Electromagnetic Induction Previous Year Questions — JEE Main Physics
28 past-year Electromagnetic Induction questions from JEE Main (Physics) — page 5.
Q45 (2026)
Three identical coils $C_{1}$, $C_{2}$ and $C_{3}$ are closely placed such that they share a common axis. $C_{2}$ is exactly midway. $C_{1}$ carries current I in anti-clockwise direction while $C_{3}$ carries current I in clockwise direction. An induced current flows through $C_{2}$ will be in clockwise direction when
{{IMG1}}
- $C_{1}$ and $C_{3}$ move with equal speeds away from $C_{2}$
- $C_{1}$ moves towards $C_{2}$ and $C_{3}$ moves away from $C_{2}$
- $\mathrm{C}_1$ moves away from $\mathrm{C}_2$ and $\mathrm{C}_3$ moves towards $\mathrm{C}_2$
- $C_{1}$ and $C_{3}$ move with equal speeds towards $C_{2}$
### Related Formula
$$\vec{B}_{\text{net}} = \vec{B}_{C_2} - \vec{B}_{C_1}, \quad \varepsilon = -\frac{d\phi}{dt}$$
### Core Logic
{{SOL_IMG1}}
Applying Lenz's law and magnetic field superposition: for induced current in $C_2$ to be clockwise, the net magnetic flux through $C_2$ must change accordingly. When $C_1$ moves towards $C_2$ and $C_3$ moves away from $C_2$, the net field change induces the specified clockwise current.
### Pattern Recognition
Sees: Coaxial current-carrying coils in relative motion.
Shortcut: Analyze net magnetic field variation at middle coil $C_2$ using Lenz's law.
Check: Matches option (2). ✓
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q50 (2026)
Inductance of a coil with $10^{4}$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of $10~\Omega$. The energy density in the inductor when the current reaches $\left(\frac{1}{\mathrm{e}}\right)$ of its maximum value is $\alpha \pi \times \frac{1}{\mathrm{e}^2} \mathrm{~J} / \mathrm{m}^3$. The value of $\alpha$ is \_\_\_\_. ($\mu_0 = 4\pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}$)
### Related Formula
$$E_d = \frac{B^2}{2\mu_0}, \quad B = \mu_0 n I$$
### Core Logic
{{SOL_IMG1}}
Maximum current:
$$I_0 = \frac{10}{10} = 1 \text{ A}$$
Current at given instant:
$$I = \frac{I_0}{e} = \frac{1}{e}$$
Energy density expression:
$$E_d = \frac{\mu_0 n^2 I^2}{2} = \frac{4\pi \times 10^{-7} \times (10^4)^2 \times (1/e)^2}{2} = \frac{20\pi}{e^2} \text{ J/m}^3$$
Therefore, $\alpha = 20$.
### Pattern Recognition
Sees: Energy density in inductor at exponential current growth stage.
Shortcut: Express magnetic field $B$ in terms of turns density $n$ and current $I$, substitute into energy density formula $E_d = \frac{B^2}{2\mu_0}$.
Check: Numerical answer is 20. ✓
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q38 (2026)
Figure shows the circuit that contains three resistances (9 $\Omega$ each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is ____ A.
{{IMG1}}
- $1$
- zero
- $3$
- $2$
### Related Formula
$$I(t=0^+) \text{ for an ideal inductor} = 0 \quad (\text{Open Circuit})$$
### Core Logic
At $t = 0$ (immediately after closing switch K), inductors oppose any instant change in current and behave as open circuits ($I_L = 0$).
Removing the branches containing the $4 \mathrm{~mH}$ inductors leaves only the middle branch containing a single $9 \Omega$ resistor connected to the $9 \mathrm{~V}$ battery.
Calculating total initial current $I$ measured by the ammeter:
$$I = \frac{V}{R} = \frac{9 \mathrm{~V}}{9 \Omega} = 1 \mathrm{~A}$$
{{SOL_IMG1}}
### Step 1: Final Conclusion
The reading of the ammeter at the moment switch K is turned ON is $1 \mathrm{~A}$.
### Pattern Recognition
LR Circuit Transient Rule: At $t=0$, Replace Inductor $\rightarrow$ Open Circuit. At $t=\infty$, Replace Inductor $\rightarrow$ Short Circuit wire.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q46 (2026)
A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is ____ mm.
(Take $\pi = \frac{22}{7}$)
### Related Formula
$$\mathcal{E} = B A \omega \sin(\omega t)$$
### Core Logic
Given $B = 0.5 \mathrm{~T}, \omega = 100 \mathrm{~rad/s}, \theta = \omega t = 30^\circ$ and $\mathcal{E} = 15.4 \times 10^{-3} \mathrm{~V}$:
$$15.4 \times 10^{-3} = B (\pi r^2) \omega \sin(30^\circ)$$
$$15.4 \times 10^{-3} = 0.5 \times \left(\frac{22}{7} r^2\right) \times 100 \times \frac{1}{2}$$
$$15.4 \times 10^{-3} = \frac{550}{7} r^2$$
$$r^2 = \frac{15.4 \times 10^{-3} \times 7}{550} = \frac{107.8 \times 10^{-3}}{550} = 1.96 \times 10^{-4} \mathrm{~m}^2$$
$$r = \sqrt{1.96 \times 10^{-4}} = 1.4 \times 10^{-2} \mathrm{~m} = 14 \mathrm{~mm}$$
### Step 1: Final Conclusion
The radius of the circular loop is $14 \mathrm{~mm}$.
### Pattern Recognition
AC Generator induced EMF formula: $\mathcal{E} = B A \omega \sin\theta$.
Substitute $\sin 30^\circ = 1/2$ and solve directly for radius $r$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q41 (2026)
A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g = 10 m/s$^{2}$) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire it travels a vertical distance of 200 m is ____ mV.
- $0.2\sqrt{10}$
- $20\sqrt{10}$
- $2 \sqrt{10}$
- $200\sqrt{10}$
### Related Formula
$v = \sqrt{2gh}$
$$\varepsilon = Bvl$$
### Step 1: Calculate Velocity
Falling freely under gravity for $200\text{ m}$:
$v = \sqrt{2gh}$
$$v = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10} \text{ m/s}$$
### Step 2: Calculate Motional EMF
Convert magnetic field to Tesla: $0.5 \text{ Gauss} = 0.5 \times 10^{-4} \text{ T}$.
The velocity, magnetic field, and length are mutually perpendicular.
$$\varepsilon = Bvl$$
$$\varepsilon = (0.5 \times 10^{-4}) \times (20\sqrt{10}) \times 20$$
$$\varepsilon = (10 \times 10^{-4}) \times 20\sqrt{10}$$
$$\varepsilon = 200\sqrt{10} \times 10^{-4} \text{ V}$$
$$\varepsilon = 20\sqrt{10} \times 10^{-3} \text{ V} = 20\sqrt{10} \text{ mV}$$
### Pattern Recognition
Sees: "wire falling in horizontal magnetic field" → find velocity using kinematics $v = \sqrt{2gh}$, then just plug into standard motional EMF $\varepsilon = Bvl$. Remember to convert Gauss to Tesla.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 11 Physics: Motion in a Straight Line
