JEE Advanced 2025 Paper 2, Physics Section 3 Q1: Torque and Work Done in a Magnetic Field

JEE Advanced2025Paper 2Physics Section 3

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 _______.

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Solution

The charge on a conductor sits on its surface with uniform density $σ = Q / (4 π R^{2})$ . Consider a thin ring on the sphere at polar angle $θ$ , of radius $r = R sin θ$ and width $R d θ$ . The charge on this ring is $d q = σ (2 π R sin θ) (R d θ)$ and it circulates with period $T = 2 π / ω$ , giving current

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

The magnetic moment contribution is $d M = d I \cdot π r^{2} = σ π R^{4} ω sin^{3} θ d θ$ . Integrating with $\int_{0}^{π} sin^{3} θ d θ = 4/3$ :

$M = σ π R^{4} ω \cdot \frac{4}{3} = \frac{Q}{4 π R ^{2}} \cdot π R^{4} ω \cdot \frac{4}{3} = \frac{Q R ^{2} ω}{3}$

For a uniform solid sphere, $L = \frac{2}{5} M R^{2} ω$ . The required ratio is

$\frac{M}{L} = \frac{Q R ^{2} ω /3}{( 2/5 ) M R ^{2} ω} = \frac{5 Q}{6 M} = \frac{5}{3} \cdot \frac{Q}{2 M}$

$∴ α = \frac{5}{3} \approx 1.67$ .

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JEE Advanced 2025 Paper 2, Physics Section 3 Q1: Torque and Work Done in a Magnetic Field

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