JEE Main · PhysicsHard
A square loop of side length $L$ and resistance $R$ is placed in a time-varying magnetic field $B = B_0 \sin(\omega t)$ directed perpendicular to the plane of the loop. What is the maximum power dissipated in the loop?
- A.$\frac{B_0^2 \omega^2 L^4}{2R}$
- B.$\frac{B_0^2 \omega^2 L^4}{R}$
- C.$\frac{B_0^2 \omega^2 L^2}{2R}$
- D.$\frac{B_0^2 \omega L^4}{2R}$
Show correct answer & step-by-step solution
Correct answer: A — $\frac{B_0^2 \omega^2 L^4}{2R}$
Solution
The induced emf is given by Faraday's law as . Calculating the derivative of the flux gives , and the power is , which has a maximum value when .
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