Does a time varying electric field always generate a Magnetic field?

Question

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

Answer 1

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.

Answer 2

The correct answer (magnetic field vanishing everywhere) can be reached on the high school level most simply by the argument using the central symmetry of the capacitor, or in a more complicated way, using this symmetry and also the Ampere law and the Gauss law of magnetism.

The main idea (assumption) is that electric and magnetic field of a charge and current distribution with central symmetry should not imply force that breaks this symmetry. Since EM force on a test charge is proportional to electric and magnetic field, then these fields themselves should not break that symmetry. This idea is actually a difficult topic to explain and convincingly argue for, especially when the EM field is not static everywhere, but let's assume it is true.

Let's consider an imaginary sphere concentric with the capacitor, with arbitrary radius $r$, and a test charge on this sphere.

First of all, let's notice that at any point P of the sphere, transversal component of electric field $\mathbf{E}_t$ (the component of $\mathbf E$ in the plane touching the sphere at the point P) has to vanish, that is, $E_t$ vanishes everywhere on the sphere. This is because any non-zero transversal component would mean test charge there would be pushed by electric force $q\mathbf E_t$ transversally, and such force can't come from the bound electric field of the symmetric charge and current distribution of the capacitor. Thinking of the capacitor as a globe with test charge being at the north pole, equally sized sectors of the capacitor at the same latitude but different longitudes act on the test charge in all different directions in the tangent plane with the same strength, and cancel each other. Thus the net electric force can, at this point of the argument, have non-zero component only in the radial direction. The radial component $E_r$ has to have the same value at any point of our imaginary sphere, again due to symmetry of the capacitor and its radial current. Using the Gauss law, we can relate $E_r$ to net charge inside the imaginary sphere: $$ 4\pi r^2 E_r = \frac{Q_{in}}{\epsilon_0}. $$

Similar argument could be made for why the transversal component of magnetic field $\mathbf B_t$ has to vanish everywhere. If it was non-zero, then a test charge moving in radial direction would experience magnetic force $q\mathbf v \times \mathbf B_t$ in transversal direction. Such transversal force would manifest asymmetry with respect to the charge and current distribution in the capacitor, because there is nothing in those that would imply single specific direction of that force. Equally sized elements of the capacitor at the same geographic latitude but different longitude contribute with equally strong forces in all different directions equally, and cancel each other. Thus net magnetic field component in the tangential plane has to be zero.

There remains a possibility that radial component of the magnetic field $B_r$ isn't zero. Again, due to symmetry, this quantity should be the same everywhere on the imaginary sphere. So we can express magnetic flux through this imaginary sphere as $$ \Phi_B = 4\pi r^2 B_r. $$ However, the Gauss law of magnetism states that magnetic flux through any closed surface (which the imaginary sphere is), is zero. Thus we have

$$ \Phi_B = 4\pi r^2 B_r = 0 $$ from which we conclude $B_r=0$ everywhere. Thus both the tangential and the radial components of $\mathbf B$ vanish everywhere, thus $\mathbf B$ vanishes everywhere.

We can make a different kind of argument for $\mathbf B_t$ being zero everywhere, using the EM laws. The Ampere law states that circulation of magnetic field around a closed path equals net real current $I_\Sigma$ (due to conduction and change of polarization of the medium) that passes through the surface $\Sigma$ that the closed path is a boundary for. Formally:

$$ \oint_{boundary~of~\Sigma} \mathbf B \cdot d\mathbf s = \mu_0 I_{\Sigma}. $$

We will use this law for the configuration where $\Sigma$ is any imaginary disk whose center coincides with the center of the capacitor, and has arbitrary radius (smaller or greater than the capacitor).

This law, as almost all physical laws, has limited applicability. Usually it is said this law holds only when electric field is constant in time. But it also holds while electric field changes in time, provided the flux integral of $\partial_t \mathbf E$ through the surface $\Sigma$ remains zero. This follows from the Maxwell-Ampere law; unfortunately, I don't know how to show this on the high-school level.

In between the capacitor plates, although electric field changes in time, its radial character is preserved, so rate of change ($\partial_t \mathbf E$) is radial too. Thus flux of $\partial_t \mathbf E$ through the disk is zero, and the Ampere law holds. Due to symmetry of the current flow (which we assume to be everywhere radial), we expect magnetic field component in direction of the circle element $B_t$, at any point of the imaginary circle, to have the same value (tangential component). Then circulation of magnetic field on the circle is

$$ C=2\pi r B_t. $$ According to the Ampere law, this should be equal to $\mu_0$ times the net current passing through the disk. But that current is, due to net current density on the disk pointing everywhere in the plane of the disk, zero! Thus we have

$$ C=2\pi r B_t = 0, $$ from which we conclude $B_t$ is zero. Orientation of the circle is arbitrary, thus at any point, magnetic field component lying in the tangent plane to any point of the imaginary sphere is zero.

This result (magnetic field vanishing everywhere during the whole discharge) raises the question, how on earth does energy flow in this system during the discharge, when the Poynting vector $\mathbf E\times\mathbf B/\mu_0$ vanishes everywhere all the time. It looks like there is no EM energy flow during the discharge, which is a little unexpected. EM energy in the capacitor does not flow in space at all, but just stays at its place and dissipates there.

Answer 3

Here is a different way, from @JEB 's, of looking at this problem. The current $\bf J$ is cancelled by the displacement current ${\dot {\bf D}}$, so ${\bf B}=0$. The reason is that $${\bf J} = \frac{{\dot q}{\hat r}}{4\pi r^2} = {\dot {\bf D}} ~.$$ If a dielectric constant is taken into account Since $\bf D$, so $\bf E$ is rotation free also ${\dot {\bf B}}=0$. It does not matter what the conduction mechanism is, as long as its is isotropic.

A current does not always generate a magnetic field, apparently!

Here are the relevant Maxwell equations for reference.

\begin{align} \nabla \times \mathbf{E} &= - {\dot {\bf B}} \\ \nabla \times \mathbf{B} &= \mu_0 {\bf J} + {\dot {\bf D}} \end{align}

I edited this answer to reflect part of the discussion. The conclusion is unaffected.

Answer 4

The zero magnetic field can be understood intuitively from the symmetry of the situation. If we work with geographic-like coordinates on the sphere, a radial current, at a given longitude and latitude, would contribute to a B-field say one second North. But that B-field would be neutralized by the contribution from a radial current with the same magnitude two seconds North.

More deeply, it is a consequence of Ampère-Maxwell equation. The (radial) current here is proportional to the decrease of the build up of charges with respect to time. So, the rate of change of the (radial) electric field compensates the current, resulting in zero curl for the magnetic field everywhere.

Answer 5

More conceptually, this comes from the fact that 1D spatial dimensions, electromagnetism (and one time dimension) has no magnetism. There is simply not enough spatial dimensions. More generally, in $D$ spatial dimensions, you have $D$ components of electric fields but $D(D-1)/2$ components of magnetic field. In particular, you do not have waves, hence no energy flux.

Due to spherical, your problem is 1D along the radial direction, so the previous analysis applies. Note that there are many other situations where this applies. Replace spherical symmetry with cylindrical symmetry and you get concentric cylindrical capacitors, or replace it with 2D translations and you get two parallel plane capacitors. In all those cases, you have no magnetic field and no energy flux.

Hope this helps.