A square loop PQRS having 10 turns, area $3.6 \times 10^{-3} \mathrm{~m}^2$ and resistance $100 \Omega$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $\mathrm{B}=0.5 \mathrm{~T}$ as shown. Work done in pulling the loop out of the field in $1.0 \mathrm{~s}$ is ______ $\times 10^{-6} \mathrm{~J}$.

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Aicharya · Accepted Answer

$\begin{aligned} & \epsilon=N B \ell v \ & i=\frac{\epsilon}{R}=\frac{N B \ell v}{R} \ & F=N(i \ell B)=\frac{N^2 B^2 \ell^2 v}{R}\end{aligned}$ $\begin{aligned} & \mathrm{W}=\mathrm{F} 	imes \ell=\frac{\mathrm{N}^2 \mathrm{~B}^2 \ell^3}{\mathrm{R}}\left(\frac{\ell}{\mathrm{t}}ight) \ & \mathrm{A}=\ell^2 \ & \mathrm{~W}=\frac{(10 	imes 10)(0.5)^2 	imes\left(3.6 	imes 10^{-3}ight)^2}{100 	imes 1} \ & \mathrm{~W}=3.24 	imes 10^{-6} \mathrm{~J}\end{aligned}$

A square loop PQRS having 10 turns, area $3.6 \times 10^{-3} \mathrm{~m}^2$ and resistance $100 \Omega$ is…

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