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Magnetic field at the centre of a circular loop of area 'A' is ' $B^{\prime}$ '. The magnetic moment
of the loop will be $\left(\mu_{0}=\right.$ permeability of free space)
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of the loop will be $\left(\mu_{0}=\right.$ permeability of free space)
Solution:
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Verified Answer
The correct answer is:
$\frac{2 \mathrm{~B} A^{\frac{3}{2}}}{\mu_{0} \pi^{\frac{1}{2}}}$
If $\mathrm{r}$ is the radius of the circular loop them the area of the loop $\mathrm{A}=\pi \mathrm{r}^{2}$ $\therefore \mathrm{r}=\sqrt{\frac{\mathrm{A}}{\pi}}$
$\therefore$ The magnetic field at the centre is given by
$$
\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2 \sqrt{\frac{\mathrm{A}}{\pi}}}
$$
$$
\therefore I=\frac{2 B}{\mu_{0}} \cdot \sqrt{\frac{A}{\pi}}
$$
Magnetic moment
$$
\begin{array}{c}
\mathrm{M}=\mathrm{IA}=\frac{2 \mathrm{~B}}{\mu_{0}} \cdot \sqrt{\frac{\mathrm{A}}{\pi}} \cdot \mathrm{A} \\
=\frac{2 \mathrm{BA}^{3 / 2}}{\mu \pi^{\frac{1}{2}}}
\end{array}
$$
$\therefore$ The magnetic field at the centre is given by
$$
\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2 \sqrt{\frac{\mathrm{A}}{\pi}}}
$$
$$
\therefore I=\frac{2 B}{\mu_{0}} \cdot \sqrt{\frac{A}{\pi}}
$$
Magnetic moment
$$
\begin{array}{c}
\mathrm{M}=\mathrm{IA}=\frac{2 \mathrm{~B}}{\mu_{0}} \cdot \sqrt{\frac{\mathrm{A}}{\pi}} \cdot \mathrm{A} \\
=\frac{2 \mathrm{BA}^{3 / 2}}{\mu \pi^{\frac{1}{2}}}
\end{array}
$$
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