A hollow, right circular cone of base radius R and height h, with its tip at the origin is rotating

Q: Why can we treat the cone as a magnetic dipole?

Because the problem specifies $z \gg R$ and $z \gg h$. At distances much larger than the dimensions of the current distribution, the higher-order multipole moments become negligible, and the system behaves predominantly like a simple magnetic dipole.

Q: What is the gyromagnetic ratio and when can I use it?

The gyromagnetic ratio ($\frac{M}{L} = \frac{q}{2m}$) is a shortcut that applies to any rigid body where the charge distribution is geometrically identical to the mass distribution. Since the cone has uniform surface charge, assuming a uniform surface mass allows us to use this powerful relation.

Q: Why is the moment of inertia of a hollow cone $\frac{mR^2}{2}$?

If you slice a hollow cone parallel to its base, you get rings. The mass is distributed at varying radii from $0$ to $R$. Integrating the moment of inertia of these rings ($dm \cdot r^2$) over the surface of the cone yields $\frac{mR^2}{2}$, which is interestingly the same as a solid disk!

Q: What if the cone was solid instead of hollow?

If the cone were solid and the charge was uniformly distributed throughout its volume, the moment of inertia would be $\frac{3mR^2}{10}$. The gyromagnetic ratio would still apply, but the angular momentum $L$ would change, leading to a different magnetic moment and a different value for $n$.

Q: Why didn't we need the height $h$ in the final formula?

The height $h$ affects the slant length and the total area, but when calculating the moment of inertia in terms of total mass $m$ and base radius $R$, $h$ cancels out. The rotational inertia depends only on how far the mass is from the axis of rotation, which is governed entirely by $R$.

Question

A hollow, right circular cone of base radius $R$ and height $h$, with its tip at the origin is rotating about the Z-axis with an angular velocity $\omega$, as shown in the figure. The cone carries a total charge $Q$ uniformly distributed on its curved surface. The magnitude of magnetic field at a point $(0, 0, z)$, where $z \gg R$ and $z \gg h$, is $\frac{n\mu_0 Q R^2 \omega}{4\pi z^3}$. The value of $n$ is:

Sigma Percentile · Accepted Answer

### The Magic of the Gyromagnetic Ratio: Magnetic Field of a Rotating Cone

Imagine a hollow cone, uniformly coated with an electric charge $Q$, spinning rapidly around its central axis. The moving charges create tiny current loops all over the surface, generating a complex magnetic field. Calculating this field at an arbitrary point would require a terrifying surface integral.

However, physics often rewards us with elegant shortcuts when we look at the bigger picture. The problem asks for the magnetic field at a point $P(0, 0, z)$ where $z \gg R$ and $z \gg h$. This condition is our golden ticket.

### The Dipole Approximation

When you observe a localized current distribution from a distance much larger than its own dimensions, the intricate details of the shape blur away. The entire rotating cone effectively behaves as a single, tiny magnetic dipole located at the origin.

For a magnetic dipole with moment $M$, the magnetic field on its axis at a distance $z$ is given by the standard formula:

$$B = \frac{\mu_0}{4\pi} \frac{2M}{z^3}$$

Our entire mission now boils down to finding $M$, the magnetic dipole moment of this spinning cone.

### The Gyromagnetic Ratio Shortcut

We could find $M$ by integrating the magnetic moments of infinitesimal rotating rings. But there is a much faster, more profound method: the **Gyromagnetic Ratio**.

For any rigid body where the charge distribution is geometrically identical to the mass distribution, the ratio of its magnetic dipole moment $M$ to its angular momentum $L$ is a constant:

$$\frac{M}{L} = \frac{Q}{2m}$$

Here, $Q$ is the total charge and $m$ is the total mass. Since our cone has a uniform surface charge, we can imagine it having a uniform surface mass $m$ to exploit this relation.

### Calculating Angular Momentum

To use the gyromagnetic ratio, we need the angular momentum $L = I\omega$. This requires the moment of inertia $I$ of a hollow cone about its central axis.

If you slice a hollow cone parallel to its base, you get a series of rings. Integrating the moment of inertia of these rings over the surface yields a surprisingly simple result, identical to that of a solid disk:

$$I = \frac{mR^2}{2}$$

Therefore, the angular momentum is:

$$L = \frac{mR^2}{2}\omega$$

### Bringing It All Together

Now, let's plug this angular momentum back into our gyromagnetic ratio equation. Notice how beautifully the assumed mass $m$ cancels out, leaving a purely electromagnetic result:

$$M = \frac{Q}{2m} \left( \frac{mR^2}{2}\omega \right) = \frac{QR^2\omega}{4}$$

We have our magnetic moment! The final step is to substitute this $M$ back into our dipole magnetic field equation:

$$B = \frac{\mu_0}{4\pi} \frac{2 \left( \frac{QR^2\omega}{4} \right)}{z^3}$$

Simplifying the numerator, the $2$ partially cancels the $4$, giving us:

$$B = \frac{0.5 \mu_0 Q R^2 \omega}{4\pi z^3}$$

Comparing this with the expression given in the problem, $\frac{n\mu_0 Q R^2 \omega}{4\pi z^3}$, it is crystal clear that the value of $n$ is exactly **$0.5$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Dipole Approximation

Magnetic Field of a Dipole

The Gyromagnetic Ratio

Moment of Inertia

Angular Momentum

Calculating Magnetic Moment

Substituting into Magnetic Field Equation

Simplifying the Expression

Final Answer

The Sigma Insight: Magnetic Moment of Current Loop

The Magic of the Gyromagnetic Ratio

The Dipole Approximation

The Gyromagnetic Ratio Shortcut

Calculating Angular Momentum

Bringing It All Together

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Dipole Approximation

Magnetic Field of a Dipole

The Gyromagnetic Ratio

Moment of Inertia

Angular Momentum

Calculating Magnetic Moment

Substituting into Magnetic Field Equation

Simplifying the Expression

Final Answer

The Sigma Insight: Magnetic Moment of Current Loop

The Magic of the Gyromagnetic Ratio

The Dipole Approximation

The Gyromagnetic Ratio Shortcut

Calculating Angular Momentum

Bringing It All Together

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A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is