Electromagnetic Induction JEE Main previous year questions with solutions

4 solved JEE Main questions on Electromagnetic Induction, free to read — no sign-in needed. The full chapter has 173 questions; sign in to attempt the remaining 169 in the exam simulator.

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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 (29 Jan, Shift 2)Force and Torque on Current Carrying Conductor
   A square loop of area $25{cm}^{2}$ has a resistance of $10\Omega$. The loop is placed in uniform magnetic field of magnitude $40.0T$. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1.0$ sec, will be
   1. A.$2.5\times {10}^{-3}J$
   2. B.$1.0\times {10}^{-3}J$
   3. C.$1.0\times {10}^{-4}J$
   4. D.$5\times {10}^{-3}J$
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   Answer: (B)

   As ${l}^{2}=25{cm}^{2}$, then $l=5cm=0.05m$. Given: $t=1s$ Velocity of the square loop, $v=\dfrac{0.05}{1}=0.05m{s}^{-1}$. Induced current, $i=\dfrac{V}{R}=\dfrac{Blv}{R} =\dfrac{40\times 0.05\times 0.05}{10}=0.01A$ Now force acting on the side of the square loop, $F=Bil=40\times 0.01\times 0.05$ $\Rightarrow F=0.02N$ Therefore, work done $W=Fl =0.02\times l =0.02\times 0.05$ $\Rightarrow W=1\times {10}^{-3}J$