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Electromagnetic Induction · Question 19

JEE Main 2026 (Online) 21st January Morning ShiftMCQEasy📈 Topic in 19% papers+4−1
1

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

Answer

(2) 7.5 10-3

Solution

Motional emf in the rod: = Bvl = 0.10 1.5 1 = 0.15 V

Current: I = R = 0.152 = 0.075 A

Magnetic force on the rod (opposes the motion): FB = BIl = 0.10 0.075 1 = 7.5 10-3 N

For constant speed, the applied force balances FB.

F = 7.5 10-3 N

Answer: (2)

JEE Main 2026 (Online) 21st January Morning Shift

Keep practising — Electromagnetic Induction

2

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
3

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
4

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
5

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
6

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 rad s-1 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? ( = 3.14)

  1. 0.5024 V
  2. 0.0628 V
  3. 0.2512 V
  4. 0.1256 V
JEE Main 2025 (Online) 28th January Evening Shift
7

[figure] 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 E tn, then value of n is .

JEE Main 2025 (Online) 28th January Evening Shift

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