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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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  • Maybe the relevant source would be a dipole? But is there a simple convincing reason that all possible fields in that sphere of vacuum can me modelled by a distribution of point dipoles outside the sphere? – PPenguin Nov 29 '23 at 19:25
  • Related question, considering instead only sources inside the sphere https://physics.stackexchange.com/questions/601075/average-magnetic-field-inside-a-sphere – PPenguin Nov 29 '23 at 19:26

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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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  • YES, this makes way more sense now. Previously everytime I've seen the mean value property used, they dropped to the potential (like in that linked electric field case). But noting that the fields themselves are harmonic, and thus (if I'm understanding correctly) the cartesian field components obey the mean value property (and thus the resulting vector), is indeed MUCH easier that trying to follow carefully applying a vector calc operation to the integral bounds, etc. – PPenguin Nov 29 '23 at 20:33