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Electromagnetic Induction

20 JEE Main questions · Physics

Induction laws

  1. PhysicsJEE Main 21 Jan 2026, Shift 1·Q3·MCQ·Moderate
    Open
    A conducting circular loop of area 1.0 m21.0\ \mathrm{m^2} is placed perpendicular to a magnetic field which varies as B=sin⁡(100 t)B = \sin(100\,t) Tesla. If the resistance of the loop is 100 Ω100\ \Omega, then the average thermal energy dissipated in the loop in one period is J.
    1. (1)π2\dfrac{\pi}{2}
    2. (2)2π2\pi
    3. (3)π\pi
    4. (4)π2\pi^2
  2. PhysicsJEE Main 21 Jan 2026, Shift 1·Q16·MCQ·Easy–moderate
    Open
    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 Ω2\ \Omega then the force needed to move the rod towards right with constant speed (vv) of 1.5 m/s is N.
    1. (1)7.5×10−27.5\times10^{-2}
    2. (2)5.7×10−35.7\times10^{-3}
    3. (3)5.7×10−25.7\times10^{-2}
    4. (4)7.5×10−37.5\times10^{-3}
  3. PhysicsJEE Main 22 Jan 2026, Shift 1·Q4·MCQ·Easy–moderate
    Open
    XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is ll. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L>lL > 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. (gg = acceleration due to gravity)
    1. (1)2mgRB2l2\dfrac{2mgR}{B^2l^2}
    2. (2)8mgRB2l2\dfrac{8mgR}{B^2l^2}
    3. (3)2mgRB2L2\dfrac{2mgR}{B^2L^2}
    4. (4)mgRB2l2\dfrac{mgR}{B^2l^2}
  4. PhysicsJEE Main 22 Jan 2026, Shift 1·Q20·MCQ·Moderate
    Open
    Three identical coils C1C_1, C2C_2 and C3C_3 are closely placed such that they share a common axis. C2C_2 is exactly midway. C1C_1 carries current II in anti-clockwise direction while C3C_3 carries current II in clockwise direction. An induced current flowing through C2C_2 will be in clockwise direction when
    1. (1)C1C_1 and C3C_3 move with equal speeds away from C2C_2
    2. (2)C1C_1 moves towards C2C_2 and C3C_3 moves away from C2C_2
    3. (3)C1C_1 moves away from C2C_2 and C3C_3 moves towards C2C_2
    4. (4)C1C_1 and C3C_3 move with equal speeds towards C2C_2
    We corrected a printing error in the original paper

    Source printed “An induced current flows through C2C_2 will be”.

  5. PhysicsJEE Main 23 Jan 2026, Shift 1·Q16·MCQ·Easy–moderate
    Open
    A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g=10 m/s2g = 10\ \mathrm{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 when it travels a vertical distance of 200 m is mV.
    1. (1)0.2100.2\sqrt{10}
    2. (2)201020\sqrt{10}
    3. (3)2102\sqrt{10}
    4. (4)20010200\sqrt{10}
    We corrected a printing error in the original paper

    Source printed “across the wire it travells”.

  6. PhysicsJEE Main 23 Jan 2026, Shift 1·Q21·Numerical·Moderate
    Open
    A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of 60∘60^\circ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is mV. (Take g=10 m/s2g = 10\ \mathrm{m/s^2})
  7. PhysicsJEE Main 23 Jan 2026, Shift 2·Q16·MCQ·Easy–moderate
    Open
    A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed perpendicular to plane of loop. The loop is converted into a square loop in 0.5 s. The EMF induced in the loop is mV.
    1. (1)6.6
    2. (2)13.2
    3. (3)8.25
    4. (4)1.32
  8. PhysicsJEE Main 23 Jan 2026, Shift 2·Q17·MCQ·Moderate
    Open
    Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I=10sin⁡(ωt)I = 10\sin(\omega t) A, where ω=1000\omega = 1000 rad./s. A circular conducting loop (B) of radius 1 cm coaxially slid through the solenoid at a speed v=1v = 1 cm/s. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is α/2 μ\alpha/\sqrt2\ \muA. The value of α\alpha is .
    [Resistance of the loop =10 Ω= 10\ \Omega]
    1. (1)197
    2. (2)80
    3. (3)280
    4. (4)100
    We corrected a printing error in the original paper

    Source printed “coaxially slided”.

  9. PhysicsJEE Main 2 Apr 2026, Shift 1·Q13·MCQ·Moderate
    Open
    When a coil is placed in a time dependent magnetic field the power dissipated in it is PP. The number of turns, area of the coil and radius of the coil wire are NN, AA and rr respectively. For a second coil's number of turns, area of the coil and radius of the coil wire are 2N2N, 2A2A and 3r3r respectively. When the first coil is replaced with second coil the power dissipated in it is 2 αP\sqrt2\,\alpha P. The value of α\alpha is .
    1. (1)36
    2. (2)1282128\sqrt2
    3. (3)16
    4. (4)64
    We corrected a printing error in the original paper

    Source printed “For a second coils number of turns”.

  10. PhysicsJEE Main 5 Apr 2026, Shift 2·Q11·MCQ·Moderate
    Open
    A metal rod of length LL rotates about one end at origin with a uniform angular velocity ω\omega. The magnetic field radially falls off as B(r)=B0e−λrB(r) = B_0e^{-\lambda r}; λ\lambda being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:
    1. (1)B0ω[1λ2−e−λL(1λ2+Lλ)]B_0\omega\left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]
    2. (2)B0ω[1λ2+e−λL(1λ2+Lλ)]B_0\omega\left[\dfrac{1}{\lambda^2} + e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]
    3. (3)B0ω[4λ2−e−2λL(1λ2+2Lλ)]B_0\omega\left[\dfrac{4}{\lambda^2} - e^{-2\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{2L}{\lambda}\right)\right]
    4. (4)B0ω[3λ2−e−3λL(3λ2+Lλ)]B_0\omega\left[\dfrac{3}{\lambda^2} - e^{-3\lambda L}\left(\dfrac{3}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]
    We corrected a printing error in the original paper

    Source printed “rotates about on end at origin”.

  11. PhysicsJEE Main 5 Apr 2026, Shift 2·Q25·Numerical·Easy–moderate
    Open
    A circular loop of radius 20 cm and resistance 2 Ω2\ \Omega is placed in a time varying magnetic field B⃗=(2t2+2t+3)\vec B = (2t^2 + 2t + 3) T. At t=0t = 0, for the plane of the loop being perpendicular to the magnetic field, the induced current in the loop at t=3t = 3 s is α50\dfrac{\alpha}{50} A. The value of α\alpha is (Take π=22/7\pi = 22/7)
    We corrected a printing error in the original paper

    Source printed “perpendicular to the magnetic field and, the induced current”.

  12. PhysicsJEE Main 6 Apr 2026, Shift 2·Q14·MCQ·Easy–moderate
    Open
    A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(300t)B = 0.4\sin(300t) Tesla. The normal to the plane of loop makes an angle of 60∘60^\circ with the field. The maximum induced emf produced in the loop is mV.
    1. (1)12
    2. (2)18
    3. (3)21
    4. (4)24

Inductance

  1. PhysicsJEE Main 22 Jan 2026, Shift 1·Q25·Numerical·Moderate
    Open
    Inductance of a coil with 10410^{4} turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 Ω10\ \Omega. The energy density in the inductor when the current reaches (1e)\left(\dfrac{1}{e}\right) of its maximum value is απ×1e2 J/m3\alpha\pi\times\dfrac{1}{e^2}\ \mathrm{J/m^3}. The value of α\alpha is . (μ0=4π×10−7\mu_0 = 4\pi\times10^{-7} Tm/A). (Take the length of the coil as 1 m)
    We corrected a printing error in the original paper

    Assumption “Take the length of the coil as 1 m” added; the source gives neither the length nor the turns per unit length, and the key 20 needs n=104n = 10^4 turns/m (consistent with L=10L = 10 mH for a long solenoid).

  2. PhysicsJEE Main 22 Jan 2026, Shift 2·Q13·MCQ·Easy–moderate
    Open
    Figure shows the circuit that contains three resistances (9 Ω9\ \Omega each) and two inductors (4 mH each). The reading of ammeter at the moment switch KK is turned ON, is A.
    1. (1)1
    2. (2)zero
    3. (3)3
    4. (4)2
  3. PhysicsJEE Main 22 Jan 2026, Shift 2·Q21·Numerical·Easy–moderate
    Open
    A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a vertical magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by 30∘30^\circ from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is mm. (Take π=227)\left(\text{Take } \pi = \dfrac{22}{7}\right)
    We corrected a printing error in the original paper

    Assumption “vertical” (magnetic field) added; the source leaves the field direction relative to the horizontal position unstated, and the key 14 needs e=BAωsin⁡30∘e = BA\omega\sin30^\circ.

  4. PhysicsJEE Main 28 Jan 2026, Shift 2·Q23·Numerical·Moderate
    Open
    An inductor stores 16 J of magnetic field energy and dissipates 32 W of thermal energy due to its resistance when an a.c. current of 2 A (rms) and frequency 50 Hz flows through it. The ratio of inductive reactance to its resistance is . (π=3.14\pi = 3.14)
  5. PhysicsJEE Main 2 Apr 2026, Shift 2·Q18·MCQ·Moderate
    Open
    A circular current loop of radius R is placed inside square loop of side length L (L≫RL \gg R) such that they are co-planar and their centers coincide. The permeability of free space is μ0\mu_0. The mutual inductance between circular loop and square loop is .
    1. (1)22 μ0L2R2\sqrt2\,\dfrac{\mu_0 L^2}{R}
    2. (2)2 μ0L2R\sqrt2\,\dfrac{\mu_0 L^2}{R}
    3. (3)2 μ0R2L\sqrt2\,\dfrac{\mu_0 R^2}{L}
    4. (4)22 μ0R2L2\sqrt2\,\dfrac{\mu_0 R^2}{L}
  6. PhysicsJEE Main 5 Apr 2026, Shift 1·Q25·Numerical·Easy–moderate
    Open
    In the given circuit below inductance values of L1L_1, L2L_2 and L3L_3 are same. The magnetic energy stored in the entire circuit is (UtU_t) and that stored in the L2L_2 inductor is (UlU_l). (Ut/UlU_t/U_l) is . (Ignore the mutual inductance if any)
  7. PhysicsJEE Main 6 Apr 2026, Shift 1·Q18·MCQ·Easy–moderate
    Open
    An inductor of inductance 10 mH having resistance of 100 Ω100\ \Omega is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous induced emfs across the inductor when the current passing through it is 2 mA and 4 mA is .
    1. (1)4/3
    2. (2)3/4
    3. (3)5/3
    4. (4)3/5
    We corrected a printing error in the original paper

    Source printed “the ratio of instantaneous voltages across the inductor” (the terminal voltage of the resistive inductor equals the 1.0 V emf at all times, ratio 1); corrected to the intended “induced emfs” (L di/dtL\,di/dt: 0.8 V and 0.6 V).

  8. PhysicsJEE Main 8 Apr 2026, Shift 2·Q15·MCQ·Easy–moderate
    Open
    A 30 cm long solenoid has 10 turns per cm and area of 5 cm25\ \mathrm{cm^2}. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is α×10−5\alpha\times10^{-5} V. The value of α\alpha is .
    1. (1)60
    2. (2)12
    3. (3)120
    4. (4)34
    We corrected a printing error in the original paper

    Source printed “varies from 2 A to 4 A is 3.14 s”; “The value α\alpha is”.