Test: Electromagnetic Induction - 2 - CUET Humanities MCQ

L = 2 mH, i = t²e^-t
E = -L(di/dt)= -L [-t²e^-t + 2 te^-t]
When E = 0 - t²e^-t + 2te^-t = 0
Or 2te^-t = t²e^-t
Thus, t = 2 s

If a coil is not moved in a magnetic field, the magnetic flux does not change. Hence, no emf or current is induced in the coil. So, the correct choice is (4).

Electric field will be induced in both AD and BC since both are moving perpendicular to the direction of the magnetic field and the flux linked with them is changing with time. Hence, the correct choice is (4).

This answer is correct because the process of producing induced current in a coil due to relative motion between a magnet and a coil is known as electromagnetic induction.

If the coil is not moved in a magnetic field, then the magnetic flux linked with the coil does not change. Hence, no emf or current is induced in the coil. Thus, the correct choice is (4).

As the rheostat is being shifted, the resistance in series with the coil is decreasing and the current in the coil is increasing. Hence, the magnetic flux through the neighbouring coil increases. By Lenz's law, the induced current in the coil must oppose the increase in flux. Therefore, the induced current must flow in the clockwise direction, i.e. along CBA. Hence, the correct choice is (2).

Electromagnetic induction occurs because of the change in the flux through the coil. Flux through the coil can be changed by changing the current or by magnetic field or by changing the area of the coil.

A circular loop of radius R, carrying current I, lies in xy-plane with its centre at origin. The total magnetic flux through coil will be zero as the number of magnetic field lines entering the coil is equal to the number of magnetic field lines leaving the coil. Hence, the total magnetic flux linked with the coil is zero.

A rectangular coil made of copper is rotated in a magnetic field. The direction of the induced current changes once every half revolution.

Power VI = 10 x 10^-3 x (5+16t)
At t = 1s, power = 10 x 10^-3 x (5 + 16) = 0.21W
Hence the correct choice is (2).

E = 12 V, L₁ = 10 H, L₂ = 10 × 10^-3 H and R = 48 Ω.
Steady state current is I₀ = E/R and is independent of the inductance.
Hence, the value of the steady state current is the same for both circuits.

I = 2 A
L = self-inductance
 = 2 Weber
dl/dt = 1 A/s


L = 1 wb
Induced emf,
e = -L(di/dt)
e = –1  1
e = –1
The magnitude of e is 1 V.

Michael Faraday is generally credited with the discovery of induction in 1831, and James Clerk Maxwell mathematically described it as Faraday's law of induction.

Induced emf in the secondary is 

Hence the correct choice is (2).