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A straight wire carrying a current (I) is turned into a circular loop. If the magnitude of the magnetic moment associated with it is ' $\mathrm{M}$ ', then the length of the wire will be
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Verified Answer
The correct answer is:
$\left[\frac{4 \mathrm{M} \pi}{\mathrm{I}}\right]^{\frac{1}{2}}$
Magnetic moment is given as $\mathrm{M}=\mathrm{IA}$
$$
\therefore \quad \mathrm{M}=\mathrm{I}\left(\pi \mathrm{R}^2\right)
$$
Now, length of the wire is, $L=2 \pi R$
$$
\begin{aligned}
& \therefore \quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi} \\
& \Rightarrow \quad \mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \\
& \therefore \quad \mathrm{L}=\sqrt{\frac{4 \mathrm{M} \pi}{\mathrm{I}}}
\end{aligned}
$$
$$
\therefore \quad \mathrm{M}=\mathrm{I}\left(\pi \mathrm{R}^2\right)
$$
Now, length of the wire is, $L=2 \pi R$
$$
\begin{aligned}
& \therefore \quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi} \\
& \Rightarrow \quad \mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \\
& \therefore \quad \mathrm{L}=\sqrt{\frac{4 \mathrm{M} \pi}{\mathrm{I}}}
\end{aligned}
$$
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