Loading Solutions...
📘 Electromagnetic Induction — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Electromagnetic Induction for JEE Main with previous year questions on Faraday's law, Lenz's law, and inductance. Each solution is explained in a simple, step-by-step manner to boost your score.
Electromagnetic Induction — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (28 Jan Shift 2)
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 \(10\pi \, \text{rad/s}\) 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) 0.5024 V (2) 1.256 V (3) 0.2512 V (4) 0.1256 V
(1) 0.5024 V (2) 1.256 V (3) 0.2512 V (4) 0.1256 V
✅ Answer & Explanation:
Correct Option: (3)
- For a rotating disc, the potential difference between the center and the rim is:
- \(\epsilon = \frac{1}{2} B \omega R^2\).
- Substituting values: \(B = 0.4 \, \text{T}\), \(\omega = 10\pi\), \(R = 0.2 \, \text{m}\).
- \(\epsilon = 0.5 \times 0.4 \times (10 \times 3.14) \times (0.2)^2 = 0.2 \times 31.4 \times 0.04 = 0.2512 \, \text{V}\).
Q2. JEE Main 2025 (28 Jan Shift 2)
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time \(t = 0\) with a constant velocity. If the induced EMF is \(\epsilon \propto t^n\), then value of n is _____.
✅ Answer & Explanation:
Correct Answer: 1
- The area of the triangular loop formed is \(A = \frac{1}{2} \times \text{base} \times \text{height}\).
- Since velocity \(v\) is constant, height \(h = vt\) and base \(b \propto vt\).
- Thus, area \(A \propto t^2\).
- Magnetic flux \(\phi = BA \propto t^2\).
- Induced emf \(\epsilon = \frac{d\phi}{dt} \propto \frac{d(t^2)}{dt} \propto t^1\).
- Therefore, n = 1.
Q3. JEE Main 2025 (22 Jan Shift 2)
A rectangular metallic loop is moving out of a uniform magnetic field region to a field free region with a constant speed. When the loop is partially inside the magnetic field, the plot of magnitude of induced emf (\(\epsilon\)) with time (t) is given by:
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (4)
As the loop moves out at constant speed \(v\), the rate of change of flux is constant: \(\epsilon = B l v\). Since \(B\), \(l\) (width), and \(v\) are constant, the induced emf magnitude remains constant until the loop has completely exited. Graph (4) represents a horizontal line.
As the loop moves out at constant speed \(v\), the rate of change of flux is constant: \(\epsilon = B l v\). Since \(B\), \(l\) (width), and \(v\) are constant, the induced emf magnitude remains constant until the loop has completely exited. Graph (4) represents a horizontal line.
Q4. JEE Main 2024 (29 Jan Shift 1)
A square loop of side 10 cm and resistance \(0.7 \, \Omega\) is placed vertically in the east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, the magnitude of induced emf is \(\sqrt{x} \times 10^{-3} \, \text{V}\). The value of \(x\) is _______.
✅ Answer & Explanation:
Correct Answer: 2
- Initial flux \(\phi_i = B A \cos\theta\).
- Area \(A = (0.1)^2 = 0.01 \, \text{m}^2\). Angle between normal (north or south) and field (north-east) is \(45^\circ\).
- \(\phi_i = 0.20 \times 0.01 \times \frac{1}{\sqrt{2}} = \frac{0.002}{\sqrt{2}} \, \text{Wb}\).
- \(\epsilon = \left| \frac{\Delta\phi}{\Delta t} \right| = \frac{0.002}{\sqrt{2} \times 1} = \frac{2 \times 10^{-3}}{\sqrt{2}} = \sqrt{2} \times 10^{-3} \, \text{V}\).
- Comparing with \(\sqrt{x} \times 10^{-3}\), we get x = 2.
Q5. JEE Main 2024 (08 Apr Shift 1)
A square loop PQRS having 10 turns, area \(3.6 \times 10^{-3} \, \text{m}^2\) and resistance \(100 \, \Omega\) is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B = 0.5 T as shown. Work done in pulling the loop out of the field in 1.0 s is ______ \(\times 10^{-6} \, \text{J}\).
✅ Answer & Explanation:
Correct Answer: 3
- Induced emf \(\epsilon = N \frac{\Delta\phi}{\Delta t} = 10 \times \frac{0.5 \times 3.6 \times 10^{-3}}{1.0} = 0.018 \, \text{V}\).
- Work done \(W = \text{Power} \times \text{time} = \frac{\epsilon^2}{R} t = \frac{(0.018)^2}{100} \times 1 = \frac{3.24 \times 10^{-4}}{100} \approx 3.24 \times 10^{-6} \, \text{J}\).
- Rounding to the nearest integer, value is 3.
Q6. JEE Main 2024 (04 Apr Shift 2)
A rod of length 60 cm rotates with a uniform angular velocity \(20 \, \text{rad/s}\) about its perpendicular bisector, in a uniform magnetic field 0.5 T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is _____V.
✅ Answer & Explanation:
Correct Answer: 0
For each half of the rod, potential difference from the center is \(\frac{1}{2} B \omega (\frac{L}{2})^2\). Since the rod rotates about its midpoint, the potentials at both ends relative to the center have the same magnitude and sign. Thus, the potential difference between the two ends is 0 V.
For each half of the rod, potential difference from the center is \(\frac{1}{2} B \omega (\frac{L}{2})^2\). Since the rod rotates about its midpoint, the potentials at both ends relative to the center have the same magnitude and sign. Thus, the potential difference between the two ends is 0 V.
Q7. JEE Main 2023 (10 Apr Shift 1)
A 1 m long metal rod XY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is \(5 \, \Omega\), the force needed to move the rod in direction, as indicated, with a constant speed of \(4 \, \text{m/s}\) will be _____ \(\times 10^{-3} \, \text{N}\).
✅ Answer & Explanation:
Correct Answer: 18
- Motional emf \(\epsilon = Blv = 0.15 \times 1 \times 4 = 0.6 \, \text{V}\).
- Current \(I = \epsilon/R = 0.6 / 5 = 0.12 \, \text{A}\).
- Magnetic force \(F_m = IlB = 0.12 \times 1 \times 0.15 = 0.018 \, \text{N} = 18 \times 10^{-3} \, \text{N}\).
- Applied force needed to maintain constant speed is equal to \(F_m\).
Q8. JEE Main 2022 (26 Jun Shift 2)
A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity \(5 \, \text{rad/s}\). If the horizontal component of earth's magnetic field is \(0.2 \times 10^{-4} \, \text{T}\), then emf induced between the two ends of the conductor is :
(1) \(5\mu\text{V}\) (2) \(50\mu\text{V}\) (3) 5 mV (4) 50 mV
(1) \(5\mu\text{V}\) (2) \(50\mu\text{V}\) (3) 5 mV (4) 50 mV
✅ Answer & Explanation:
Correct Option: (2)
Induced emf \(\epsilon = \frac{1}{2} B \omega L^2 = 0.5 \times (0.2 \times 10^{-4}) \times 5 \times (1)^2 = 0.5 \times 10^{-4} \, \text{V} = 50 \, \mu\text{V}\).
Induced emf \(\epsilon = \frac{1}{2} B \omega L^2 = 0.5 \times (0.2 \times 10^{-4}) \times 5 \times (1)^2 = 0.5 \times 10^{-4} \, \text{V} = 50 \, \mu\text{V}\).
Q9. JEE Main 2021 (01 Sep Shift 2)
A square loop of side 20 cm and resistance \(1 \, \Omega\) is moved towards right with a constant speed \(v_0\). The right arm of the loop is in a uniform magnetic field of 5 T. The loop is connected to a network of resistors each of value \(4 \, \Omega\). What should be the value of \(v_0\) so that a steady current of 2 mA flows in the loop?
(1) \(10^{-2} \, \text{cm/s}\) (2) \(1 \, \text{m/s}\)
(3) \(1 \, \text{cm/s}\) (4) \(10^2 \, \text{m/s}\)
(3) \(1 \, \text{cm/s}\) (4) \(10^2 \, \text{m/s}\)
✅ Answer & Explanation:
Correct Option: (3)
- The resistor network has equivalent resistance \(R_{ext} = 4 \, \Omega\).
- Total resistance \(R_{total} = R_{loop} + R_{ext} = 1 + 4 = 5 \, \Omega\).
- Motional emf \(\epsilon = B l v_0 = 5 \times 0.2 \times v_0 = 1.0 v_0\).
- Current \(I = \epsilon / R_{total} = v_0 / 5\).
- \(2 \times 10^{-3} = v_0 / 5 \implies v_0 = 10 \times 10^{-3} = 0.01 \, \text{m/s} = 1 \, \text{cm/s}\).
Q10. JEE Main 2025 (29 Jan Shift 1)
Consider \(I_1\) and \(I_2\) are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If \(L_1 =\) self inductance of coil 1, \(M_{12} =\) mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :
(1) \(\epsilon_1 = -L_1 \frac{dI_2}{dt} - M_{12} \frac{dI_1}{dt}\) (2) \(\epsilon_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt}\)
(3) \(\epsilon_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_1}{dt}\) (4) \(\epsilon_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt}\)
(1) \(\epsilon_1 = -L_1 \frac{dI_2}{dt} - M_{12} \frac{dI_1}{dt}\) (2) \(\epsilon_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt}\)
(3) \(\epsilon_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_1}{dt}\) (4) \(\epsilon_1 = -L_1 \frac{dI_1}{dt} + M_{12} \frac{dI_2}{dt}\)
✅ Answer & Explanation:
Correct Option: (2)
- The total magnetic flux linked with coil 1 is \(\phi_1 = L_1 I_1 + M_{12} I_2\).
- By Faraday's Law, induced emf \(\epsilon_1 = -\frac{d\phi_1}{dt} = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt}\).
Q11. JEE Main 2025 (23 Jan Shift 1)
Regarding self-inductance:
A. The self-inductance of the coil depends on its geometry.
B. Self-inductance does not depend on the permeability of the medium.
C. Self-induced e.m.f. opposes any change in the current in a circuit.
D. Self-inductance is electromagnetic analogue of mass in mechanics.
E. Work needs to be done against self-induced e.m.f. in establishing the current.
Choose the correct answer from the options given below:
(1) A, B, C, E only (2) B, C, D, E only (3) A, C, D, E only (4) A, B, C, D only
A. The self-inductance of the coil depends on its geometry.
B. Self-inductance does not depend on the permeability of the medium.
C. Self-induced e.m.f. opposes any change in the current in a circuit.
D. Self-inductance is electromagnetic analogue of mass in mechanics.
E. Work needs to be done against self-induced e.m.f. in establishing the current.
Choose the correct answer from the options given below:
(1) A, B, C, E only (2) B, C, D, E only (3) A, C, D, E only (4) A, B, C, D only
✅ Answer & Explanation:
Correct Option: (3)
- A is true: \(L = \mu_0 n^2 A l\).
- B is false: It depends directly on relative permeability \(\mu_r\).
- C is true: This is Lenz's Law.
- D is true: Inductance provides "electrical inertia".
- E is true: Energy stored is \(U = \int \epsilon I dt\).
Q12. JEE Main 2024 (27 Jan Shift 1)
Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation \(i = i_0 \sin\omega t\), where \(i_0 = 5 \, \text{A}\) and \(\omega = 50\pi \, \text{rad/s}\). The maximum value of emf in the second coil is \(\frac{\pi}{\alpha} \, \text{V}\). The value of \(\alpha\) is _______.
✅ Answer & Explanation:
Correct Answer: 2
- Induced emf \(\epsilon = M \frac{di}{dt} = M \frac{d}{dt}(i_0 \sin\omega t) = M i_0 \omega \cos\omega t\).
- Maximum emf \(\epsilon_{max} = M i_0 \omega = 0.002 \times 5 \times 50\pi = 0.5\pi = \frac{\pi}{2} \, \text{V}\).
- Comparing with \(\pi/\alpha\), we find \(\alpha = 2\).
Q13. JEE Main 2024 (09 Apr Shift 1)
When a coil is connected across a 20 V dc supply, it draws a current of 5 A. When it is connected across 20 V, 50 Hz ac supply, it draws a current of 4 A. The self inductance of the coil is ______ mH. (Take \(\pi = 3\))
✅ Answer & Explanation:
Correct Answer: 10
- Resistance \(R = V/I_{dc} = \frac{20}{5} = 4 \, \Omega\).
- Impedance \(Z = V/I_{ac} = \frac{20}{4} = 5 \, \Omega\).
- Reactance \(X_L = \sqrt{Z^2 - R^2} = \sqrt{25 - 16} = 3 \, \Omega\).
- \(X_L = 2\pi f L \implies 3 = 2 \times 3 \times 50 \times L \implies L = \frac{3}{300} = 0.01 \, \text{H} = 10 \, \text{mH}\).
Q14. JEE Main 2024 (01 Feb Shift 1)
A rectangular loop of sides 12 cm and 5 cm moves with a velocity of 5 cm/s in the positive x-direction, in a space containing a variable magnetic field in the positive z direction. The field has a gradient of \(10^{-3} \, \text{T/cm}\) along the negative x direction and it is decreasing with time at the rate of \(10^{-3} \, \text{T/s}\). If the resistance of the loop is \(6 \, \text{m}\Omega\), the power dissipated by the loop as heat is ______ \(\times 10^{-9} \, \text{W}\).
✅ Answer & Explanation:
Correct Answer: 216
- Total rate of change of flux \(\frac{d\phi}{dt} = A \frac{\partial B}{\partial t} + A \frac{\partial B}{\partial x} v_x\).
- Area \(A = 0.12 \times 0.05 = 0.006 \, \text{m}^2\).
- \(\frac{d\phi}{dt} = 0.006 [ -10^{-3} + (-0.1) \times 0.05 ] = 0.006 [ -0.001 - 0.005 ] = 36 \times 10^{-6} \, \text{V}\).
- Power \(P = \frac{\epsilon^2}{R} = \frac{(36 \times 10^{-6})^2}{6 \times 10^{-3}} = \frac{1296 \times 10^{-12}}{6 \times 10^{-3}} = 216 \times 10^{-9} \, \text{W}\).
Q15. JEE Main 2022 (26 Jun Shift 2)
Two coils of self inductance \(L_1\) and \(L_2\) are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be :
(1) \(\frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{M}\) (2) \(L_1 + L_2 - 2M\)
(3) \(L_1 + L_2 + M\) (4) \(L_1 + L_2 + 2M\)
(3) \(L_1 + L_2 + M\) (4) \(L_1 + L_2 + 2M\)
✅ Answer & Explanation:
Correct Option: (2)
For coils in series, \(L_{eq} = L_1 + L_2 \pm 2M\). The diagram indicates that the coils are wound in opposing senses (subtractive polarity), leading to the formula in Option (2).
For coils in series, \(L_{eq} = L_1 + L_2 \pm 2M\). The diagram indicates that the coils are wound in opposing senses (subtractive polarity), leading to the formula in Option (2).
Q16. JEE Main 2020 (08 Jan Shift 1)
At time \(t = 0\) magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, then induced EMF in the loop is:
(1) \(56\mu\text{V}\) (2) \(28\mu\text{V}\) (3) \(48\mu\text{V}\) (4) \(36\mu\text{V}\)
✅ Answer & Explanation:
Correct Option: (1)
- Loop Area \(A = \text{Rectangle Area} - 2 \times \text{Triangle Area}\).
- \(A = (16 \times 4) - 2 \times (\frac{1}{2} \times 4 \times 2) = 64 - 8 = 56 \, \text{cm}^2 = 56 \times 10^{-4} \, \text{m}^2\).
- \(\Delta B = (1000 - 500) \, \text{Gauss} = 500 \times 10^{-4} \, \text{T} = 0.05 \, \text{T}\).
- \(\epsilon = A \frac{\Delta B}{\Delta t} = 56 \times 10^{-4} \times \frac{0.05}{5} = 56 \times 10^{-4} \times 0.01 = 56 \mu\text{V}\).
Q17. JEE Main 2020 (07 Jan Shift 1)
A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0), C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is \(\vec{B} = (3\hat{i} + 4\hat{k}) \, \text{T}\). The quantity of flux through the loop (in Wb) is ______.
✅ Answer & Explanation:
Correct Answer: 175
- The loop consists of two rectangular segments:
- ABCD in the XY plane: Area vector \(\vec{A}_1 = 25 \hat{k} \, \text{m}^2\).
- ADEFA in the YZ plane: Area vector \(\vec{A}_2 = 25 \hat{i} \, \text{m}^2\).
- Total flux \(\phi = \vec{B} \cdot (\vec{A}_1 + \vec{A}_2) = (3\hat{i} + 4\hat{k}) \cdot (25\hat{i} + 25\hat{k})\).
- \(\phi = (3 \times 25) + (4 \times 25) = 75 + 100 = 175 \, \text{Wb}\).
📘 Exam Preparation Tip for Electromagnetic Induction
- Faraday's Law: \(\epsilon = -\frac{d\phi}{dt}\). The induced emf in a circuit is equal to the negative rate of change of magnetic flux through the circuit. For a coil with N turns, \(\epsilon = -N\frac{d\phi}{dt}\).
- Lenz's Law: The direction of induced current is such that it opposes the change that produces it. This is the negative sign in Faraday's Law and ensures energy conservation.
- Motional EMF: \(\epsilon = Blv\) for a rod moving perpendicular to a uniform magnetic field. For a rotating rod, \(\epsilon = \frac{1}{2}B\omega L^2\).
- Self-Inductance: \(L = \frac{\phi}{I}\). Energy stored in an inductor \(U = \frac{1}{2}LI^2\). Inductance is the electrical analogue of mass (inertia).
- Mutual Inductance: \(M = \frac{\phi_{12}}{I_1} = \frac{\phi_{21}}{I_2}\). It depends on the geometry of the coils and the medium.
- Series Inductance: For series combination, \(L_{eq} = L_1 + L_2 \pm 2M\) (the sign depends on the polarity of the magnetic coupling).
- LR Circuits: Growth of current: \(I = I_0(1 - e^{-t/\tau})\). Decay: \(I = I_0 e^{-t/\tau}\). Time constant \(\tau = L/R\).