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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.

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    $\begingroup$ 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. $\endgroup$
    – G. Smith
    Oct 2, 2019 at 16:18

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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.

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  • $\begingroup$ 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! $\endgroup$
    – stoic-santiago
    Oct 3, 2019 at 4:35
  • $\begingroup$ 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. $\endgroup$
    – G. Smith
    Oct 3, 2019 at 4:48
  • $\begingroup$ 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. $\endgroup$
    – stoic-santiago
    Oct 3, 2019 at 4:53
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    $\begingroup$ 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! $\endgroup$
    – G. Smith
    Oct 3, 2019 at 5:07

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