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Electromagnetic Induction

JEE Main 2025 (Online) 4th April Morning ShiftNumericalMedium📈 Topic in 22% papers+4
Ques. 1JEE Main 2025, 4 April (Morning Shift)

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]

Answer

10

Solution

The motional emf across a conductor of any shape is = Bv leff, where leff is the length of the conductor perpendicular to v (the projection of AE perpendicular to the velocity).

AB and DE are parallel to v, so there is no emf in them.

BC and CD each make 45 with v (angle BCD = 90 , symmetric about the velocity), so

leff = 10 45 + 10 45 = 10 2 cm = 0.1 2 m

= 1 2 0.1 0.1 2

= 0.01 V = 10 mV

Answer: 10

Keep practising — Electromagnetic Induction

Ques. 2JEE Main 2025, 2 April (Evening Shift)

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
Ques. 3JEE Main 2025, 29 January (Morning Shift)

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
Ques. 4JEE Main 2025, 29 January (Morning Shift)

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
Ques. 5JEE Main 2025, 28 January (Evening Shift)

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
Ques. 6JEE Main 2025, 28 January (Evening Shift)

[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 .

Ques. 7JEE Main 2025, 23 January (Morning Shift)

Regarding self-inductance: Choose the correct answer from the options given below:

  1. A, B, C, E only
  2. A, B, C, D only
  3. A, C, D, E only
  4. B, C, D, E only
Ques. 8JEE Main 2025, 23 January (Morning Shift)

[figure] In the given circuit the sliding contact is pulled outwards such that electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 , the value of the current in the circuit will be A.

Ques. 9JEE Main 2025, 22 January (Evening Shift)

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 ( ) with time (t) is given by

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 10JEE Main 2024, 9 April (Evening Shift)

A square loop of side 15 cm is being moved towards right at a constant speed of 2 cm/s as shown in figure. The front edge enters the 50 cm wide magnetic field at t = 0. The value of induced emf in the loop at t = 10 s will be : [figure]

  1. zero
  2. 4.5 mV
  3. 0.3 mV
  4. 3 mV
Ques. 11JEE Main 2024, 8 April (Morning Shift)

A square loop PQRS having 10 turns, area 3.6 10-3 m2 and resistance 100 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 10-6 J. [figure]

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