# 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

• Is there a reason for you to believe it is possible to answer this question without vector calculus? To me the result seems obvious with vector calculus, but impossible without it Commented May 14 at 18:05
• hint: (1) what is the direction of the discharge current? (2) what is the direction of the magnetic field of this discharge current? Commented May 14 at 18:10
• Fun fact: prediction of how charge density evolves in time in this kind of system after the discharge starts (a transient process) is a tricky problem to solve, as assuming current obeys Ohm's law implies magnitude of charge density in the medium decreasing exponentially with time to zero value, while intuitive idea of the process says charge density near the negative sphere must temporarily decrease from zero to some non-zero negative value, and only later can decay back to zero. Does anybody know of a source with detailed analysis of how ρ in the medium between the spheres evolves in time? Commented May 15 at 1:44
• (The naive use of Ohm's law $\mathbf j = \sigma \mathbf E$ with constant $\sigma$ is incorrect, as $\sigma$ increases when concentration of free charges increases, and thus $\sigma$ depends on $\rho$, thus on position and time). Commented May 15 at 7:32
• @nickolasAlves I didn't study electrodynamics using vector calculus and all such things.I am preparing for an Entrance test to college [JEE] Commented May 15 at 7:51

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.

• > charge is conserved and doesn't build up Charge is locally conserved, and can build up, that is, $\partial_t \rho$ need not vanish in the medium. In fact during the discharge it can't vanish in the medium everywhere, because the discharge process means appearance of charge carriers in the medium where $\rho$ was initially zero, so $\rho$ can't be zero when the new charge is present. If the mobile charge carriers carry only negative charge (when only electrons are mobile carriers), $\rho$ in the medium decreases from zero to negative value after the electrons appear there. Commented May 15 at 1:39
• So current density isn't simply divergence-less $k\frac{\hat{r}}{r^2}$, but a more complicated function of position, with non-zero divergence. Of course, due to symmetry of the system, it should still always be radial during the discharge. Commented May 15 at 1:59
• The argument with mirror is unclear. A mirror image of medium element with current $\vec{J}$, in general, does not produce an image medium element with current density $-\vec{J}$. Mirror image modifies only one cartesian component, e.g. mirror in $z$-plane creates a mirror image current element with current density $(J_x,J_y,-J_z)$ out of a real element with current density components $(J_x,J_y,J_z)$. Commented May 15 at 2:03
• @JánLalinský I'm talking about parity inversion, not image charges. OP requested: keep it high school level.
– JEB
Commented May 15 at 13:50
• @JánLalinský charge carriers? Holdup. That is far too practical, I never let reality ruin a thought experiment. If you're concerned about the sign, we'll make it out of antimatter. You can do that in thought experiments.
– JEB
Commented May 15 at 13:52

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.

• Should the flux equations read $B_r$, not $B_t$? Commented May 15 at 19:47
• @CameronWilliams Indeed they should, thanks. Commented May 15 at 20:34
• You seem to be assuming the radial component of d$E_r$/dt is zero in your presentation of Ampere's Law. Why? By construction, the problem has finite $E_r$ at t=0 and it decreases to over the period of concern. Commented May 16 at 13:30
• @SarahMesser Current through the disk defined by the circle is zero, because current density flows in the plane of the disk. There would have to be current density component perpendicular to the disk. Commented May 16 at 15:42
• @SarahMesser You're right that in between the capacitor plates $\partial_t E_r \neq 0$, and this is a non-zero displacement current, but Ampere's law still applies, because this displacement current does not pass through the surface considered. My statement about applicability of the Ampere law in the answer was too restrictive, I'll fix it. Commented May 16 at 21:39

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.

• Why do you think the displacement current cancels the conduction current? This does not happen in general. It happens here, but how do you know it without assuming the result, or some detailed model of the conduction and polarization? Commented May 14 at 22:48
• (I should have written, it may happen here.) Commented May 15 at 8:51
• @JánLalinský I assume a slightly conductive dielectric, so Ohm's law applies. However, as long as the assumption of a spherical symmetry holds, it does not matter what kind of conduction is involved. That is, until the quantum regime is reached and shot noise occurs. Commented May 15 at 10:00
• Ohm's law for conduction current may apply, but conductivity $\sigma$ is a function of position and time, because it is proportional to concentration of the free charge. And how would Ohm's law imply the displacement current cancels the conduction current? Commented May 15 at 10:24
• Nitpick: I would replace the line with "... does not always generate a magnetic ..." with something along the lines of: "Apparently, the induced magnetic fields all cancel each other out exactly and everywhere!" My point is, that each infinitesimal current and field change does indeed have an associated induced magnetic field, but it's the integral of all these fields that happens to be zero everywhere. Commented May 15 at 21:33

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.

• I'd reword the argument a bit, but I think this is the only correct answer posted so far. Perhaps something like "Because of spherical symmetry, B must be constant under all rotations. In particular, imagine a circle concentric with the sphere with a non-zero tangential B. That tangential B must be constant under infinitessimal rotations of the zero, so it is a constant value on the circle. Tangential B must also be constant under a rotation (by 180 deg.) about any axis through opposite sides of the circle. But having a constant value for the first rotation means the second rotation is just Commented May 16 at 13:55
• a reversed sign. So the tangential B is a constant value which is equal to its negative, i.e. $B_{tan} = - B_{tan}$ everywhere. Thus the tangential B vanishes. By spherical symmetry and the vanishing divergence of B, the $B_r$ must also be zero." Commented May 16 at 13:56

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.