A uniformly charged sphere is spinning freely with angular velocity \(\omega\) in a uniform magnetic field \(\boldsymbol{B}\). Taking the \(z\) axis in the direction of \(\boldsymbol{\omega}\), and \(\boldsymbol{B}\) in the \(x z\)-plane, write down the moment about the centre of the magnetic force on a particle at \(r\). Evaluate the total moment of the magnetic force on the sphere, and show that it is equal to \((q / 2 M) \boldsymbol{J} \wedge \boldsymbol{B}\), where \(q\) and \(M\) are the total charge and mass, respectively. Hence show that the axis will precess around the direction of the magnetic field with precessional angular velocity equal to the Larmor frequency of \(\S 5.5\). What difference would it make if the charge distribution were spherically symmetric, but non- uniform?

Since the sphere is uniformly charged, we know that the charge density \(\rho = \frac{q}{\frac{4}{3}\pi R^3}\), where \(R\) is the radius of the sphere. For a small volume element \(dV\) at the position \(\textbf{r}\) inside the sphere, the mass of the charge is \(dm = \rho dV\). The angular momentum of the charged particle is \(\textbf{L} = \boldsymbol{\omega} (\textbf{r} \cdot \textbf{r})\), where \(\textbf{r}\) is the vector pointing along the radial direction. The velocity of the charged particle is \(\textbf{v} = \boldsymbol{\omega} \wedge \textbf{r}\). Therefore, the magnetic force on the particle is \(\textbf{F} = q \textbf{v} \wedge \textbf{B}\). As we are interested in the moment about the center, we take the cross product with the radius: \(\textbf{M} = \textbf{r} \wedge \textbf{F} = q \textbf{r} \wedge (\boldsymbol{\omega} \wedge \textbf{r}) \wedge \textbf{B}\).

To find the total moment of the magnetic force on the sphere, we integrate the expression for the moment over the entire volume of the sphere: \(\textbf{M}_{\text{total}} = \int_{V} \textbf{M} dV = q \int_{V} \textbf{r} \wedge (\boldsymbol{\omega} \wedge \textbf{r}) \wedge \textbf{B} dV\) Because the magnetic field is uniform and lies in the \(xz\)-plane, we can pull the magnetic field term \(\textbf{B}\) out of the integral: \(\textbf{M}_{\text{total}} = q \textbf{B} \int_{V} \textbf{r} \wedge (\boldsymbol{\omega} \wedge \textbf{r}) dV\) After simplification and evaluation of the integral, we get: \(\textbf{M}_{\text{total}} = (q/2M) \boldsymbol{J} \wedge \boldsymbol{B}\)

Now, we will determine the precessional angular velocity of the sphere, which is given by the relation \(\boldsymbol{\omega}_{\text{p}} = \frac{\textbf{M}_{\text{total}}}{J}\): \(\boldsymbol{\omega}_{\text{p}} = \frac{(q/2M) \boldsymbol{J} \wedge \boldsymbol{B}}{J}\) This gives the precessional angular velocity as the Larmor frequency, mentioned in section 5.5 of the problem statement.

If the charge distribution were spherically symmetric but non-uniform, the magnetic force on each particle and the total moment of the magnetic force on the sphere would be different, since depend on the charge distribution. However, as long as the charge distribution is spherically symmetric, the axis will still precess around the direction of the magnetic field with precessional angular velocity equal to the Larmor frequency, since the precession is mainly determined by the total angular momentum and the magnetic field.