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 \(\alpha \frac{Q}{2 M}\). The value of \(\alpha\) is ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Magnetic Effects of Current

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 ____.

Answer: 1.67

Consider a thin ring element at angle

θ

from the axis. The current due to rotation is:

d I = σ R^{2} ω sin θ d θ

The magnetic moment contribution:

d M = d I \cdot A = σ R^{2} ω sin θ d θ \cdot π R^{2} sin^{2} θ

Integrating:

M = σ R^{4} ω π \int_{0}^{π} sin^{3} θ d θ = σ R^{4} ω π \cdot \frac{4}{3}

With

σ = \frac{Q}{4 π R ^{2}}

M = \frac{Q R ^{2} ω}{3}

Angular momentum:

L = \frac{2}{5} M R^{2} ω

Ratio:

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

Therefore

α = \frac{5}{3} = 1.67

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer