A square loop of area \(25 \mathrm{~cm}^2\) has a resistance of \(10 \Omega\). The loop is placed in uniform magnetic field of magnitude 40 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1 second, will be

Area of square is \(25 \mathrm{~cm}^2\)
\(\therefore \quad\) length \(=5 \mathrm{~cm}=0.05 \mathrm{~m}\)
Velocity, \(\mathrm{v}=\frac{l}{\mathrm{t}}=\frac{0.05}{1}=0.05 \mathrm{~m} / \mathrm{s}\)
\(\mathrm{e}=\mathrm{B} / \mathrm{v} \quad \Rightarrow \mathrm{IR}=\mathrm{B} / \mathrm{v}\)
\(\therefore \quad \mathrm{I}=\frac{\mathrm{B} / \mathrm{v}}{\mathrm{R}}=\frac{40 \times 0.05 \times 0.05}{10}=0.01 \mathrm{~A}\)
Since magnetic field is perpendicular to plane of loop, \(\theta=90^{\circ}\)
\(\begin{aligned}
\therefore \quad \mathrm{F} & =\mathrm{BI} l \sin \theta \\
& =40 \times 0.01 \times 0.05 \times \sin 90=0.02 \mathrm{~N} \\
\therefore \quad \mathrm{~W} & =\mathrm{F} \times l \\
& =0.02 \times 0.05=1 \times 10^{-3} \mathrm{~J}
\end{aligned}\)