MHT CET · Physics · Magnetic Effects of Current
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?
- A \(\frac{\mathrm{M}}{\mathrm{L}}\)
- B \(\frac{\mathrm{M}}{\pi}\)
- C \(\frac{\mathrm{M}}{2 \pi}\)
- D \(\frac{2 \mathrm{M}}{\pi}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \mathrm{M}}{\pi}\)
Step-by-step Solution
Detailed explanation
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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