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Electromagnetic Induction · Question 23

JEE Main 2025 (Online) 29th January Morning ShiftMCQEasy📈 Topic in 6% papers+4−1
1

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

Answer

(4) e1 = -L1 dI1dt - M12 dI2dt

Solution

Flux linked with coil 1 (own current + current of coil 2):

1 = L1 I1 + M12 I2

By Faraday's law, e1 = - d 1dt

e1 = -L1 dI1dt - M12 dI2dt

Answer: (4)

JEE Main 2025 (Online) 29th January Morning Shift

Keep practising — Electromagnetic Induction

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 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
3

[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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