Electromagnetic Induction — NEET UG practice

32 questions

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Sample questions with solutions

Q1 · 2026

Two identical inductors are connected in two different configurations $P$ and $Q$ , where a time varying current $I(t)$ is flowing, as shown in the figure. The induced emf between points $a$ and $b$ for configuration $P$ is $E_P$ and that for configuration $Q$ is $E_Q$ . The ratio $E_P/E_Q$ is:

[Neglect the effect of mutual inductance.]

A.
$2$
B.
$\frac{1}{4}$
C.
$\frac{1}{2}$
D.
$1$

Answer: A

For any inductor, the induced emf opposes the change in current and is given by: $\varepsilon = -L\frac{dI}{dt}$
In configuration $P$ , the two identical inductors combine to give an effective inductance $L$ across $a$ and $b$ . So: $E_P = -L\frac{dI(t)}{dt}$
In configuration $Q$ , the way the two inductors are connected gives an effective inductance of $L/2$ . So: $E_Q = -\frac{L}{2}\frac{dI(t)}{dt}$
Dividing the two expressions to compare them: $\frac{E_P}{E_Q} = \frac{L}{L/2} = 2$

Hence, the answer is option A.

Q2 · 2026

A conducting loop of finite resistance lies on the $x-y$ plane. There is a constant magnetic field in the $z$ direction. The area of the loop varies with time $t$ , as $A=A_0(1+\sin t)$ in appropriate units. The figure that correctly indicates the qualitative behaviour of the power $P$ dissipated in the loop as a function of time is:

Answer: C

The area of the loop is given as: $A = A_0(1+\sin t)$
Faraday's law states the induced emf equals the negative rate of change of magnetic flux, and since $\phi = BA$ : $\varepsilon = -\frac{d\phi}{dt} = -BA_0\cos t$
By Ohm's law, the induced current in the loop is: $I = \frac{\varepsilon}{R} = -\frac{BA_0}{R}\cos t$
Power dissipated in a resistor is $P = I^2R$ . Substituting the current: $P = \frac{B^2A_0^2}{R}\cos^2 t$
So $P \propto \cos^2 t$ , which is always non-negative and oscillates with twice the frequency of $\sin t$ .

Hence, the correctly matching graph is option C.

Q3 · 2026

A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is $2\text{ cm s}^{-1}$ , in a direction normal to the shorter side of the loop, will be :

A.
$4.8\times10^{-4}$ volt
B.
$1.2\times10^{-4}$ volt
C.
$1.3\times10^{-4}$ volt
D.
$1.8\times10^{-4}$ volt

Answer: D

When a conducting rod of length $l$ moves with velocity $v$ perpendicular to a magnetic field $B$ , the motional emf induced across it is: $\varepsilon = Bvl$
Here the loop moves in a direction normal to the shorter side, so the emf is induced across that shorter side, $l = 3\text{ cm}$ , with $v = 2\text{ cm/s}$ and $B = 0.3\text{ T}$ . Converting to SI units and substituting: $\varepsilon = 0.3 \times (2\times10^{-2}) \times (3\times10^{-2})$
Calculating the product: $\varepsilon = 1.8\times10^{-4}\text{ V}$

Hence, the answer is option D.

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