Are these the official NEET questions on Electromagnetic Induction?

Yes. They are the questions NTA set in each paper, in its wording. The solutions are written by BrainIt.

Electromagnetic Induction NEET PYQs with Solutions (2026) PDF

The questions come from NEET 2026 and Re-NEET 2026. Each shows its answer and solution when you open it, so you can attempt it first. For what the chapter covers and how NEET tests it, see the Electromagnetic Induction guide.

Practise on the CBT screen4 questions, 4 minutes, marked +4 / −1 Download PDF3 pages, 0.7 MB, with answers and solutions

NEET 2026 · the full paper

NEET 2026 · Q30
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:
1. (1)
  $4.8 \times 10^{-4}$ volt
2. (2)
  $1.2 \times 10^{-4}$ volt
3. (3)
  $1.3 \times 10^{-4}$ volt
4. (4)
  $1.8 \times 10^{-4}$ volt
Answer and solution
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.

Re-NEET 2026 · the full paper

Re-NEET 2026 · Q3
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.]
1. (1)
  $\frac{1}{4}$
2. (2)
  $\frac{1}{2}$
3. (3)
  1
4. (4)
  2
Answer and solution
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
Re-NEET 2026 · Q5
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 :
1. (1)
  $\mu_0\pi n^2 r^2 l$
2. (2)
  $\mu_0 n^2 r^2 l$
3. (3)
  $(\mu_0/2\pi) n^2 r^2 l$
4. (4)
  $2\mu_0\pi n^2 r^2 l$
Answer and solution
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).
Re-NEET 2026 · Q32
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:
1. (1)
2. (2)
3. (3)
4. (4)
Answer and solution
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

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Electromagnetic Induction: NEET previous year questions

NEET 2026 · the full paper

Re-NEET 2026 · the full paper

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