JEE Advanced 2025 Paper 2 · Q09 · Magnetic Dipole & Moment
A conducting solid sphere of radius $R$ and mass $M$ carries a charge $Q$. The sphere is rotating about an axis passing through its center with a uniform angular speed $\omega$. The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as $\alpha\,Q/(2M)$. The value of $\alpha$ is __________.
Reveal answer + step-by-step solution
Correct answer:1.67
Solution
On a conducting sphere all the charge resides on the surface ($\sigma = Q/(4\pi R^2)$). Consider a circular ring at polar angle $\theta$ of width $R\,d\theta$. The ring carries charge $dq = \sigma \cdot 2\pi R \sin\theta \cdot R\,d\theta = (Q/2)\sin\theta\,d\theta$. It rotates with period $2\pi/\omega$, giving current $dI = dq\,\omega/(2\pi)$. The ring's area is $\pi(R\sin\theta)^2$. Magnetic moment $dM = \pi R^2 \sin^2\theta \cdot dq\,\omega/(2\pi) = (Q\omega R^2/4)\sin^3\theta\,d\theta$. $M_{\text{total}} = (Q\omega R^2/4)\int_0^{\pi}\sin^3\theta\,d\theta = (Q\omega R^2/4)(4/3) = Q\omega R^2/3$. Angular momentum of solid sphere of mass $M$ about its axis: $L = (2/5) M R^2 \omega$. Ratio $M/L = (Q\omega R^2/3)/((2 M R^2 \omega/5)) = 5Q/(6M) = (5/3)\,Q/(2M)$. Hence $\alpha = 5/3 \approx 1.67$.
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