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A regular hexagon of side $m$ which is a wire of length $24 \mathrm{~m}$ is coiled on that hexagon. If current in hexagon is $I$, then the magnetic moment,
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Verified Answer
The correct answer is:
$6 \sqrt{3} \operatorname{Im}^2$
Let number of turns of the regular hexagon $=n$
Now, $n \times 6 m=24 m$
$$
\begin{aligned}
& \therefore \quad n=4 \\
& \text { Magnetic moment, } M=n I A=4 I A \\
& \therefore \text { Area of hexagon } \\
& \quad=\frac{1}{2} m^2 \sin ^2 60^{\circ}+m \cdot 2 m \sin 60^{\circ}+\frac{1}{2} m^2 \sin 120^{\circ} \\
& =\frac{\sqrt{3}}{4} m^2+\sqrt{3} m^2+\frac{\sqrt{3} m^2}{4} \\
& =\frac{6 \sqrt{3} m^2}{4}=\frac{3 \sqrt{3} m^2}{2}
\end{aligned}
$$
$$
\text { Hence, } \begin{aligned}
\text { magnetic moment } & =4 I\left(\frac{3 \sqrt{3} m^2}{2}\right) \\
& =6 \sqrt{3} \mathrm{Im}^2
\end{aligned}
$$
Now, $n \times 6 m=24 m$
$$
\begin{aligned}
& \therefore \quad n=4 \\
& \text { Magnetic moment, } M=n I A=4 I A \\
& \therefore \text { Area of hexagon } \\
& \quad=\frac{1}{2} m^2 \sin ^2 60^{\circ}+m \cdot 2 m \sin 60^{\circ}+\frac{1}{2} m^2 \sin 120^{\circ} \\
& =\frac{\sqrt{3}}{4} m^2+\sqrt{3} m^2+\frac{\sqrt{3} m^2}{4} \\
& =\frac{6 \sqrt{3} m^2}{4}=\frac{3 \sqrt{3} m^2}{2}
\end{aligned}
$$
$$
\text { Hence, } \begin{aligned}
\text { magnetic moment } & =4 I\left(\frac{3 \sqrt{3} m^2}{2}\right) \\
& =6 \sqrt{3} \mathrm{Im}^2
\end{aligned}
$$
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