Electromagnetic Induction NEET PYQs (2005–2026) with Solutions

10 solved Electromagnetic Induction PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumSelf-inductance and combination of inductors

   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. AOption A: $\frac{1}{4}$
   2. BOption B: $\frac{1}{2}$
   3. COption C: $1$
   4. DOption D: $2$

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   Correct answer: (D) $2$

   Explanation

   Let each inductor have inductance $L$; the emf across an inductance is $E = L_{\text{eff}}\frac{dI}{dt}$. In configuration $P$ the two inductors are in series, but point $a$ is at the top of the first inductor and point $b$ is at the junction between the two. So $a$ and $b$ span only one inductor: $E_P = L\frac{dI}{dt}$. In configuration $Q$ the two inductors are in parallel and $a$, $b$ are the two junctions. The equivalent inductance is $\frac{L}{2}$ and the full current $I(t)$ flows through the combination, so $E_Q = \frac{L}{2}\frac{dI}{dt}$. $\frac{E_P}{E_Q} = \frac{L}{L/2} = 2$, option (D).

2. Question 2 (NEET 2026)

   NEET 2026EasySelf-inductance of a long solenoid

   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. AOption A: $\mu_0 \pi n^2 r^2 l$
   2. BOption B: $\mu_0 n^2 r^2 l$
   3. COption C: $(\mu_0/2\pi) n^2 r^2 l$
   4. DOption D: $2\mu_0 \pi n^2 r^2 l$

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   Correct answer: (A) $\mu_0 \pi n^2 r^2 l$

   Explanation

   Field inside a long solenoid: $B = \mu_0 n I$. Total number of turns $N = nl$, and the flux through each turn is $B\pi r^2$. Flux linkage: $N\Phi = (nl)(\mu_0 n I)(\pi r^2)$. $L = \frac{N\Phi}{I} = \mu_0 \pi n^2 r^2 l$, option (A).

3. Question 3 (NEET 2026)

   NEET 2026MediumFaraday's Law

   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. AOption A: 4.8 × 10⁻⁴ volt
   2. BOption B: 1.2 × 10⁻⁴ volt
   3. COption C: 1.3 × 10⁻⁴ volt
   4. DOption D: 1.8 × 10⁻⁴ volt

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   Correct answer: (D) 1.8 × 10⁻⁴ volt

   Explanation

   Induced emf ε = Bℓv. Here B=0.3 T, ℓ=3×10⁻² m (shorter side), v=2×10⁻² m s⁻¹. Therefore ε=0.3×3×10⁻²×2×10⁻²=1.8×10⁻⁴ V. Hence option (D) is correct.

4. Question 4 (NEET 2026)

   NEET 2026MediumFaraday's law and induced emf

   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. AOption A: Graph A
   2. BOption B: Graph B
   3. COption C: Graph C
   4. DOption D: Graph D

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   Correct answer: (B) Graph B

   Explanation

   The graphs ($P$ against $t$): (A) $P$ starts at zero at $t = 0$ and forms repeated humps that return to zero; (B) $P$ starts at a maximum at $t = 0$, falls to zero, then rises again; (C) $P$ constant; (D) $P$ increases steadily from zero. Flux $\Phi = BA_0(1 + \sin t)$, so the induced emf is $\varepsilon = \left|\frac{d\Phi}{dt}\right| = BA_0|\cos t|$. Power dissipated: $P = \frac{\varepsilon^2}{R} = \frac{B^2A_0^2}{R}\cos^2 t$. At $t = 0$, $P$ is maximum; it falls to zero at $t = \frac{\pi}{2}$ and then rises again. This is the shape in option (B).

5. Question 5 (NEET 2025)

   NEET 2025MediumSelf & Mutual Inductance

   AB is a part of an electrical circuit (see figure). The potential difference $V_A - V_B$, at the instant when current $i = 2\,\text{A}$ and is increasing at a rate of 1 amp / second is:

   1. AOption A: 5 volt
   2. BOption B: 6 volt
   3. COption C: 9 volt
   4. DOption D: 10 volt

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   Correct answer: (D) 10 volt

   Explanation

   Given: $$L = 1\,\text{H}, \quad i = 2\,\text{A}, \quad \frac{di}{dt} = 1\,\text{A s}^{-1}, \quad R = 2\,\Omega$$ Potential drop across the inductor: $$V_L = L\frac{di}{dt} = 1 \times 1 = 1\,\text{V}$$ Potential drop across the resistor: $$V_R = iR = 2 \times 2 = 4\,\text{V}$$ Battery emf is: $$E = 5\,\text{V}$$ Therefore, $$V_A - V_B = V_L + E + V_R = 1 + 5 + 4 = 10\,\text{V}$$ Hence, the correct option is (4).

6. Question 6 (NEET 2024)

   NEET 2024MediumLenz's Law

   In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively are through the directions:

   1. AOption A: AB and CD
   2. BOption B: AB and DC
   3. COption C: BA and DC
   4. DOption D: BA and CD

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   Correct answer: (C) BA and DC

   Explanation

   By Lenz's law, the induced current opposes the relative motion between the magnet and the solenoids. In solenoid-1, the magnet is moving away, so the near face becomes south to attract the magnet, giving current direction BA. In solenoid-2, the magnet approaches, so the near face becomes south to oppose the approach, giving current direction DC. Hence, option (C) is correct.

7. Question 7 (NEET 2023)

   NEET 2023EasySelf & Mutual Inductance

   The magnetic energy stored in an inductor of inductance 4 μH carrying a current of 2 A is :

   1. AOption A: 4 μJ
   2. BOption B: 4 mJ
   3. COption C: 8 mJ
   4. DOption D: 8 μJ

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   Correct answer: (D) 8 μJ

   Explanation

   Energy stored in an inductor is: U = (1/2)LI² Given: L = 4 × 10⁻⁶ H I = 2 A U = (1/2)(4 × 10⁻⁶)(2²) = 8 × 10⁻⁶ J = 8 μJ Hence, the correct answer is 8 μJ.

8. Question 8 (NEET 2023)

   NEET 2023EasyMagnetic flux

   The net magnetic flux through any closed surface is :

   1. AOption A: Zero
   2. BOption B: Positive
   3. COption C: Infinity
   4. DOption D: Negative

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   Correct answer: (A) Zero

   Explanation

   According to Gauss’s law for magnetism, magnetic monopoles do not exist. Hence, the net magnetic flux through any closed surface is zero. Therefore, option (A) is correct.

9. Question 9 (NEET 2022)

   NEET 2022HardFaraday's Law

   A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad s⁻¹. If the vertical component of earth's magnetic field at that place is 2 × 10⁻⁵ T and electrical resistance of the coil is 12.56 Ω, then the maximum induced current in the coil will be:

   1. AOption A: 2 A
   2. BOption B: 0.25 A
   3. COption C: 1.5 A
   4. DOption D: 1 A

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   Correct answer: (D) 1 A

   Explanation

   Maximum induced emf: Eₘ = NBAω N = 1000, B = 2 × 10⁻⁵ T, A = π(10)² = 100π m², ω = 2 rad/s Eₘ = 1000 × 2×10⁻⁵ × 100π × 2 = 4π V Maximum current: Iₘ = Eₘ/R = 4π/12.56 ≈ 1 A Hence option (D) is correct.

10. Question 10 (NEET 2021)

    NEET 2021MediumLCR Series Circuit, Reactance and Impedance

    An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to a source of potential difference 'V' volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10√2 A. The impedance of the circuit is:

    1. AOption A: 5 Ω
    2. BOption B: 4√2 Ω
    3. COption C: 5/√2 Ω
    4. DOption D: 4 Ω

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    Correct answer: (A) 5 Ω

    Explanation

    Given: VL = 40 V VC = 10 V VR = 40 V For a series LCR circuit: Vrms = √[VR² + (VL - VC)²] Vrms = √[(40)² + (40 - 10)²] Vrms = √(1600 + 900) Vrms = √2500 = 50 V Current amplitude: I0 = 10√2 A Irms = I0/√2 = 10 A Using: Z = Vrms / Irms Z = 50 / 10 = 5 Ω Therefore, the correct answer is option (A).