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NEETPhysicsElectromagnetic Induction

A stationary conducting loop of resistance R is placed in a time-varying magnetic field B(t) perpendicular to its plane. Match the time dependence of the magnetic field in List-I with the corresponding qualitative behaviour of the electrical power P(t) dissipated in the loop in List-II. List-I ( B(t) ) List-II ( P(t) ) (A) B(t) = k t (I) P(t) is constant (B) B(t) = k t^2 (II) P(t) is proportional to t^2 (C) B(t) = B₀

Options

  1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. B(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
  3. C(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
  4. D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Correct answer

A. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Step-by-step solution

The magnetic flux through the stationary loop of area A is = B(t) A . According to Faraday's law, the induced EMF is = - d dt = -A dB dt . The power dissipated is P(t) = ^2 R = A^2 R ( dB dt )^2 . (A) B(t) = k t dB dt = k . Thus, P(t) k^2 , which is a constant. Matches (I). (B) B(t) = k t^2 dB dt = 2kt . Thus, P(t) (2kt)^2 t^2 . Matches (II). (C) B(t) = B₀ ( t) dB dt = B₀ ( t) . Thus, P(t) ^2( t) , which oscillates between zero and a maximum. Matches (III). (D) B(t) = B₀ e^ - t dB dt = -B₀ e^ - t . Thus, P(t) e^ -2

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