Does a time varying electric field always generate a Magnetic field


I have come across this problem from 200 Puzzling physics problems

A charged spherical capacitor slowly discharges as a result of the slight conductivity of the dielectric between its concentric plates. What are the magnitude and direction of the magnetic field caused by the resulting electric current?

My attempt:
Since the situation is radially symmetric the energy flow (pointing vector) should be along the radial direction
But S can't parallel to E
This implies both S and B are zero ?
Is my reasoning correct? Does it imply time varying electric field need not produce magnetic field always

Please answer avoiding higher level vector analysis and all…I am a high school student

Best Answer

That's a fun problem. Vector analysis with $\vec \nabla$ can be used to verify an answer that doesn't need it in the derivation.

So you have basically described a spherically symmetric shell of radially directed current.

With that, at any point in the shell $(r, \theta, \phi)$ (standard spherical coordinates), you have an electric field:

$$ \vec E = E_r\hat r + E_{\theta}\hat\theta + E_{\phi}\hat\phi $$

Symmetry considerations tell you $E_{\theta} = E_{\phi}=0$, so the field looks like:

$$ \vec E \propto \hat r $$

Then you need a current density, which is clear parallel to $\vec E$:

$$ \vec J \propto \hat r / r^2 $$

where I threw in the areal factor of $1/r^2$ so it's divergence is zero (i.e.: charge is conserved and doesn't build up, except of course in the conducting shells).

But none of that is totally necessary...I'm just covering all the bases.

Now the magnetic field $\vec B$ cannot be proportion to $\vec J$, because it is an axial-vector, and current is a vector. If you take mirror image of a system, then:

$$ \vec J \rightarrow -\vec J$$ $$ \vec B \rightarrow +\vec B$$

so any proportionality would violate parity symmetry, which is a fundamental symmetry of electromagnetism (and newtonian dynamics).

So at this point, you can confidently say

$$\vec B=0$$

but we still have the vector potential, $\vec A$, and ofc

$$ \vec B = \vec\nabla \times \vec A $$

and there could be a non-zero $\vec A$. So let's check.

By symmetry, it also has to be radial:

$$ \vec A = A_r(r)\hat r $$

where it may be a function of $r$ (but not the angular coordinates, again: by symmetry).

Now we take the curl of it:

$$ \vec\nabla \times A \equiv \frac 1 {r\sin\theta}\Big( \frac{\partial}{\partial\theta}(A_{\phi}\sin\theta) - \frac{\partial}{\partial\phi}A_{\theta} \Big)\hat r + \frac 1 r\Big(\frac 1 {\sin\theta} \frac{\partial A_r}{\partial\phi} - \frac{\partial}{\partial r}(rA_{\phi}) \Big)\hat\theta + \frac 1 r \Big( \frac{\partial}{\partial r}(rA_{\theta})- \frac{\partial A_r}{\partial\theta} \Big)\hat\phi $$

and the only terms depending on $A_r$ involve angular derivatives, so it is zero.

Related Question