# Surprisingly uniform magnetic field inside a rotating charged sphere

Consider this:

A spherical shell of radius R, carrying a uniform surface charge $$σ$$, is set spinning at an angular velocity $$ω$$. What can be said about the magnetic field inside the sphere?

I found the magnetic vector potential* at an arbitrary point in space, and used $$B = curl(A)$$ to find the magnetic field. To my surprise, the magnetic field inside the sphere is uniform!

$$\mathbf{B}=\mathbf{\nabla}\times\mathbf{A} =\frac{2}{3}\mu_0\sigma R\mathbf{\omega}.$$

Why is this so? Is there a way to predict the same (uniformity of B inside the sphere) without going through the mathematical derivation? I'd love to get some more insight into this problem, after all, there is more to physics than just mathematics!

P.S.

*The calculation, if you want to go through it, can be found in the Electrodynamics text by Griffiths.

• The electric field inside a charged sphere is uniform. (It’s zero.) Coulomb’s Law and the Biot-Savart Law are both inverse square, so it shouldn’t be too surprising that the magnetic field is uniform. Commented Oct 2, 2019 at 16:18

There are equivalences you can use here. You can do the precise mathematical checks if you want to, I'll just list the steps to take:

1. The surface current of the rotating surface charge is equivalent to the surface current of a uniform distribution of magnetic dipoles in the entire volume of the sphere, represented by infinitesimal current loops. The current of neighboring loops cancels everywhere except at the surface, see [researchgate 373623441 drawing]
1. The infinitesimal magnetic dipoles can also be represented by positive and negative magnetic monopoles (with infinitesimal separation) instead of current loops.
1. This is of course equivalent to the case with electric charges instead of monopoles, which is treated in many places and can be shown to have a uniform field inside the entire sphere and a pure dipole field outside it.

The electric field inside a charged sphere is uniform. (It’s zero.) Coulomb’s Law and the Biot-Savart Law both have inverse-square spatial dependence, so it shouldn’t be too surprising that the magnetic field is uniform.

• There's one difference, though. In the integration we go through in Coulomb's Law, the unit vector is radial, while in Biot-Savart it is along the direction of current. How do you explain that? The r² dependence is fine, just the direction part bothers me! Commented Oct 3, 2019 at 4:35
• Magnetic field is caused by moving charge (also by spin) so it ought to depend on both the direction of motion of the charge and the direction to the field point. And the only way to get a vector from two vectors (other than addition, which makes no sense) is by the vector product of these two things. Commented Oct 3, 2019 at 4:48
• That's alright; but since the direction used in both the laws are different, how can you use an electrostatic argument to justify that field is uniform here? The r² dependence is still there in the magnitude (while finding B), but the direction is not the same as it would have been in an analogous electrostatics case. Commented Oct 3, 2019 at 4:53
• I didn’t justify it. I said that you shouldn’t find it too surprising. You should not expect to be able to intuit too much in physics. We would get nowhere without lots of math. Math is the secret sauce of the universe! Commented Oct 3, 2019 at 5:07
##### How to show that there is a uniform magnetic field inside a rotating charged sphere
Start from [Maxwell's Vector Equations][1] $$\overset{\rightarrow}{\boldsymbol \nabla} \bullet \overset{\rightarrow}{\boldsymbol D}= \rho \tag{Eq. 1}$$ $$\overset{\rightarrow}{\boldsymbol \nabla} \bullet \overset{\rightarrow}{\boldsymbol B}= 0 \tag{Eq. 2}$$ $$\overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol E}=\frac{\partial}{\partial t} \overset{\rightarrow}{\boldsymbol B} \tag{Eq. 3}$$ $$\overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol H}= \overset{\rightarrow}{\boldsymbol J}\text{ }{\boldsymbol +}\text{ } \frac{\partial}{\partial t} \overset{\rightarrow}{\boldsymbol D} \tag{Eq. 4}$$ In free space (inside the shell or outside the shell) with $$\mu=\mu_0$$ and $$\epsilon = \epsilon_0$$ being constants: $$\overset{\rightarrow}{\boldsymbol D} =\epsilon \overset{\rightarrow}{\boldsymbol E} \text{ , } \overset{\rightarrow}{\boldsymbol B} =\mu \overset{\rightarrow}{\boldsymbol H} \tag{Eqs. 5}$$ Since (from Equation 2) $$\overset{\rightarrow}{\boldsymbol \nabla} \bullet \overset{\rightarrow}{\boldsymbol B}= 0$$, then $$\overset{\rightarrow}{\boldsymbol B}\text{ }$$ is divergent-less and so it can be written as the curl of a vector-potential $$\overset{\rightarrow}{\boldsymbol A}\text{ }$$: $$\overset{\rightarrow}{\boldsymbol B}= \overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol A} \text{ , } \overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol B}= \overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol A} \underset{implies} \implies \overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \overset{\rightarrow}{\boldsymbol A}= \overset{\rightarrow}{\boldsymbol \nabla}{\boldsymbol \times} \mu \left( \overset{\rightarrow}{\boldsymbol J}\text{ }{\boldsymbol +}\text{ } \frac{\partial}{\partial t} \overset{\rightarrow}{\boldsymbol D} \right) \tag{Eqs. 6}$$ In a differential area element in the shell element namely $$(\partial r)(r \partial \phi)$$, the current density $$\overset{\rightarrow}{\boldsymbol J}$$ is (where Q is the total charge in the shell): $$\overset{\rightarrow}{\boldsymbol J}= \frac{Q}{4\pi r^2}\overset{\rightarrow}{\boldsymbol \omega} \times \overset{\rightarrow}{\boldsymbol r} \text{ }(\partial r) (r \partial \phi) \tag{Eqs. 7}$$

I am in the process of updating this answer further but I am saving it now so that I can print it out to review it more carefully.