Electromagnetic Induction Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Electromagnetic Induction with concise, step-by-step solutions covering induced emf, mutual inductance, LCR circuits and RL transients.
A curated set of JEE Main 2026 Electromagnetic Induction problems with clean, worked solutions to help you master induced emf, inductance, and AC circuit analysis.
Solutions are AI-generated and pending review.
Solution
Power factor gives the impedance:
$$\cos\phi = \frac{R}{Z} = 0.5 \implies Z = \frac{R}{0.5} = \frac{100}{0.5} = 200\ \Omega$$For a series LCR circuit:
$$Z^2 = R^2 + (X_L - X_C)^2$$$$(X_L - X_C)^2 = Z^2 - R^2 = 200^2 - 100^2 = 30000$$$$|X_L - X_C| = \sqrt{30000} = 100\sqrt{3}\ \Omega$$Comparing with $\sqrt{3}\,\alpha$:
$$\sqrt{3}\,\alpha = 100\sqrt{3} \implies \alpha = 100$$Answer: 100
Solution
By Kirchhoff’s voltage law, the emf equals the drop across the coil’s resistance plus the voltage across the ideal inductor. The voltage appearing across the inductor (the back-emf term) is:
$$V_L = E - iR$$At $i = 2$ mA:
$$V_{L,1} = 1.0 - (2\times10^{-3})(100) = 1.0 - 0.2 = 0.8\ \text{V}$$At $i = 4$ mA:
$$V_{L,2} = 1.0 - (4\times10^{-3})(100) = 1.0 - 0.4 = 0.6\ \text{V}$$$$\frac{V_{L,1}}{V_{L,2}} = \frac{0.8}{0.6} = \frac{4}{3}$$Answer: 4/3
Solution
The induced emf and coil resistance are:
$$\varepsilon = N A \frac{dB}{dt}, \qquad R_{coil} = \rho\,\frac{\ell}{a_{wire}} = \rho\,\frac{N(2\sqrt{\pi A})}{\pi r^2}$$Here the wire length $\ell \propto N\sqrt{A}$ (perimeter $\propto \sqrt{A}$) and cross-section $a_{wire}=\pi r^2$.
Power dissipated:
$$P = \frac{\varepsilon^2}{R_{coil}} \propto \frac{(NA)^2}{\dfrac{N\sqrt{A}}{r^2}} = N\,A^{3/2}\,r^2$$Taking the ratio (second to first coil) with $N\to2N$, $A\to2A$, $r\to3r$:
$$\frac{P_2}{P_1} = \frac{(2N)(2A)^{3/2}(3r)^2}{N\,A^{3/2}\,r^2} = 2\cdot 2^{3/2}\cdot 9 = 18\cdot 2\sqrt{2} = 36\sqrt{2}$$Comparing with $\sqrt{2}\,\alpha$:
$$\sqrt{2}\,\alpha = 36\sqrt{2} \implies \alpha = 36$$Answer: 36
Solution
Send current $I$ through the square loop. The field at the centre of a square of side $L$ (sum of four finite wires, each a distance $L/2$ from the centre) is:
$$B = 4\cdot\frac{\mu_0 I}{4\pi (L/2)}\big(\sin45^\circ + \sin45^\circ\big) = \frac{2\sqrt{2}\,\mu_0 I}{\pi L}$$Since $L \gg R$, this field is essentially uniform over the small circular loop, so the flux linking it is:
$$\Phi = B\cdot \pi R^2 = \frac{2\sqrt{2}\,\mu_0 I}{\pi L}\cdot \pi R^2 = \frac{2\sqrt{2}\,\mu_0 I R^2}{L}$$Mutual inductance:
$$M = \frac{\Phi}{I} = 2\sqrt{2}\,\frac{\mu_0 R^2}{L}$$Answer: $2\sqrt{2}\,\dfrac{\mu_0 R^2}{L}$
Solution
Flux through the loop:
$$\Phi = B A \cos\theta = 0.4\sin(300t)\cdot A \cos60^\circ$$Induced emf:
$$\varepsilon = -\frac{d\Phi}{dt} = -0.4\cdot 300\cos(300t)\cdot A\cos60^\circ$$Maximum value ($|\cos(300t)|=1$), with $A=(0.02)^2 = 4\times10^{-4}\ \text{m}^2$ and $\cos60^\circ = 0.5$:
$$\varepsilon_{max} = 0.4\times 300\times 4\times10^{-4}\times 0.5 = 0.024\ \text{V} = 24\ \text{mV}$$Answer: 24
Solution
Let the reactance at frequency $\omega$ be $X_C = \dfrac{1}{\omega C}$. At $\omega/4$ the reactance becomes $4X_C$.
At $\omega$:
$$I = \frac{V}{\sqrt{R^2 + X_C^2}}$$At $\omega/4$:
$$\frac{I}{3} = \frac{V}{\sqrt{R^2 + (4X_C)^2}}$$Dividing:
$$3 = \frac{\sqrt{R^2 + 16X_C^2}}{\sqrt{R^2 + X_C^2}} \implies 9(R^2 + X_C^2) = R^2 + 16X_C^2$$$$8R^2 = 7X_C^2 \implies \frac{R^2}{X_C^2} = \frac{7}{8} \implies \frac{R}{X_C} = \sqrt{\frac{7}{8}}$$Answer: $\sqrt{\dfrac{7}{8}}$
Solution
Let each inductance be $L$ and the total current from the source be $I$ (this flows fully through $L_1$).
The parallel combination $L_2 \parallel L_3 = \dfrac{L}{2}$, and since $L_2 = L_3$ the current splits equally, so each carries $I/2$.
Total energy:
$$U_t = \frac{1}{2}L_1 I^2 + \frac{1}{2}\left(\frac{L}{2}\right)I^2 = \frac{1}{2}L I^2 + \frac{1}{4}L I^2 = \frac{3}{4}L I^2$$Energy in $L_2$ (carrying $I/2$):
$$U_l = \frac{1}{2}L\left(\frac{I}{2}\right)^2 = \frac{1}{8}L I^2$$Ratio:
$$\frac{U_t}{U_l} = \frac{\tfrac{3}{4}L I^2}{\tfrac{1}{8}L I^2} = \frac{3/4}{1/8} = 6$$Answer: 6
Solution
A small element $dr$ at radius $r$ moves with speed $v = \omega r$, contributing motional emf $d\varepsilon = B(r)\,v\,dr$:
$$\varepsilon = \int_0^L B_0 e^{-\lambda r}\,(\omega r)\,dr = B_0\omega \int_0^L r\,e^{-\lambda r}\,dr$$Using $\displaystyle\int r\,e^{-\lambda r}\,dr = -\frac{r}{\lambda}e^{-\lambda r} - \frac{1}{\lambda^2}e^{-\lambda r}$:
$$\int_0^L r\,e^{-\lambda r}\,dr = \left[-\frac{r}{\lambda}e^{-\lambda r} - \frac{1}{\lambda^2}e^{-\lambda r}\right]_0^L = \frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{L}{\lambda} + \frac{1}{\lambda^2}\right)$$Therefore:
$$\varepsilon = B_0\omega\left[\frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{1}{\lambda^2} + \frac{L}{\lambda}\right)\right]$$Answer: $B_0 \omega\left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
Solution
At resonance $X_L = X_C$, and the resonant angular frequency is $\omega_0 = \dfrac{1}{\sqrt{LC}}$. Hence:
$$X_L = \omega_0 L = \frac{L}{\sqrt{LC}} = \sqrt{\frac{L}{C}}$$$$X_L = \sqrt{\frac{1.6}{40\times10^{-6}}} = \sqrt{4\times10^{4}} = 200\ \Omega$$Answer: 200
Solution
With the plane perpendicular to $\vec B$, the normal is parallel to $\vec B$, so $\Phi = B A$. The induced emf:
$$\varepsilon = A\left|\frac{dB}{dt}\right| = A\,(4t + 2)$$Area of the loop:
$$A = \pi r^2 = \frac{22}{7}(0.2)^2 = \frac{22}{7}\times 0.04 = 0.12571\ \text{m}^2$$At $t = 3$ s, $\dfrac{dB}{dt} = 4(3) + 2 = 14$ T/s:
$$\varepsilon = 0.12571\times 14 = 1.76\ \text{V}$$Induced current:
$$I = \frac{\varepsilon}{R} = \frac{1.76}{2} = 0.88\ \text{A} = \frac{44}{50}\ \text{A}$$Comparing with $\dfrac{\alpha}{50}$: $\alpha = 44$.
Answer: 44
Solution
Self-inductance of a solenoid:
$$L = \mu_0 n^2 A \ell$$with $n = 10\ \text{turns/cm} = 1000\ \text{turns/m}$, $A = 5\times10^{-4}\ \text{m}^2$, $\ell = 0.3\ \text{m}$:
$$L = (4\pi\times10^{-7})(1000)^2(5\times10^{-4})(0.3) = 1.885\times10^{-4}\ \text{H}$$Induced emf with $\dfrac{dI}{dt} = \dfrac{4-2}{3.14} = \dfrac{2}{3.14}$ A/s:
$$\varepsilon = L\frac{dI}{dt} = 1.885\times10^{-4}\times\frac{2}{3.14} \approx 12\times10^{-5}\ \text{V}$$Comparing with $\alpha\times10^{-5}$: $\alpha = 12$.
Answer: 12