radius r.
Hence,magnetic flux linked with the smaller coil will be
μ0 I2
2
ϕ1 = B2 A1 = ⋅ πr
2R

As, by definition ϕ 1 = M12 I2

2
μ0 π r
Now mutual inductance, M 12 = ϕ1 /I2 =
2R

Say M 12 = M21 = M ,
2
μ0 π r
∴ M =
2R

11. i. The impedance of a series L-C-R circuit is given by

−−−−−−−−−−−−−−−
2

Z =√R 2
+ (ωL −
ωC
1
)

where R, ωL and 1

ωC
are resistance, inductive reactance and capacitive reactance of the circuit respectively. Z will be
minimum when ωL = 1/ωC , i.e. when the circuit is under resonance with resonant frequency ω 0 =
1
. Hence, in this
√LC

condition Z will be minimum and equal to R, resistance of the resistor.

ii. P = E I cos ϕ ....(i)
av V V

For wattless current, the power dissipated through the circuit should be zero.
i.e. From equation (i), cos ϕ = 0
π π
⇒ cos ϕ = cos ⇒ ϕ =
2 2

Hence, the condition for wattless current is that the phase difference between the current and the voltage will be equal to π /2.
i.e. the circuit should be purely inductive or purely capacitive in which the voltage and current differ by a phase angle of
π/2, i.e. ϕ = ±
π

Section C
12. i. Total charge passed through the loop (Q)
Q = area under the I-t graph
= × 0.4 × 1 coulomb = 0.2 C
1

ii. Change in magnetic flux

change in magnetic flux
Total charge passing = ( R
)

Change in magnetic flux = [R× 0.2C]

= [10× 0.2] Wb
= 2 Wb
iii. Magnitude of magnetic field applied
Let B be the magnitude of the magnetic field applied
Initial magnetic flux = B× (10Change in magnetic flux10-4) Wb
Final magnetic flux = zero
Change in magnetic flux = (B× 10-3-0) = 2
⇒ B = 2× 103 Wb/m2
hence, magnitude of the magnetic field applied is 2× 103 Wb/m2.

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13. i. ε rms =220 V, R = 40 Ω
εrms 220
∴ Irms = = = 5.5 A
R 40

ii. Maximum instantaneous current,

–
I = √2I
0
= 1.414 × 5.5 = 7.8 A
rms

iii. Let the alternating current be given by

I = I0 sin ωt

Let the a.c. take its maximum and rms values at instants t1 and t2 respectively. Then
I0 = I0 sin ω t1 ,
I0
which implies ω t 1 =
π

2
and I rms = = I0 sin ω t2 ,
√2

which implies ω t 2
=
π

2
+
π

4
π π
∴ t2 − t1 = =
4ω 4×2πf

=
π

4×2π×50
=
400
1
s = 2.5 ms
14. i. N1 = 400, N2 = 2000, ε = 1100 V 2

N1
ε1 = ε2 ⋅
N2
= 1100 ×
400

2000
= 220V
ii. Resistance of primary, = 0.2 Ω
Output power = ε I = 12.1 kW = 12100 W
2 2

∴ Current in the secondary,

ε2 I2 12100
I2 = = = 11 A
ε2 1100

Output power
As Efficiency =
Input power

90 12100 W
=
100
Input power
or Input power,
12100×100 3
ε1 I1 = = 13.44 × 10 W
90

Current in the primary,

ε1 I1 3
13.44×10
I1 = = = 61.1 A
ε1 220

Power loss in the primary

= I
2
1
R1 = (61.1) = 746.61 W2
× 0.2

Power loss in the secondary

= I
2
2
R2 = (11)
2
× 2.0 = 242 W
15. a. Gamma rays, use: it is used in cancer treatment.
b. Ultraviolet rays use: it is used to sterilize surgical instruments.
c. Infrared radiation uses: it is used in thermal imaging cameras.
Section D
16. Read the text carefully and answer the questions:
Lenz's law states that the direction of induced current in a circuit is such that it opposes the change which produces it. Thus, if the
magnetic flux linked with a closed circuit increases, the induced current flows in such a direction that magnetic flux is created in
the opposite direction of the original magnetic flux. If the magnetic flux linked with the closed circuit decreases, the induced
current flows in such a direction so as to create magnetic flux in the direction of the original flux.

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 (i) (d) The induced e.m.f is not in the direction opposing the change in magnetic flux so as to oppose the cause which
produces it.
Explanation:
The spark coil designed on the principle of electromagnetic induction was the heart of the earliest radio transmitters.
Marconi used a spark coil designed by Heinrich Rhumkorff in his early experiments.
(ii) (d) Lenz's law
Explanation:
A spark coil is a type of electrical transformer used to produce high-voltage pulses from a low-voltage (d.c.) supply.
To create the flux changes necessary to induce voltage in the secondary coil, the direct current in the primary coil is
repeatedly interrupted by a vibrating mechanical contact called interrupter.
(iii) (c) energy
Explanation:
The spark coil consists of two coils of insulated wire wound around a common iron core. One coil, called the primary
coil, is made from relatively few (tens or hundreds) turns of coarse wire. The other coil, the secondary coil typically
consists of up to a million turns of fine wire (up to 40 gauge).
(iv) (d) is repelled
Explanation:
The spark coil designed on the principle of electromagnetic induction was the heart of the earliest radio transmitters.
Marconi used a spark coil designed by Heinrich Rhumkorff in his early experiments.
Section E
17. i. Labelled diagram:

Flux created
= NBA cos ωt
−dϕ
emf induced E = dt
−d
=
dt
(NBA cos ωt)
E = NBA ω sin ωt
The instantaneous value of emf induced.
Expression: If at any moment, t be the perpendicular vector to the plane of coil that makes an angle θ with the direction of the
magnetic field (B), so flux passing through the coil will be :
ϕ = nBA cos θ

θ = ωt

[ω = angular velocity of the coil]

Now, ϕ = nBA cos ωt
If e is instantaneous induced emf produced in the coil at the instant t, then
dρ
e= −
dt

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 d(nAB cos ωt)
= −
dt

= -nABω (-sin ωt)

Maximum value or peak value of instantaneous induced emf e which is attained when sin ωt = ±1
∴ emax = nω AB

So, e = emax sin ωt

As the value of sine function varies from +1 to -1, the polarity of emf changes with time. Also, the output voltage is sinusoidal
in nature.
ii. The maximum value of emf is given by : No of turns = 20 , B = 3× 10-2 T and area A= 200 × 10-4 m2
e0 = NBAω also sinωt = 1 hence ,

= 20 × 200 × 10-4 × 3 × 10-2 × 50 V

= 600 mV
Maximum induced current in the coil ,
e0
i =
0 =
R
mA 600

18. i. Let at any instant, the current and voltage in an L-C-R series AC circuit is given by
V = V sin ωt and
m

I = Im sin(ωt + ϕ)

where Vm and Im are the peak values of the ac voltage and ac current respectively.
The instantaneous power is given by
P = V I = Im sin(ωt + ϕ)Vm sin ωt
Vm Im
⇒ P = [2 sin ωt sin(ωt + ϕ)]
2
Vm Im
∴ P = VI = [cos ϕ − cos(2ωt + ϕ)] … (i)
2

[∵ 2 sin A sin B = cos(A − B) − cos(A + B)]

Work done for a very small time interval dt is given by

dW = P dt

⇒ dW = V I dt

∴ Total work done over a complete cycle i.e. from 0 to T is given by,
T
W = ∫ V I dt
0
T
∫ V Idt
W
But P
0

av = =
T T
1 T
⇒ Pav = ∫ V I dt
T 0

1 T Vm Im
= ∫ [cos ϕ − cos(2ωt + ϕ)]dt
T 0 2

Vm Im T T
= [∫ cosϕdt − ∫ cos(2ωt + ϕ)dt]
2T 0 0

Vm Im
=
2T
[cos ϕ(t)]
T
0
− 0 (By trigonometry)
Vm Im Vm Im
= cos ϕ × T = cos ϕ
2T 2
Vm Im
= × cos ϕ
√2 √2

⇒ Pav = Vrms Irms cos ϕ

This is the required expression.

R
ii. Power factor, cosϕ = Z

where, R = resistance and Z = impedance of the circuit.

Low power factor (cos ϕ) implies lower ohmic resistance which implies larger power loss in power system (transmission line),
because in power system power, P ∝ . 1

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