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 \fracQ2 M . The value of \alpha is \underline\hspace2cm .

Magnetism PYQ — JEE Advanced 2025 · Shift 2 · Q41 with Solution

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Correct answer: 1.65TO1.67

Given: A conducting solid sphere of radius $R$ , mass $M$ , total charge $Q$ , rotating with angular speed $\omega$ about a diameter.

We need the ratio

\frac{\mu}{L} = \alpha \frac{Q}{2M}

and must find $\alpha$ .

Magnetic dipole moment of the rotating charged sphere

Since the sphere is a conductor, all charge resides on the surface. So it behaves like a spherical shell carrying charge $Q$ uniformly over its surface.

Surface charge density:

\sigma = \frac{Q}{4\pi R^2}

Consider a thin ring on the sphere at polar angle $\theta$ of width $R\,d\theta$ .

Radius of ring:

r = R\sin\theta

Area of ring:

dA = 2\pi R^2 \sin\theta\, d\theta

Charge on ring:

dq = \sigma \, dA = \frac{Q}{4\pi R^2}(2\pi R^2\sin\theta\, d\theta)=\frac{Q}{2}\sin\theta\, d\theta

The ring rotates with angular speed $\omega$ , so current in the ring is

dI = \frac{dq}{T} = \frac{\omega}{2\pi}dq

Magnetic moment of a current loop is current times area:

d\mu = dI \cdot \pi r^2

Thus,

d\mu = \frac{\omega}{2\pi}dq \cdot \pi (R\sin\theta)^2 = \frac{\omega}{2} dq \, R^2 \sin^2\theta

Substitute $dq$ :

d\mu = \frac{\omega}{2} \left(\frac{Q}{2}\sin\theta\, d\theta\right) R^2 \sin^2\theta = \frac{Q\omega R^2}{4} \sin^3\theta\, d\theta

Integrate from $\theta=0$ to $\pi$ :

\mu = \frac{Q\omega R^2}{4} \int_0^\pi \sin^3\theta\, d\theta

Using

\int_0^\pi \sin^3\theta\, d\theta = \frac{4}{3}

we get

\mu = \frac{Q\omega R^2}{4}\cdot \frac{4}{3} = \frac{1}{3}Q\omega R^2

So,

\boxed{\mu = \frac{1}{3}Q\omega R^2}

Angular momentum of the rotating solid sphere

For a solid sphere, moment of inertia about a diameter is

I = \frac{2}{5}MR^2

Hence angular momentum is

L = I\omega = \frac{2}{5}MR^2\omega

So,

\boxed{L = \frac{2}{5}MR^2\omega}

Take the ratio $\mu/L$

\frac{\mu}{L} = \frac{\frac{1}{3}Q\omega R^2}{\frac{2}{5}MR^2\omega}

Cancel $R^2\omega$ :

\frac{\mu}{L} = \frac{1}{3}Q \cdot \frac{5}{2M} = \frac{5Q}{6M}

Now compare with the given form:

\frac{\mu}{L} = \alpha \frac{Q}{2M}

Thus,

\alpha \frac{Q}{2M} = \frac{5Q}{6M}

So,

\alpha = \frac{5}{3}

Hence,

\boxed{\alpha = \frac{5}{3}}

Numerically,

\frac{5}{3} = 1.666\ldots

Comparison with stored answer

Stored correct answer: $1.65$ to $1.67$

Our answer:

\alpha = \frac{5}{3} \approx 1.667

This lies in the given range.

So the derived answer agrees with the stored correct answer.

More from Magnetism

View written solutionFree

Correct answer: 1.65TO1.67

Given: A conducting solid sphere of radius $R$ , mass $M$ , total charge $Q$ , rotating with angular speed $\omega$ about a diameter.

We need the ratio

\frac{\mu}{L} = \alpha \frac{Q}{2M}

and must find $\alpha$ .

Magnetic dipole moment of the rotating charged sphere

Since the sphere is a conductor, all charge resides on the surface. So it behaves like a spherical shell carrying charge $Q$ uniformly over its surface.

Surface charge density:

\sigma = \frac{Q}{4\pi R^2}

Consider a thin ring on the sphere at polar angle $\theta$ of width $R\,d\theta$ .

Radius of ring:

r = R\sin\theta

Area of ring:

dA = 2\pi R^2 \sin\theta\, d\theta

Charge on ring:

dq = \sigma \, dA = \frac{Q}{4\pi R^2}(2\pi R^2\sin\theta\, d\theta)=\frac{Q}{2}\sin\theta\, d\theta

The ring rotates with angular speed $\omega$ , so current in the ring is

dI = \frac{dq}{T} = \frac{\omega}{2\pi}dq

Magnetic moment of a current loop is current times area:

d\mu = dI \cdot \pi r^2

Thus,

d\mu = \frac{\omega}{2\pi}dq \cdot \pi (R\sin\theta)^2 = \frac{\omega}{2} dq \, R^2 \sin^2\theta

Substitute $dq$ :

d\mu = \frac{\omega}{2} \left(\frac{Q}{2}\sin\theta\, d\theta\right) R^2 \sin^2\theta = \frac{Q\omega R^2}{4} \sin^3\theta\, d\theta

Integrate from $\theta=0$ to $\pi$ :

\mu = \frac{Q\omega R^2}{4} \int_0^\pi \sin^3\theta\, d\theta

Using

\int_0^\pi \sin^3\theta\, d\theta = \frac{4}{3}

we get

\mu = \frac{Q\omega R^2}{4}\cdot \frac{4}{3} = \frac{1}{3}Q\omega R^2

So,

\boxed{\mu = \frac{1}{3}Q\omega R^2}

Angular momentum of the rotating solid sphere

For a solid sphere, moment of inertia about a diameter is

I = \frac{2}{5}MR^2

Hence angular momentum is

L = I\omega = \frac{2}{5}MR^2\omega

So,

\boxed{L = \frac{2}{5}MR^2\omega}

Take the ratio $\mu/L$

\frac{\mu}{L} = \frac{\frac{1}{3}Q\omega R^2}{\frac{2}{5}MR^2\omega}

Cancel $R^2\omega$ :

\frac{\mu}{L} = \frac{1}{3}Q \cdot \frac{5}{2M} = \frac{5Q}{6M}

Now compare with the given form:

\frac{\mu}{L} = \alpha \frac{Q}{2M}

Thus,

\alpha \frac{Q}{2M} = \frac{5Q}{6M}

So,

\alpha = \frac{5}{3}

Hence,

\boxed{\alpha = \frac{5}{3}}

Numerically,

\frac{5}{3} = 1.666\ldots

Comparison with stored answer

Stored correct answer: $1.65$ to $1.67$

Our answer:

\alpha = \frac{5}{3} \approx 1.667

This lies in the given range.

So the derived answer agrees with the stored correct answer.

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