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Electromagnetic Induction · Question 13

JEE Main 2026 (Online) 22nd January Morning ShiftMCQEasy📈 Topic in 19% papers+4−1
1

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) [figure]

  1. mgRB2l2
  2. 2mgRB2L2
  3. 2mgRB2l2
  4. 8mgRB2l2

Answer

(1) mgRB2l2

Solution

Only the part of the rod between the rails (length l) is in the circuit.

At speed v: motional emf = Blv

Current I = BlvR

Magnetic force on the rod (upward, opposing the motion by Lenz's law): F = BIl = B2l2vR

At terminal speed the net force is zero: mg = B2l2vtR

vt = mgRB2l2

Answer: (1)

JEE Main 2026 (Online) 22nd January Morning Shift

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2

Three identical coils C1, C2 and C3 are closely placed such that they share a common axis. C2 is exactly midway. C1 carries current I in anti-clockwise direction while C3 carries current I in clockwise direction. An induced current flows through C2 will be in clockwise direction when

  1. C1 and C3 move with equal speeds away from C2
  2. C1 moves towards C2 and C3 moves away from C2
  3. C1 and C3 move with equal speeds towards C2
  4. C1 moves away from C2 and C3 moves towards C2
JEE Main 2026 (Online) 22nd January Morning Shift
3

Inductance of a coil with 104 turns per metre is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 . The energy density in the inductor when the current reaches ( 1e ) of its maximum value is 1e2 J/m3. The value of is . ( 0 = 4 10-7 T m/A).

JEE Main 2026 (Online) 22nd January Morning Shift
4

Figure shows the circuit that contains three resistances (9 each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is A. [figure]

  1. 2
  2. 3
  3. 1
  4. zero
JEE Main 2026 (Online) 22nd January Evening Shift
5

A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5 T 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 = 227)

JEE Main 2026 (Online) 22nd January Evening Shift
6

A conducting circular loop of area 1.0 m2 is placed perpendicular to a magnetic field which varies as B = (100t) Tesla. If the resistance of the loop is 100 , then the average thermal energy dissipated in the loop in one period is J.

  1. 2
  2. 2
  3. 2
JEE Main 2026 (Online) 21st January Morning Shift
7

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 then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is N. [figure]

  1. 5.7 10-2
  2. 7.5 10-3
  3. 5.7 10-3
  4. 7.5 10-2
JEE Main 2026 (Online) 21st January Morning Shift
8

Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of 1 2 Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s, induced emf between points A and E is mV. [figure]

JEE Main 2025 (Online) 4th April Morning Shift
9

A solenoid having area A and length 'l' is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

  1. B2Al4 0
  2. B2Al2 0
  3. B2Al 0
  4. B2Al
JEE Main 2025 (Online) 2nd April Evening Shift
10

A coil of area A and N turns is rotating with angular velocity in a uniform magnetic field B about an axis perpendicular to B. Magnetic flux and induced emf across it, at an instant when B is parallel to the plane of coil, are :

  1. = AB, = NAB
  2. = AB, = 0
  3. = 0, = 0
  4. = 0, = NAB
JEE Main 2025 (Online) 29th January Morning Shift
11

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :

  1. e1 = -L1 dI2dt - M12 dI1dt
  2. e1 = -L1 dI1dt + M12 dI2dt
  3. e1 = -L1 dI1dt - M12 dI1dt
  4. e1 = -L1 dI1dt - M12 dI2dt
JEE Main 2025 (Online) 29th January Morning Shift

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