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JEE Main 2019 (10 Jan, Shift 1) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from JEE Main 2019 (10 Jan, Shift 1)

Q74JEE Main 2019 · 10 JAN · SHIFT 1Difficult

Current Loop as a Magnetic Dipole

An insulating thin rod of length $l$ has a linear charge density $\rho(x) = \rho_0\frac{x}{l}$ on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is -

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Answer: (b)

Element $dx$ carries $dq = \rho_0\frac{x}{l}dx$ ; rotating at $n$ rev/s it is a current $dI = n\,dq$ .
$dm = dI \cdot \pi x^2 = \frac{n\pi\rho_0}{l}x^3dx$
$m = \frac{n\pi\rho_0}{l}\int_0^l x^3dx = \frac{\pi}{4}n\rho_0 l^3$
So the answer is (b) $\frac{\pi}{4}n\rho l^3$ .

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