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A thin rod of length $\mathrm{L}$ has magnetic moment $\mathrm{M}$ when magnetised. If rod is bent in a semicircular arc what is magnetic moment in new shape?
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Verified Answer
The correct answer is:
$\frac{2 \mathrm{M}}{\pi}$
Magnetic moment of $\operatorname{rod}=\mathrm{M}$
Let $\mathrm{r}$ be the radius after bending it into a semicircular arc.
$\therefore \quad$ The separation between the two ends is 2 r. Here length $=$ circumference of the semicircle i.e., $\mathrm{L}=\pi \mathrm{r}$
$\therefore \quad r=\frac{L}{\pi}$
Also $\mathrm{M}=\mathrm{m} \times \mathrm{L}$ and $\mathrm{m}=\frac{\mathrm{M}}{\mathrm{L}}$
$\begin{aligned}
\therefore \quad \mathrm{M}_{\text {new }}=\mathrm{m}(2 \mathrm{r}) & =\frac{\mathrm{M}}{\mathrm{L}} \times \frac{2 \mathrm{~L}}{\pi} \\
& =\frac{2 \mathrm{M}}{\pi}
\end{aligned}$
Let $\mathrm{r}$ be the radius after bending it into a semicircular arc.
$\therefore \quad$ The separation between the two ends is 2 r. Here length $=$ circumference of the semicircle i.e., $\mathrm{L}=\pi \mathrm{r}$
$\therefore \quad r=\frac{L}{\pi}$
Also $\mathrm{M}=\mathrm{m} \times \mathrm{L}$ and $\mathrm{m}=\frac{\mathrm{M}}{\mathrm{L}}$
$\begin{aligned}
\therefore \quad \mathrm{M}_{\text {new }}=\mathrm{m}(2 \mathrm{r}) & =\frac{\mathrm{M}}{\mathrm{L}} \times \frac{2 \mathrm{~L}}{\pi} \\
& =\frac{2 \mathrm{M}}{\pi}
\end{aligned}$
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