Electromagnetic Induction — JEE Main practice

28 questions

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Sample questions with solutions

Q1 · 2026

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is $2\Omega$ then the force needed to move the rod towards right with constant speed $(v)$ of 1.5 m/s is $\_\_\_\_$ N.

A.
$5.7\times10^{-2}$
B.
$5.7\times10^{-3}$
C.
$7.5\times10^{-2}$
D.
$7.5\times10^{-3}$

Answer: D

When a rod of length $l$ moves with velocity $v$ perpendicular to a field $B$ , it develops a motional emf.

\varepsilon = Bvl

Substitute $B=0.10$ T, $v=1.5$ m/s, $l=1$ m.

\varepsilon = (0.10)\times(1.5)\times(1) = 0.15\text{ V}

Use Ohm's Law to find the induced current, given total resistance $R=2\Omega$ .

I = \frac{\varepsilon}{R} = \frac{0.15}{2} = 0.075\text{ A}

A current-carrying rod in a magnetic field experiences a force opposing its motion (by Lenz's Law), given by $F_m=IlB$ since the rod is perpendicular to the field.

F_m = (0.075)\times(1)\times(0.10) = 0.0075\text{ N} = 7.5\times10^{-3}\text{ N}

Since the rod moves at constant speed, the net force is zero, so the applied external force must exactly balance this magnetic braking force.

F_{ext} = F_m = 7.5\times10^{-3}\text{ N}

Hence, the force needed is $7.5\times10^{-3}$ N, so the answer is option D.

Q2 · 2026

A conducting circular loop of area $1.0\text{ m}^2$ is placed perpendicular to a magnetic field which varies as $B=\sin(100t)$ Tesla. If the resistance of the loop is $100\Omega$ , then the average thermal energy dissipated in the loop in one period is $\_\_\_\_$ J.

A.
$\pi^2$
B.
$\pi$
C.
$\frac{\pi}{2}$
D.
$2\pi$

Answer: B

Since the field is perpendicular to the loop, the magnetic flux is simply $\phi=BA$ . With $A=1.0\text{ m}^2$ and $B=\sin(100t)$ .

\phi = \sin(100t)

By Faraday's Law, differentiate the flux with respect to time to find the induced emf.

\varepsilon = -\frac{d}{dt}[\sin(100t)] = -100\cos(100t)\text{ V}

The instantaneous power dissipated in the resistor is

P = \frac{\varepsilon^2}{R} = \frac{[100\cos(100t)]^2}{100} = 100\cos^2(100t)\text{ W}

Find the time period of the oscillation from the angular frequency $\omega=100$ .

T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \frac{\pi}{50}\text{ s}

Integrate power over one period to find the total energy dissipated, using the identity $\cos^2\theta=\frac{1+\cos2\theta}{2}$ .

E = \int_0^T 100\cos^2(100t)\,dt = 50\int_0^T[1+\cos(200t)]\,dt

Evaluate the integral.

E = 50\left[t+\frac{\sin(200t)}{200}\right]_0^{\pi/50} = 50\left[\frac{\pi}{50}+\frac{\sin(4\pi)}{200}\right] = 50\times\frac{\pi}{50} = \pi\text{ J}

Hence, the average thermal energy dissipated in one period is π J, so the answer is option B.

Q3 · 2026

Three identical coils $C_1, C_2$ and $C_3$ are closely placed such that they share a common axis. $C_2$ is exactly midway. $C_1$ carries current $I$ in anti-clockwise direction while $C_3$ carries current $I$ in clockwise direction. An induced current flows through $C_2$ will be in clockwise direction when

A.
$C_1$ moves towards $C_2$ and $C_3$ moves away from $C_2$
B.
$C_1$ and $C_3$ move with equal speeds away from $C_2$
C.
$C_1$ and $C_3$ move with equal speeds towards $C_2$
D.
$C_1$ moves away from $C_2$ and $C_3$ moves towards $C_2$

Answer: A

Looking along the common axis, coil $C_1$ 's anti-clockwise current produces a field (by the Right-Hand Grip Rule) pointing towards the observer — call this positive flux. Coil $C_3$ 's clockwise current produces a field pointing away from the observer — call this negative flux. [IMAGE: Three coaxial coils C1, C2, C3 with C2 in the middle, showing the direction of magnetic fields produced by C1 and C3]
Since $C_1$ and $C_3$ are identical and carry equal current, at the midpoint $C_2$ their fields have equal magnitude but opposite sign, so they exactly cancel, giving zero net initial flux through $C_2$ .
If $C_2$ develops a clockwise induced current, it creates its own field pointing away from the observer (negative direction). By Lenz's Law, this induced field must oppose an increasing positive flux through $C_2$ .
To increase the net positive flux through $C_2$ , we can either increase the positive contribution from $C_1$ or decrease the negative contribution from $C_3$ — this happens if $C_1$ moves closer to $C_2$ (strengthening its positive flux) while $C_3$ moves farther away (weakening its negative flux). Hence, $C_1$ moves towards $C_2$ and $C_3$ moves away from $C_2$ , so the answer is option A.

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