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JEE MainPhysicsElectromagnetic Induction

A rectangular conducting loop of resistance R lies in the xy -plane with its vertices at (0,0) , (a,0) , (a,b) , and (0,b) . A time-varying magnetic field is applied perpendicular to the plane of the loop, given by B = (C x) ( t) k , where C and are positive constants. The average thermal power dissipated in the loop is:

Options

  1. AC^2 a^4 b^2 ^2 2R
  2. BC^2 a^4 b^2 ^2 8R
  3. CC^2 a^4 b^2 ^2 4R
  4. DC^2 a^4 ^2 8R

Correct answer

B. C^2 a^4 b^2 ^2 8R

Step-by-step solution

The magnetic field varies with the position x . The magnetic flux d through an elemental strip of width dx and length b at position x is given by d = B dA = (C x ( t))(b dx) . The total flux through the loop is obtained by integrating from x = 0 to x = a : = ₀^ a C b x ( t) dx = C b ( t) [ x^2 2 ]₀^ a = 1 2 C a^2 b ( t) . By Faraday's law, the induced EMF is = - d dt = 1 2 C a^2 b ( t) . The instantaneous power dissipated is P_ inst = ^2 R = C^2 a^4 b^2 ^2 ^2( t) 4R . The average power over a full cycle is obtained

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