MHT CET · Physics · Magnetic Effects of Current
A current ' \(\mathrm{I}\) ' is flowing in a conductor of length ' \(\mathrm{L}\) ' when it is bent in the form of a circular loop, its magnetic moment will be
- A \(\frac{\mathrm{IL}}{4 \pi^2}\)
- B \(4 \pi \mathrm{IL}^2\)
- C \(\frac{4 \pi}{\mathrm{IL}^2}\)
- D \(\frac{\mathrm{IL}^2}{4 \pi}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{IL}^2}{4 \pi}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{L}=2 \pi \mathrm{r} \\
& \therefore \mathrm{r}=\frac{\mathrm{L}}{2 \pi}
\end{aligned}
\)
Area of the loop, \(\mathrm{A}=\pi \mathrm{r}^2=\pi \frac{\mathrm{L}^2}{4 \pi^2}=\frac{\mathrm{L}^2}{4 \pi}\)
Magnetic moment, \(\mathrm{M}=\mathrm{IA}=\frac{\mathrm{IL}^2}{4 \pi}\)
\begin{aligned}
& \mathrm{L}=2 \pi \mathrm{r} \\
& \therefore \mathrm{r}=\frac{\mathrm{L}}{2 \pi}
\end{aligned}
\)
Area of the loop, \(\mathrm{A}=\pi \mathrm{r}^2=\pi \frac{\mathrm{L}^2}{4 \pi^2}=\frac{\mathrm{L}^2}{4 \pi}\)
Magnetic moment, \(\mathrm{M}=\mathrm{IA}=\frac{\mathrm{IL}^2}{4 \pi}\)
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