MHT CET · Physics · Magnetic Effects of Current
A straight wire carrying a current (I) is turned into a circular loop. If the magnitude of the magnetic moment associated with it is ' \(\mathrm{M}\) ', then the length of the wire will be
- A \(\frac{\mathrm{M} \pi}{4 \mathrm{I}}\)
- B \(\left[\frac{4 \pi \mathrm{I}}{\mathrm{M}}\right]^{\frac{1}{2}}\)
- C \(\left[\frac{4 \mathrm{M} \pi}{\mathrm{I}}\right]^{\frac{1}{2}}\)
- D \(4 \pi \mathrm{MI}\)
Answer & Solution
Correct Answer
(C) \(\left[\frac{4 \mathrm{M} \pi}{\mathrm{I}}\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
Magnetic moment is given as \(\mathrm{M}=\mathrm{IA}\)
\(
\therefore \mathrm{M}=\mathrm{I}\left(\pi \mathrm{R}^2\right)
\)
Now, length of the wire is, \(L=2 \pi R\)
\(\therefore \mathrm{R}=\frac{\mathrm{L}}{2 \pi} \)
\( \Rightarrow \mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \)
\( \therefore \mathrm{L}=\sqrt{\frac{4 \mathrm{M} \pi}{\mathrm{I}}}\)
\(
\therefore \mathrm{M}=\mathrm{I}\left(\pi \mathrm{R}^2\right)
\)
Now, length of the wire is, \(L=2 \pi R\)
\(\therefore \mathrm{R}=\frac{\mathrm{L}}{2 \pi} \)
\( \Rightarrow \mathrm{M}=\mathrm{I} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \)
\( \therefore \mathrm{L}=\sqrt{\frac{4 \mathrm{M} \pi}{\mathrm{I}}}\)
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