Electromagnetic Induction JEE Main previous year questions with solutions

5 solved JEE Main questions on Electromagnetic Induction, free to read — no sign-in needed. The full chapter has 151 questions; sign in to attempt the remaining 146 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Magnetic Flux, Faraday's and Lenz law
   A square loop of side $2$ cm is placed in a time varying magnetic field with magnitude as $B = 0.4\sin(300t)$ Tesla. The normal to the plane of loop makes an angle of $60°$ with the field. The maximum induced emf produced in the loop is __________ mV.
   1. A.$12$
   2. B.$18$
   3. C.$21$
   4. D.$24$
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   Answer: (D)

   Given: Side of the square loop, $a = 2$ cm $= 2 \times 10^{-2}$ m Area of the loop, $A = a^2 = 4 \times 10^{-4}$ m$^2$ Magnetic field, $B = 0.4 \sin(300t)$ T Angle between the normal to the plane and the magnetic field, $\theta = 60^\circ$ The magnetic flux through the loop is given by: $\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta$ $\Phi = (0.4 \sin(300t))(4 \times 10^{-4}) \cos(60^\circ)$ $\Phi = 1.6 \times 10^{-4} \sin(300t) \times \dfrac{1}{2}$ $\Phi = 0.8 \times 10^{-4} \sin(300t)$ Wb By Faraday's law of induction, the induced emf is: $e = -\dfrac{d\Phi}{dt}$ $e = -\dfrac{d}{dt} (0.8 \times 10^{-4} \sin(300t))$ $e = -0.8 \times 10^{-4} \times 300 \cos(300t)$ $e = -240 \times 10^{-4} \cos(300t)$ V $e = -24 \times 10^{-3} \cos(300t)$ V The maximum induced emf is the amplitude of $e$: $e_{\text{max}} = 24 \times 10^{-3}$ V $= 24$ mV Answer: $24$
2. Q2JEE Main 2025 (28 Jan, Shift 2)Motional and Rotational EMF
   A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of $10 \pi \mathrm{rad} \mathrm{s}^{-1}$ about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim ? $(\pi=3.14)$
   1. A.$0.5024 \mathrm{~V}$
   2. B.$\mathrm{~V}$
   3. C.$0.2512 \mathrm{~V}$
   4. D.$0.1256 \mathrm{~V}$
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   Answer: (C)

   $\begin{aligned} & \mathrm{B}=0.4 \mathrm{~T} \\ & \mathrm{r}=20 \mathrm{~cm} \\ & \omega=10 \pi \mathrm{rad} / \mathrm{s} \\ & \mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^2 \\ & =0.2512 \mathrm{~V}\end{aligned}$
3. Q3JEE Main 2024 (29 Jan, Shift 2)Induced Electric Field
   A horizontal straight wire $5m$ long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field $0.60\times {10}^{-4}Wb{m}^{-2}$. The instantaneous value of emf induced in the wire when its velocity is $10m{s}^{-1}$ is _______$\times {10}^{-3}V$.
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   Answer: 3

   Horizontal component of earth's magnetic field, ${B}_{H}=0.60\times {10}^{-4}Wb{m}^{-2}$. Induced $\text{emf}e={B}_{H}v\ell$ $=0.60\times {10}^{-4}\times 10\times 5$ $=3\times {10}^{-3}V$
4. Q4JEE Main 2023 (11 Apr, Shift 2)Inductance
   A coil has an inductance of $2H$ and resistance of $4\Omega$. A $10V$ is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be _____ $\times {10}^{-2}J$
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   Answer: 625

   The data given is $L=2H R=4\Omega V=10V$ Using Ohm's law, $V=IR$, $I=\dfrac{10V}{4\Omega }=\dfrac{5}{2}A$ The energy of an inductor is given by $E=\dfrac{1}{2}L{I}^{2}=\dfrac{1}{2}\times 2\times \left(\dfrac{25}{4}\right)=625\times {10}^{-2}J$
5. Q5JEE Main 2022 (26 Jul, Shift 1)LR Circuit
   In a series $LR$ circuit ${X}_{L}=R$ and power factor of the circuit is ${P}_{1}$. When capacitor with capacitance $C$ such that ${X}_{L}={X}_{C}$ is put in series, the power factor becomes ${P}_{2}$. The ratio $\dfrac{{P}_{1}}{{P}_{2}}$ is
   1. A.$\dfrac{1}{2}$
   2. B.$\dfrac{1}{\sqrt{2}}$
   3. C.$\dfrac{\sqrt{3}}{\sqrt{2}}$
   4. D.$2:1$
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   Answer: (B)

   In case of $L-R$ circuit $Z=\sqrt{{{X}_{L}}^{2}+{R}^{2}}$ Therefore, power factor ${P}_{1}=\cos \phi =\dfrac{R}{Z}$ As ${X}_{L}=R$ $\Rightarrow Z=\sqrt{2}R$ $\Rightarrow {P}_{1}=\dfrac{R}{\sqrt{2}R}\Rightarrow {P}_{1}=\dfrac{1}{\sqrt{2}}$ In case of $L-C-R$ circuit $Z=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}$ According to the question ${X}_{L}={X}_{C}$ $\Rightarrow Z=R$ $\Rightarrow {P}_{2}=\cos \phi =\dfrac{R}{R}=1$ $\Rightarrow \dfrac{{P}_{1}}{{P}_{2}}=\dfrac{1}{\sqrt{2}}$