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A coil having 100 turns, area of $5 \times 10^{-3} \mathrm{~m}^2$, carrying current of $1 \mathrm{~mA}$ is placed in uniform magnetic field of $0.20 \mathrm{~T}$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $90^{\circ}$ is _______ $\mu \mathrm{J}$.
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The correct answer is:
100
$\begin{aligned} & W=\Delta U=U_f-U_i \\ & W=(-\vec{\mu} . \vec{B})_f-(-\vec{\mu} . \vec{B})_i \\ & =0+(\vec{\mu} \cdot \vec{B})_i \\ & =\left(100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}\right) \times 0.2 \mathrm{~J} \\ & =1 \times 10^{-4} \mathrm{~J}=100 \mu \mathrm{J}\end{aligned}$
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