# Magnetic field due to a spinning sphere

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1. Oct 22, 2016

### Jezza

1. The problem statement, all variables and given/known data

We consider a sphere of radius $R$ which carries a uniform surface charge density $\sigma$ and spins with angular velocity $\omega$ around a diameter. We use spherical coordinates $(r, \theta, \phi)$ with origin at the centre of the sphere and the $z$-axis along the rotation axis.

a) Find the surface current density $\vec{K}(\vec{r})$ as a function of $\theta$.

b) Justify that the magnetic field produced by the surface current can be written under the form $\vec{B} = \vec{\nabla}\Psi$ where $\Psi$ is a scalar function. Show that $\Psi$ satisfies Laplace's equation inside and outside the sphere. Write the boundary conditions on $\Psi$. Caution: $\Psi$ is not continuous at $r=R$ (explain why).

2. Relevant equations

Laplace's equation:
$$\nabla ^2 \Psi = 0$$
Boundary condition on magnetic field at a surface current:
$$\vec{B_{2}}-\vec{B_{1}} = \mu_0 \vec{K} \times \hat{n}_{12}$$
Where $\hat{n}_{12}$ is the unit vector locally normal to the surface current, pointing from side 1 to side 2.

3. The attempt at a solution

I think I'm alright with part (a); I mainly provided it for context. The solution is simple enough by taking a ring of charge on the surface of the sphere, perpendicular to the rotation axis, from some $\theta$ to $\theta + \mathrm{d}\theta$. We have current $I=\Delta Q/\Delta t$ with $\Delta Q$ the amount of charge passing a fixed point in time $\Delta t$.

We can pick $\Delta t = 2\pi/\omega$ so that $\Delta Q = 2 \pi R \sigma \sin{\theta} \mathrm{d}\theta$.
This gives $I = \omega R \sigma \sin{\theta} \mathrm{d}\theta$. Writing $I = K(\vec{r}) \mathrm{d} \theta$, we have:
$$K(\vec{r}) = \omega R \sigma \sin{\theta}$$
Clearly, $\vec{K}$ is directed tangentially to the sphere, along the direction of rotation of the sphere. In other words, perpendicular to both $\vec{R}$ and $\vec{\omega}$. We can therefore write:
$$\vec{K}(\vec{r}) = \sigma \vec{\omega} \times \vec{R}$$
I'm also comfortable with justifying why $\Psi$ satisfies Laplace's equation. We begin with two of Maxwell's laws for steady currents:
$$\vec{\nabla} \cdot \vec{B} = 0$$
$$\vec{\nabla} \times \vec{B} = \mu_0 \vec{J}$$
Except at the surface, we have $\vec{J} = \vec{0}$. Hence, we have a discontinuity in $\vec{B}$ at $r=R$, but otherwise we have that $\vec{\nabla} \times \vec{B} = \vec{0} \implies \vec{B} = \vec{\nabla} \Psi$.

Then from the first of the above equations, we have $\vec{\nabla} \cdot (\nabla \Psi ) = 0$, giving:
$$\nabla ^2 \Psi = 0$$
Here's where my knowledge gets a bit sketchy. I can tentatively suggest two boundary conditions; the first derived from the boundary condition on a magnetic field at a surface current (listed in relevant equations):
$$\vec{\nabla} \Psi_{\mathrm{out}} - \vec{\nabla} \Psi_{\mathrm{in}} = \mu_0 \frac{\vec{K} \times \vec{R}}{|\vec{R}|}$$
The second I've guessed is that as $r \to \infty$, $\vec{B} \to \vec{0}$ then $\Psi \to 0$ also.

As for why $\Psi$ is not continuous at the boundary (as opposed to the electric equivalent, which would be), my thinking is that it's something to do with the fact that $\vec{\nabla}\Psi$ is not normal to the surface where it would be in the electric equivalent. Of course we also have the fact that, since the magnetic scalar potential has no physical interpretation like the electric potential energy does, it doesn't need to be continuous to conserve energy.

The later parts of the question ask me to proceed to solve Laplace's equations given my boundary conditions. Of course, given the two regions I'm solving in, I'm going to need a third condition, and I have no idea what that might be. I'm also not even sure the two I've come up with are the right ones!

2. Oct 22, 2016

### TSny

The right side does not have the dimensions of charge.
The right side does not have the dimensions of current.
I believe this is correct.

OK
Yes, your discussion of showing $\Psi$ satisfies Laplace's equation looks good to me.
OK, but note that the equation involving the gradients is a vector equation. You can get two boundary conditions from this by looking at components of this equation.

See if you can prove that $\Psi$ is not continuous at the boundary by considering the equation
$$\vec{\nabla} \Psi_{\mathrm{out}} - \vec{\nabla} \Psi_{\mathrm{in}} = \mu_0 \frac{\vec{K} \times \vec{R}}{|\vec{R}|}$$
applied in the direction of increasing $\theta$.

Last edited: Oct 22, 2016