A square loop of area \(25 \mathrm{~cm}^2\) has a resistance of \(10 \Omega\). This loop is placed in a uniform magnetic field of magnitude \(40 \mathrm{~T}\). 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 one second, will be

Given: area of square loop \(=25 \mathrm{~cm}\)
\(\therefore l=\sqrt{25}=5 \mathrm{~cm} \Rightarrow 0.05 \mathrm{~m} \)
\( \mathrm{R}=10 \Omega, \mathrm{t}=1 \mathrm{sec}, \mathrm{B}=40 \mathrm{~T} \)
\( \therefore \text {Velocity } \mathrm{v}=\frac{l}{\mathrm{t}}=\frac{0.05}{1}=0.05 \mathrm{~m} / \mathrm{s} \)
\( \text {Motional emf } \varepsilon_{\max }=\mathrm{B} / \mathrm{v} \)
\( \therefore \mathrm{I}=\frac{\varepsilon}{\mathrm{R}}=\frac{\mathrm{B} / \mathrm{v}}{\mathrm{R}} \)
\( \therefore \mathrm{I}=\frac{40 \times 0.05 \times 0.05}{10}=0.01 \mathrm{~A}\)
We know, Force acting on loop
\(\therefore\mathrm{F}=\mathrm{BI} l =40 \times 0.01 \times 0.05 \)
\( =0.02 \mathrm{~N} \)
\( \text { Using W }=\mathrm{F} . \mathrm{s}, \)
\( \text { Work done } \mathrm{W} =\mathrm{BI} l \times l \)
\( =0.02 \times 0.05 \)
\( =1 \times 10^{-3} \mathrm{~J}\)