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JEE Main 2019 (10 Apr, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from JEE Main 2019 (10 Apr, Shift 2)

Q71JEE Main 2019 · 10 APR · SHIFT 2Medium

Current Loop as a Magnetic Dipole

A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

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Answer: (d)

Wire length $L$ : square $m = I\frac{L^2}{16}$ .
Circle: $m' = I\frac{L^2}{4\pi}$ .
$\frac{m'}{m} = \frac{16}{4\pi} = \frac{4}{\pi}$ , so the answer is (d) $\frac{4m}{\pi}$ .

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