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Electromagnetic Induction · Question 24

JEE Main 2025 (Online) 28th January Evening ShiftMCQEasy📈 Topic in 19% papers+4−1
1

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

Answer

(3) 0.2512 V

Solution

An element dr at distance r moves with speed r:

d = B( r) dr

= 0R B r dr = 12B R2

= 12 0.4 10 (0.2)2

= 0.08 = 0.08 3.14 = 0.2512 V

Answer: (3)

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2

[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 .

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