A conducting solid sphere of radius R and mass M carries a charge Q. The sphere is rotating about an

Q: Why is the charge only on the surface?

Because the sphere is conducting! In electrostatics, any excess charge on a conductor resides entirely on its outer surface to minimize repulsion.

Q: Why do we use an elemental ring instead of a disc?

Since the sphere is rotating about an axis, every point on a ring at a specific angle $\theta$ moves with the same speed and has the same radius from the axis. This makes a ring the perfect symmetrical element to calculate current.

Q: How did we get the current $dI$?

Current is the rate of flow of charge. For a rotating ring, the charge $dq$ passes a given point once every full rotation. So, $dI = dq / T$, where $T = 2\pi/\omega$ is the time period.

Q: Why is the moment of inertia $\frac{2}{5}MR^2$ and not $\frac{2}{3}MR^2$?

Even though the *charge* is only on the surface, the *mass* is distributed throughout the entire solid sphere. The moment of inertia for a uniform solid sphere is $\frac{2}{5}MR^2$.

Q: What is the significance of the ratio $\frac{Q}{2M}$?

This is the standard gyromagnetic ratio for a system where mass and charge are distributed identically. Because our mass is solid but our charge is hollow, we get a deviation from this standard value!

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

Sigma Percentile · Accepted Answer

### The Setup: A Spinning Charged Sphere

Imagine a solid sphere of radius $R$ and mass $M$, carrying a total charge $Q$. The problem states that this is a **conducting** solid sphere. This single word is the most critical trap in the entire question!

In electrostatics, any excess charge placed on a conductor immediately repels itself and migrates entirely to the outer surface. Therefore, even though the mass is distributed uniformly throughout the solid volume, the charge $Q$ is distributed purely on the surface, exactly like a hollow spherical shell.

As this sphere rotates with an angular velocity $\omega$, the surface charge moves with it, creating a series of circular current loops. Our goal is to find the total magnetic dipole moment generated by these loops and compare it to the sphere's mechanical angular momentum.

### Slicing the Sphere: The Elemental Ring

To find the total magnetic moment, we cannot just use a single formula. We must build it from the ground up using calculus.

Let's slice the sphere and look at an elemental ring at an angle $	heta$ from the axis of rotation, subtending a tiny angle $d	heta$.

The radius of this ring is $r = R \sin	heta$.

The width of this ring along the surface is $R d	heta$.

The area of this elemental ring is simply its circumference multiplied by its width:

$$dA = (2\pi r)(R d	heta) = 2\pi R^2 \sin	heta d	heta$$

Since the total charge $Q$ is spread over the surface area $4\pi R^2$, the surface charge density is $\sigma = \frac{Q}{4\pi R^2}$. The charge $dq$ on our tiny ring is:

$$dq = \sigma dA = \sigma (2\pi R^2 \sin	heta d	heta)$$

### From Charge to Current

A rotating charge is equivalent to an electric current. Current is defined as the rate of flow of charge.

For our spinning ring, the entire charge $dq$ passes a given point once every full rotation. The time taken for one full rotation is the time period $T = \frac{2\pi}{\omega}$.

Therefore, the equivalent current $dI$ of our elemental ring is:

$$dI = \frac{dq}{T} = \frac{\sigma (2\pi R^2 \sin	heta d	heta)}{\frac{2\pi}{\omega}}$$

Notice how beautifully the $2\pi$ cancels out, leaving us with a clean expression for the current:

$$dI = \sigma R^2 \omega \sin	heta d	heta$$

### The Magnetic Dipole Moment

Every current loop acts like a tiny magnet, characterized by its magnetic dipole moment. The magnetic moment $dM$ of a flat loop is simply the current multiplied by the area enclosed by the loop.

The area enclosed by our ring is $A_{	ext{loop}} = \pi r^2 = \pi (R \sin	heta)^2$.

Multiplying this area by our current $dI$, we get the magnetic moment of the elemental ring:

$$dM = dI 	imes A_{	ext{loop}} = (\sigma R^2 \omega \sin	heta d	heta) 	imes (\pi R^2 \sin^2	heta)$$

$$dM = \sigma \pi R^4 \omega \sin^3	heta d	heta$$

To find the total magnetic moment $M$ of the entire sphere, we integrate this expression from the top of the sphere ($	heta = 0$) to the bottom ($	heta = \pi$):

$$M = \int_0^\pi \sigma \pi R^4 \omega \sin^3	heta d	heta = \sigma \pi R^4 \omega \int_0^\pi \sin^3	heta d	heta$$

The integral of $\sin^3	heta$ from $0$ to $\pi$ is a standard calculus result equal to $\frac{4}{3}$. Substituting this, along with our earlier definition of $\sigma$:

$$M = \left(\frac{Q}{4\pi R^2}ight) \pi R^4 \omega \left(\frac{4}{3}ight)$$

Simplifying this expression, we arrive at the total magnetic moment:

$$M = \frac{Q R^2 \omega}{3}$$

### The Mechanics: Angular Momentum

Now, let's switch our physics hats from electromagnetism to mechanics. The sphere is a rotating mass, which means it possesses angular momentum $L$.

The angular momentum is the product of the moment of inertia $I$ and the angular velocity $\omega$.

Because the mass $M$ is distributed uniformly throughout the solid sphere, its moment of inertia is the standard value:

$$I = \frac{2}{5} M R^2$$

Therefore, the total angular momentum is:

$$L = \frac{2}{5} M R^2 \omega$$

### The Grand Finale: The Gyromagnetic Ratio

We are finally ready to find the ratio of the magnetic dipole moment to the angular momentum. This ratio is famously known in physics as the **gyromagnetic ratio**.

Dividing our expression for $M$ by our expression for $L$:

$$\frac{M}{L} = \frac{\frac{Q R^2 \omega}{3}}{\frac{2}{5} M R^2 \omega}$$

Watch how the $R^2$ and $\omega$ terms elegantly cancel out, leaving only the mass and charge:

$$\frac{M}{L} = \frac{Q}{3} 	imes \frac{5}{2M} = \frac{5}{3} \left(\frac{Q}{2M}ight)$$

The problem states that this ratio is equal to $\alpha \left(\frac{Q}{2M}ight)$. By direct comparison, we can see that:

$$\alpha = \frac{5}{3} \approx 1.67$$

**A Final Insight:** If the mass and charge were distributed in the exact same way (for example, a non-conducting solid sphere with uniform volume charge, or a hollow sphere with surface charge), the gyromagnetic ratio would be exactly $\frac{Q}{2M}$, meaning $\alpha$ would be $1$. The fact that $\alpha$ is $\frac{5}{3}$ is a direct consequence of the mass being solid while the charge is hollow!

Enter Numerical Value:

Visualized Solution

Elemental Ring Setup

Charge on Elemental Ring

Current of the Ring

Magnetic Moment of the Ring

Total Magnetic Moment

Angular Momentum

Gyromagnetic Ratio

The Way Forward

The Sigma Insight: Magnetic Moment of Current Loop

The Setup

Slicing the Sphere

From Charge to Current

The Magnetic Dipole Moment

The Mechanics

The Grand Finale

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