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For this problem, consider the electric and magnetic fields constant.

In this scenario, the average electromagnetic field inside a sphere of vacuum is equal to the electric field at its center. $$ \mathbf{E}_{\text{average}}(\mathbf{r}) = \frac{1}{\frac{4}{3} \pi R^3} \int_{\text{Sphere}(\mathbf{r}, \ R)} \mathbf{E}(\mathbf{r}') \ \ d^3r' = \mathbf{E}(\mathbf{r}) $$ This can be shown "directly" by showing it is true for the case of a single charge outside the sphere, and then because Maxwell's equations are linear, this is true for any constant charge distribution outside the sphere.

I am curious about the equivalent for the magnetic field. Is the following true? $$ \mathbf{B}_{\text{average}}(\mathbf{r}) = \frac{1}{\frac{4}{3} \pi R^3} \int_{\text{Sphere}(\mathbf{r}, \ R)} \mathbf{B}(\mathbf{r}') \ \ d^3r' = \mathbf{B}(\mathbf{r}) $$ Since the sources of the magnetic field are currents instead of point charges, what should I use for an equivalent "direct" proof of the magnetic field case?

By "direct" here, I mean that I am hoping to derive the result by solving an integral "directly" instead of using vector calc identities and swapping order of integration, derivatives, etc. (For example, this approach to a related average in the electric field case: Average electric field calculated from average potential There is nothing wrong with such approaches, but I am just looking for a more direct way, even if the integral is a bit harder.) However if switching to the potentials is the only reasonable way forward here, so be it.

PPenguin
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Electrostatic and magnetostatic obey the same equations in vacuum: $$ \begin{align} \nabla\cdot B &= 0 & \nabla\cdot E &= 0 \\ \nabla\times B &= 0 & \nabla\times E &= 0 \end{align} $$ The fastest way is to notice that in both cases, the fields are harmonic: $$ \begin{align} \Delta B &= 0 & \Delta E &= 0 \end{align} $$ and use the mean value property.

For your approach, yes, you can express any current as a superposition of magnetic dipoles, so you can apply the same method. I can’t say that this “direct” method is simpler, as you still need to do some involved calculations in those specific cases.

Hope this helps.

LPZ
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