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 . . . . . . $\mu \mathrm{J}$.
JEE MAIN 2024, Diffcult
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$W=\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$
$\mathrm{W}=(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_{\mathrm{f}}-(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_i$
$=0+(\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_i$
$=\left(100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}\right) \times 0.2 \mathrm{~J}$
$=1 \times 10^{-4} \mathrm{~J}=100 \mu \mathrm{J}$
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