Electromagnetic Induction Previous Year Questions — JEE Main Physics
28 past-year Electromagnetic Induction questions from JEE Main (Physics) — page 4.
Q44 (2024)
A coil is placed perpendicular to a magnetic field of $5000 \mathrm{~T}$. When the field is changed to $3000 \mathrm{~T}$ in $2\mathrm{s}$, an induced emf of $22 \mathrm{~V}$ is produced in the coil. If the diameter of the coil is $0.02 \mathrm{~m}$, then the number of turns in the coil is:
- $7$
- $70$
- $35$
- $140$
### Related Formula
$$\varepsilon = N \left| \frac{\Delta\phi}{\Delta t} \right|$$
$$\Delta\phi = (\Delta B) A \cos\theta$$
### Core Logic
Given data:
Initial Magnetic Field, $B_i = 5000\mathrm{\,T}$
Final Magnetic Field, $B_f = 3000\mathrm{\,T}$
Time interval, $\Delta t = 2\mathrm{\,s}$
Diameter, $d = 0.02\mathrm{\,m} \Rightarrow r = 0.01\mathrm{\,m}$
Induced emf, $\varepsilon = 22\mathrm{\,V}$
Change in magnetic field magnitude $|\Delta B| = 5000 - 3000 = 2000\mathrm{\,T}$.
Area of the coil $A = \pi r^2 = \pi (0.01)^2 = 10^{-4}\pi \mathrm{\,m^2}$.
### Step 2: Equation Evaluation
$$\Delta\phi = |\Delta B| A = (2000) \pi (0.01)^2 = 0.2\pi$$
Using Faraday's Law:
$$22 = N \left( \frac{0.2\pi}{2} \right)$$
$22 = N (0.1\pi)$
Taking $\pi \approx 22/7$:
$$22 = N \left( 0.1 \times \frac{22}{7} \right)$$
$$1 = \frac{N}{70} \Rightarrow N = 70$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q2086 (2026)
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is $2\Omega$ then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ____ N.
{{IMG1}}
- $7.5 \times 10^{-2}$
- $5.7 \times 10^{-3}$
- $5.7 \times 10^{-2}$
- $7.5 \times 10^{-3}$
### Related Formula
$E = B l v$
$$i = \frac{E}{R}$$
$$F_{B} = i l B = \frac{B^2 l^2 v}{R}$$
### Core Logic
To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current.
$$F_{\text{ext}} = F_{B}$$
### Step 1: Calculate External Force
Given values:
$B = 0.10\text{ T}$
$l = 1\text{ m}$
$v = 1.5\text{ m/s}$
$R = 2\, \Omega$
$$F_{\text{ext}} = \frac{B^2 l^2 v}{R}$$
$$F_{\text{ext}} = \frac{(0.10)^2 \times (1)^2 \times 1.5}{2}$$
$$F_{\text{ext}} = \frac{0.01 \times 1.5}{2} = \frac{0.015}{2} = 0.0075$$
$$F_{\text{ext}} = 7.5 \times 10^{-3}\text{ N}$$
{{SOL_IMG1}}
### Pattern Recognition
Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always $F = \frac{B^2 L^2 v}{R}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q28 (2026)
A conducting circular loop of area $1.0 \, \mathrm{m}^{2}$ is placed perpendicular to a magnetic field which varies as $B = \sin(100 \, t)\text{ Tesla}$. If the resistance of the loop is $100 \, \Omega$, then the average thermal energy dissipated in the loop in one period is ____ J.
- $\frac{\pi}{2}$
- $2\pi$
- $\pi$
- $\pi^{2}$
### Related Formula
$$\phi = B \cdot A$$
$$E = -\frac{d\phi}{dt}$$
$$P = \frac{E^2}{R}$$
### Core Logic
Given area of the loop, $A = 1\text{ m}^2$ and magnetic field $B = \sin(100t)$.
The magnetic flux passing through the loop is:
$$\phi = B \cdot A = \sin(100t) \times 1 = \sin(100t)$$
Induced EMF $E = \left| \frac{d\phi}{dt} \right| = 100\cos(100t)$
### Step 1: Calculating Power and Energy
Instantaneous power $P = \frac{E^2}{R} = \frac{100^2 \cos^2(100t)}{100} = 100\cos^2(100t)$.
Thermal energy dissipated in one time period $T$:
$$Q = \int_{0}^{T} P \, dt = \int_{0}^{T} 100\cos^2(100t) \, dt$$
The angular frequency $\omega = 100\text{ rad/s}$, so time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \frac{\pi}{50}\text{ sec}$.
$$Q = 100 \int_{0}^{\pi/50} \cos^2(100t) \, dt = 100 \int_{0}^{\pi/50} \frac{1 + \cos(200t)}{2} \, dt$$
$$Q = 50 \left[ t + \frac{\sin(200t)}{200} \right]_{0}^{\pi/50}$$
$$Q = 50 \left( \frac{\pi}{50} - 0 \right) = \pi\text{ Joules}$$
### Pattern Recognition
For a sinusoidal signal, the integral of $\cos^2(\omega t)$ over one full period $T$ is always $T/2$. Thus, $\int P \, dt = P_{\text{max}} \times \frac{T}{2}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 12 Physics: Alternating Current
Q29 (2026)
A capacitor C is first charged fully with potential difference of $V_{0}$ and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In $t \text{ s}$ 25% of the initial energy in the capacitor is transferred to the inductor. The value of t is ________ s.
- $\frac{\pi\sqrt{LC}}{3}$
- $\frac{\pi\sqrt{LC}}{6}$
- $\frac{\pi\sqrt{LC}}{2}$
- $\pi\sqrt{\frac{LC}{2}}$
### Related Formula
For LC oscillations, charge varies as:
$$Q(t) = Q_0 \cos(\omega t)$$
Where $\omega = \frac{1}{\sqrt{LC}}$
Energy in capacitor:
$$U_C = \frac{Q^2}{2C}$$
### Core Logic
Since 25% of the initial energy is transferred to the inductor, the remaining energy in the capacitor is 75% of its initial value.
$$U_{C_f} = 75\% \text{ of } U_{C_i}$$
$$\frac{Q_f^2}{2C} = \frac{3}{4} \times \frac{Q_0^2}{2C}$$
$$Q_f^2 = \frac{3}{4} Q_0^2$$
$$Q_f = \frac{\sqrt{3}}{2} Q_0$$
### Step 1: Finding Time
Using the equation for charge variation:
$$Q_0 \cos(\omega t) = \frac{\sqrt{3}}{2} Q_0$$
$$\cos(\omega t) = \frac{\sqrt{3}}{2}$$
$$\omega t = \frac{\pi}{6}$$
### Step 2: Final Conclusion
Substitute $\omega = \frac{1}{\sqrt{LC}}$:
$$\frac{1}{\sqrt{LC}} t = \frac{\pi}{6}$$
$$t = \frac{\pi \sqrt{LC}}{6}$$
### Pattern Recognition
Energy is proportional to charge squared. $75\%$ energy remaining means charge is $\sqrt{0.75} = \frac{\sqrt{3}}{2}$ of the original. Cosine of $\pi/6$ yields this exact ratio.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 12 Physics: Alternating Current
Q29 (2026)
XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is \_\_\_\_ m/s.
(g = acceleration due to gravity)
{{IMG1}}
- $\frac{2\mathrm{mgR}}{\mathrm{B}^2l^2}$
- $\frac{8\mathrm{mgR}}{\mathrm{B}^2l^2}$
- $\frac {2 \mathrm{mgR}}{\mathrm{B} ^ {2} \mathrm{L} ^ {2}}$
- $\frac{\mathrm{mgR}}{\mathrm{B}^2l^2}$
### Related Formula
$$e = Bvl, \quad i = \frac{e}{R}$$
### Core Logic
{{SOL_IMG1}}
At equilibrium (for terminal velocity):
$$mg = iBl \implies mg = \left(\frac{Bvl}{R}\right)Bl$$
$$v = \frac{mgR}{B^2l^2}$$
### Pattern Recognition
Sees: Sliding rod in magnetic field reaching terminal speed.
Shortcut: Equate gravitational force with magnetic force $iBl$ at terminal velocity.
Check: Matches option (4). ✓
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
