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The magnetic field at the centre of a current carrying circular coil of area ' $\mathrm{A}$ ' is ' $\mathrm{B}$ '. The magnetic moment of the coil is ( $\mu_0=$ permeability of free space)
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Verified Answer
The correct answer is:
$\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \sqrt{\pi}}$
$$
\begin{aligned}
& A=\pi R^2 \\
& \Rightarrow R=\sqrt{\frac{A}{\pi}} \\
& B=\frac{\mu_0 I}{2 R}
\end{aligned}
$$
Magnetic Moment,
$$
\begin{aligned}
& M=I A=\frac{2 A B R}{\mu_0} \\
& =\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \sqrt{\pi}}
\end{aligned}
$$
\begin{aligned}
& A=\pi R^2 \\
& \Rightarrow R=\sqrt{\frac{A}{\pi}} \\
& B=\frac{\mu_0 I}{2 R}
\end{aligned}
$$
Magnetic Moment,
$$
\begin{aligned}
& M=I A=\frac{2 A B R}{\mu_0} \\
& =\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \sqrt{\pi}}
\end{aligned}
$$
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