Electromagnetic Induction PYQs on the CBT Screen

Question 1 · NEET 2026 · Q30 · +4 / −1

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 cm s $^{-1}$ , in a direction normal to the shorter side of the loop, will be:

Answer and explanation

Answer: (4)

The motional electromotive force (emf) is $E = Bvl$ , where $l$ is the side perpendicular to the velocity. Since the loop moves normal to the shorter side, the effective length is the shorter side (3 cm): $E = 0.3 \times (2\times10^{-2}) \times (3\times10^{-2}) = 1.8\times10^{-4}$ V. Option (1) wrongly uses the 8 cm side.

Question 2 · Re-NEET 2026 · Q3 · +4 / −1

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.]

Answer and explanation

Answer: (4)

In configuration $P$ the two inductors are in series and the segment between $a$ and $b$ spans just ONE of them, carrying the full current, so $E_P = L\frac{dI}{dt}$ . In configuration $Q$ the points $a$ and $b$ straddle the parallel pair, whose equivalent inductance is $\frac{L}{2}$ , so $E_Q = \frac{L}{2}\frac{dI}{dt}$ . The ratio is therefore 2. NCERT Class 12, Electromagnetic Induction (self-inductance).

Why the other options are wrong:

- (1) takes $E_P$ across the series pair as $2L$ and inverts the ratio
- (2) inverts the correct answer by swapping which configuration sits in the numerator
- (3) assumes identical inductors must produce identical electromotive forces (emfs) regardless of how $a$ and $b$ are placed

Question 3 · Re-NEET 2026 · Q5 · +4 / −1

Consider a long solenoid of length $l$ and radius $r$ . If $n$ is the number of turns per unit length and $\mu_0$ is the permeability of free space, the inductance of the solenoid is :

Answer and explanation

Answer: (1)

For a long solenoid $L = \mu_0 n^2 A l$ , where $n$ is turns per unit length and $A$ the cross-sectional area. Substituting $A = \pi r^2$ gives $L = \mu_0\pi n^2 r^2 l$ . NCERT Class 12, Electromagnetic Induction (self-inductance).

Question 4 · Re-NEET 2026 · Q32 · +4 / −1

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 and explanation

Answer: (2)

Faraday's law gives $\varepsilon = -\frac{d\Phi}{dt} = -BA_0\cos t$ , so the dissipated power is $P = \frac{\varepsilon^2}{R} = \frac{B^2A_0^2}{R}\cos^2 t$ . Because $\cos^2 t$ is MAXIMUM at $t = 0$ and touches zero at $t = \pi/2$ , the correct graph starts at a peak and dips to zero, unlike a $\sin^2 t$ curve which would start at zero. $P$ is never negative because the electromotive force (emf) is squared. NCERT Class 12, Electromagnetic Induction (Faraday's law).

Why the other options are wrong:

- (1) differentiates $\sin t$ to $\sin t$ rather than $\cos t$ , producing a $\sin^2 t$ curve that starts at zero
- (3) assumes the constant field means a constant emf, forgetting that it is the CHANGING area that drives induction
- (4) confuses the growing area with growing power, ignoring that the flux change is periodic and bounded

Electromagnetic Induction, on the exam screen.