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MHT CET · Physics · Electromagnetic Induction

A square loop ABCD is moving with constant velocity ' \(\vec{v}\) ' in a uniform magnetic field ' \(\vec{B}\) ' which is perpendicular to the plane of paper and directed outward. The resistance of coil is ' \(R\) ', then the rate of production of heat energy in the loop is [ L - length of side of loop]

  1. A \(\frac{B^2 L^2 V}{R}\)
  2. B \(\frac{B^2 L^2 V^2}{R}\)
  3. C \(\frac{B^2 \mathrm{LV}^2}{\mathrm{R}}\)
  4. D \(\frac{\mathrm{BLV}^2}{\mathrm{R}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{B^2 L^2 V^2}{R}\)

Step-by-step Solution

Detailed explanation

From motional e.m.f.,
\(\begin{aligned}
& \mathrm{e}_{\max }=\mathrm{BLV} \\
\therefore \quad & \text { Heat produced }=\frac{\mathrm{e}_{\max }}{\mathrm{R}}=\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}}
\end{aligned}\)
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