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A small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\vec{B}$ such that its plane is perpendicular to $\vec{B}$. The work done in rotating it by $180^{\circ}$ about the vertical axis is
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Verified Answer
The correct answer is:
$2 \pi N A I B$
The work done in rotating the coin is given as:
\(\begin{aligned}
& \mathrm{W}=\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& =(\mathrm{NiA}) \mathrm{B}\left(\cos \mathrm{o}^{\circ}-\cos 180^{\circ}\right) \\
& =2 \mathrm{NAIB}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{W}=\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& =(\mathrm{NiA}) \mathrm{B}\left(\cos \mathrm{o}^{\circ}-\cos 180^{\circ}\right) \\
& =2 \mathrm{NAIB}
\end{aligned}\)
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