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A current ' $i$ ' is flowing through a wire of length ' $L$ '. If it is made into a circular loop of one turn, then its magnetic moment is
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Verified Answer
The correct answer is:
$\frac{\mathrm{L}^2 \mathrm{i}}{4 \pi}$
Length, $L=2 \pi r \Rightarrow r=\frac{L}{2 \pi}$
Magnetic moment, $\mathrm{M}=\mathrm{iA}=\mathrm{i} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^2$
$=\mathrm{i} \times \pi \times \frac{\mathrm{L}^2}{4 \pi^2} \quad \therefore \mathrm{M}=\frac{\mathrm{IL}^2}{4 \pi}$
Magnetic moment, $\mathrm{M}=\mathrm{iA}=\mathrm{i} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^2$
$=\mathrm{i} \times \pi \times \frac{\mathrm{L}^2}{4 \pi^2} \quad \therefore \mathrm{M}=\frac{\mathrm{IL}^2}{4 \pi}$
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