A conducting solid sphere of radius and mass carries a charge . The sphere is rotating about an axis passing through its center with a uniform angular speed . The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the sa

Question

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 $ω$ . The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as $α \frac{Q}{2 M} .$ The value of $α$ is _______. [2025]

Smart Achievers · Accepted Answer

(1.67)

Magnetic dipole moment $M$

$d M = d I A$

$α$

$d I = \frac{d q}{T} = \frac{σ (2 π r) (R d θ) ω}{2 π}$

$d I = \frac{σ 2 π R^{2} ω \sin θ d θ}{2 π}$

$d I = σ R^{2} ω \sin θ d θ$

Magnetic dipole moment:

$M = \int d M = \int_{0}^{π} σ R^{2} ω π R^{2} \sin^{3} θ d θ$

$M = σ R^{4} ω π \int_{0}^{π} \sin^{3} θ d θ$ $(∵ \int_{0}^{π} \sin^{3} θ d θ = \frac{4}{3})$

$M = (\frac{Q}{4 π R^{2}}) R^{4} ω π (\frac{4}{3}) = \frac{Q R^{2} ω}{3}$

Angular momentum, $L = I ω$

$L = (\frac{2}{5} M R^{2}) ω$

$∴$ $\frac{M}{L} = \frac{\frac{Q R^{2} ω}{3}}{\frac{2}{5} M R^{2} ω} = \frac{Q}{2 M} (\frac{5}{3})$

$∴$ $α = \frac{5}{3} = 1.67$

Solution

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