A square loop ABCD is moving with constant velocity ' $\vec{v}$ ' in a uniform magnetic field ' $\vec{B}$ '…
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]
$\frac{B^2 L^2 V}{R}$
$\frac{B^2 L^2 V^2}{R}$
$\frac{B^2 \mathrm{LV}^2}{\mathrm{R}}$
$\frac{\mathrm{BLV}^2}{\mathrm{R}}$
Solution
From motional e.m.f.,
$\begin{aligned}
& \mathrm{e}_{\max }=\mathrm{BLV} \\
\therefore \quad & \text { Heat produced }=\frac{\mathrm{e}_{\max }}{\mathrm{R}}=\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}}
\end{aligned}$