Electromagnetic Induction Chapter-Wise Test 19

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A coil of self-inductance 1.2 H has its current increased from 2 A to 6 A in 0.4 s. What is the magnitude of the induced emf?

\( \varepsilon = L \frac{\Delta I}{\Delta t} \).

\( \Delta I = 6 - 2 = 4 \, \text{A} \), \( \Delta t = 0.4 \, \text{s} \).

\( \varepsilon = 1.2 \times \frac{4}{0.4} = 1.2 \times 10 = 12 \, \text{V} \).

10 V
12 V
14 V
16 V
2

A coil with \( L = 0.5 \, \text{H} \) has its current increased from 0 to 6 A in 0.6 s. What is the energy stored?

\( W = \frac{1}{2} L I^2 = \frac{1}{2} \times 0.5 \times (6)^2 = 0.25 \times 36 = 9 \, \text{J} \).

8 V
8.5 V
8.8 V
9 V
4

A square loop of side 16 cm rotates at 18 rad/s in a 0.2 T field. What is the maximum emf induced?

\( A = (0.16)^2 = 0.0256 \, \text{m}^2 \).

\( \varepsilon_0 = N B A \omega = 1 \times 0.2 \times 0.0256 \times 18 = 0.09216 \, \text{V} \approx 0.092 \, \text{V} \).

0.07 V
0.092 V
0.1 V
0.12 V
2

A circular loop of radius 13 cm is deformed into a straight wire in a 0.18 T field in 0.6 s. What is the induced emf?

Initial flux: \( \Phi = B A = 0.18 \times \pi \times (0.13)^2 = 0.00956 \, \text{Wb} \).

Final flux = 0.

\( \varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{0.00956}{0.6} = 0.01593 \, \text{V} \approx 0.016 \, \text{V} \).

0.01 V
0.016 V
0.02 V
0.025 V
2

A coil of 75 turns and area 0.05 m² is in a 0.12 T field that drops to zero in 0.25 s. What is the induced emf?

\( \Delta \Phi = B A = 0.12 \times 0.05 = 0.006 \, \text{Wb} \).

\( \varepsilon = N \frac{\Delta \Phi}{\Delta t} = 75 \times \frac{0.006}{0.25} = 75 \times 0.024 = 1.8 \, \text{V} \).

1.2 V
1.5 V
1.8 V
2.0 V
3

A conducting loop is compressed in a uniform magnetic field. The induced current flows to maintain what quantity against the compression?

The current opposes the decrease in flux by maintaining the original flux direction, resisting the reduction in area per Lenz’s law.

Field strength
Loop resistance
Current magnitude
Magnetic flux
4

A coil of self-inductance 1.8 H has its current increased from 3 A to 7 A in 0.5 s. What is the magnitude of the induced emf?

\( \varepsilon = L \frac{\Delta I}{\Delta t} \).

\( \Delta I = 7 - 3 = 4 \, \text{A} \), \( \Delta t = 0.5 \, \text{s} \).

\( \varepsilon = 1.8 \times \frac{4}{0.5} = 1.8 \times 8 = 14.4 \, \text{V} \).

12 V
14.4 V
16 V
18 V
2

A conducting rod moves perpendicular to a uniform magnetic field with constant velocity. What is true about the induced emf across its ends?

The induced emf (\( \varepsilon = B l v \)) remains constant because the velocity, magnetic field, and rod length are constant, leading to a steady flux change rate.

It varies sinusoidally
It remains constant
It increases with time
It decreases with time
2

A conducting loop is stationary in a uniform magnetic field that increases in strength. The induced emf is generated by what field component?

A changing magnetic field induces an electric field (per Maxwell’s equations), which drives the emf in the stationary loop.

Magnetic force
Gravitational field
Electrostatic field
Induced electric field
4

A rod rotates at 10 rad/s in a 0.5 T field. If the length from the axis to the tip is 0.6 m, what is the emf induced?

\( \varepsilon = \frac{1}{2} B \omega R^2 \).

\( \varepsilon = \frac{1}{2} \times 0.5 \times 10 \times (0.6)^2 = 0.9 \, \text{V} \).

0.6 V
0.8 V
0.9 V
1.0 V
3

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