I Magnetic field inside a capacitor

[Malamala]

Hello! I have a parallel plate capacitor (we can assume that the plates are circular) and I apply a time varying voltage to it, such that the electric field inside is $E_0\sin \omega t$. If I use the Maxwell equations, I get for the magnetic field

$$B(t) = \frac{\omega E_0}{2c^2}r\hat{r}$$
so the magnetic field increases with the radius, and it is in the radial direction. However, I am not sure I understand how to define $r=0$. Naturally, I could take that at the center of the plates, such that the magnetic field at the center is zero. But I am not sure why I need to do that. If I take a small (imaginary) circular loop and apply Maxwell's equations to that, I would still get $B(t) = \frac{\omega E_0}{2c^2}r\hat{r}$, regardless of where I place that loop. Thus it seems (ignoring edge effects, which should be ok if I assume that I look at small enough loops and the plates are large) that the magnetic field is zero everywhere inside the capacitor, as I can take the center of this loop at any point inside. What am I missing? Why can't I apply this formalism to any loop in between the 2 plates?

 

[renormalize]

Hello! I have a parallel plate capacitor (we can assume that the plates are circular) and I apply a time varying voltage to it, such that the electric field inside is $E_0\sin \omega t$. If I use the Maxwell equations, I get for the magnetic field
$$B(t) = \frac{\omega E_0}{2c^2}r\hat{r}$$

Are you working in cylindrical-coordinates $r,\theta,z\,$? If so, is your electric-field vector $\vec{E}$ parallel to the $z$-direction; i.e., $\vec{E}=E_0\sin \omega t\,\hat{z}\,$?

 

[Malamala]

Are you working in cylindrical-coordinates $r,\theta,z\,$? If so, is your electric-field vector $\vec{E}$ parallel to the $z$-direction; i.e., $\vec{E}=E_0\sin \omega t\,\hat{z}\,$?

I assumed cylindrical coordinates for simplicity, as I took the plates to by disks. And yes the electric field is along the z direction (sorry for not clarifying that). My confusion is that when applying the Ampere-Maxwell law, I can take a circular loop anywhere between the 2 plates and upon solving the equation I get that the magnetic field at the center of the loop is zero. But given that I can place the loop anywhere I want, it looks like the magnetic field is zero everywhere. Basically, if I am far from the edges, the plates appear as infinite and thus I get a translational symmetry along the plane between the 2 plates.

 

[weirdoguy]

I can take a circular loop anywhere between the 2 plates

You can, but because of the symmetry of your setup, it will NOT give the same result for each placement. Circular loop which center is not alligned with the center of the plate will "break the symmetry". Think about Gauss law in electrostatics with one pointlike charge - would sphere not centered on the charge give you the same result as the one centered on the charge?

 

[Malamala]

You can, but because of the symmetry of your setup, it will NOT give the same result for each placement. Circular loop which center is not alligned with the center of the plate will "break the symmetry". Think about Gauss law in electrostatics with one pointlike charge - would sphere not centered on the charge give you the same result as the one centered on the charge?

I agree that the symmetry chooses the center as a preferential point. But I am not sure, from the point of view of applying the Ampere-Maxwell equation, why is the conclusion wrong? Again, ignoring edge effects, the position of the loop shouldn't matter. Just to clarify, by ignoring edge effects I mean that the electric field going through the loop is perfectly aligned with the z-axis and uniform (its amplitude). Of course this is not exactly the case if I have finite plates, but I use the same assumption when the loop is centered, too, so it seems like a fair comparison.

Along the same line of thought, what happens if the plates were infinite? In that case I don't have a preferential point anymore, so it seems like the magnetic field would actually be zero everywhere. However, when applying the Ampere-Maxwell formula, I would get a non-zero magnetic field along the loop I choose.

 

[renormalize]

Hello! I have a parallel plate capacitor (we can assume that the plates are circular) and I apply a time varying voltage to it, such that the electric field inside is $E_0\sin \omega t$. If I use the Maxwell equations, I get for the magnetic field$$B(t) = \frac{\omega E_0}{2c^2}r\hat{r}$$so the magnetic field increases with the radius, and it is in the radial direction.

I assumed cylindrical coordinates for simplicity, as I took the plates to by disks. And yes the electric field is along the z direction (sorry for not clarifying that).

Two of Maxwell's equations that must hold in the gap between the plates are:$$\nabla\times\vec{B}=\mu_{0}\left(\vec{J}+\varepsilon_{0}\frac{\partial\vec{E}}{\partial t}\right),\;\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}\tag{1a,b}$$Since your assumed electric-field $\vec{E}=E_0\sin \omega t\,\hat{z}$ is curl-free and $\vec{J}=0$ in the gap, eqs.(1) reduce to:$$\nabla\times\vec{B}=\frac{\omega E_{0}\cos\omega t}{c^{2}}\hat{z}\,,\;0=-\frac{\partial\vec{B}}{\partial t}\tag{2a,b}$$Can you show eqs.(2) are solved to arrive at your magnetic-field $\vec{B}=\frac{\omega E_0}{2c^2}r\,\hat{r}$ that is both time-independent and points in the radial-direction?
Edit: Also note that your $\vec{B}$ violates Gauss's law for magnetism: $\nabla\cdot\vec{B}=\frac{\omega E_{0}}{c^{2}}\neq0$.

 

[Malamala]

Two of Maxwell's equations that must hold in the gap between the plates are:$$\nabla\times\vec{B}=\mu_{0}\left(\vec{J}+\varepsilon_{0}\frac{\partial\vec{E}}{\partial t}\right),\;\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}\tag{1a,b}$$Since your assumed electric-field $\vec{E}=E_0\sin \omega t\,\hat{z}$ is curl-free and $\vec{J}=0$ in the gap, eqs.(1) reduce to:$$\nabla\times\vec{B}=\frac{\omega E_{0}\cos\omega t}{c^{2}}\hat{z}\,,\;0=-\frac{\partial\vec{B}}{\partial t}\tag{2a,b}$$Can you show eqs.(2) are solved to arrive at your magnetic-field $\vec{B}=\frac{\omega E_0}{2c^2}r\,\hat{r}$ that is both time-independent and points in the radial-direction?
Edit: Also note that your $\vec{B}$ violates Gauss's law for magnetism: $\nabla\cdot\vec{B}=\frac{\omega E_{0}}{c^{2}}\neq0$.

Sorry I am not sure what you mean by this reply. I agree with your derivations, but I still don't see why the magnetic field can't be identically zero.

 

[renormalize]

Sorry I am not sure what you mean by this reply. I agree with your derivations, but I still don't see why the magnetic field can't be identically zero.

My reply was intended to help you recognize that you made an error in calculating your magnetic field: $\vec{B}$ inside the parallel-plates of a circular capacitor of radius $a$ cannot possibly point in the radial-direction $\hat{r}$.
Instead, $\vec{B}$ points in the direction of $\hat{\theta}$. Specifically, near the center the field takes the form:$$\vec{B}(r)=\frac{\mu_0\,Ir}{2\pi a^2}\hat{\theta}$$where $I=I_0 \cos(\omega t)$ is the external current that induces the oscillating electric-field between the plates. For approximate consistency with eq.(2b) in post #6, we have to assume that the current changes slowly; i.e., the frequency $\omega$ is low enough that $\frac{\partial\vec{B}}{\partial t}\approx 0$. Also, note that as the radius $a\rightarrow\infty$ the field $\vec{B}\rightarrow 0$. So your intuition is correct: between the plates of an infinite capacitor the magnetic field is identically zero!

 

[Baluncore]

As I wave my hands, I consider a disc capacitor, with axial conductors connected to the centre of two conductive discs, with a dielectric between the inner disc surfaces. Now I consider the current flowing in one side of that circuit.

The currents in the connection wires will be surrounded by circular magnetic field lines. When the current reaches the outside centre of the disc, it will radiate outwards towards the periphery, symmetrically cancelling any external magnetic fields on the back of the disc.

On reaching the periphery, the current will flow around the edge, all in the same axial direction as the connections, forming a circular magnetic field, just outside the circumference of the disc.

Note that skin effect will cause the current to flow out, and around the conductive disc surface, not through it.

As the current turns in, to become inward radial, it begins to flow past the face of the capacitor's dielectric, where symmetry again cancels all magnetic fields. That radial current will be reduced to a zero at the centre, as the current causes charge to take up a position on the surface. It then becomes an axial displacement current, through the dielectric of the capacitor.

That displacement current will have a circular magnetic field, surrounding the dielectric, but not within. The two peripheral magnetic fields, and that due to the displacement current, will be close together and will sum, outside the dielectric, in the plane of the disc.

In summary, there will be circular magnetic fields about the connecting wires, and a circular magnetic field outside the circumference of the disc. There will be no magnetic fields on the faces of the disc electrodes, nor within the dielectric.

 

[Malamala]

As I wave my hands, I consider a disc capacitor, with axial conductors connected to the centre of two conductive discs, with a dielectric between the inner disc surfaces. Now I consider the current flowing in one side of that circuit.

The currents in the connection wires will be surrounded by circular magnetic field lines. When the current reaches the outside centre of the disc, it will radiate outwards towards the periphery, symmetrically cancelling any external magnetic fields on the back of the disc.

On reaching the periphery, the current will flow around the edge, all in the same axial direction as the connections, forming a circular magnetic field, just outside the circumference of the disc.

Note that skin effect will cause the current to flow out, and around the conductive disc surface, not through it.

As the current turns in, to become inward radial, it begins to flow past the face of the capacitor's dielectric, where symmetry again cancels all magnetic fields. That radial current will be reduced to a zero at the centre, as the current causes charge to take up a position on the surface. It then becomes an axial displacement current, through the dielectric of the capacitor.

That displacement current will have a circular magnetic field, surrounding the dielectric, but not within. The two peripheral magnetic fields, and that due to the displacement current, will be close together and will sum, outside the dielectric, in the plane of the disc.

In summary, there will be circular magnetic fields about the connecting wires, and a circular magnetic field outside the circumference of the disc. There will be no magnetic fields on the faces of the disc electrodes, nor within the dielectric.

Thanks a lot for this! But I actually got confused by the last sentence. Are you saying that indeed the magnetic field is zero in between the 2 plates (is this what you meant by "nor within the dielectric.")? And it becomes non-zero only outside?

 

[Baluncore]

Are you saying that indeed the magnetic field is zero in between the 2 plates ... and it becomes non-zero only outside?

Yes.

 

[renormalize]

Thanks a lot for this! But I actually got confused by the last sentence. Are you saying that indeed the magnetic field is zero in between the 2 plates (is this what you meant by "nor within the dielectric.")? And it becomes non-zero only outside?

Yes.

@Baluncore, that's simply wrong.
At least two experiments have successfully measured the magnetic field existing between the plates of a sinusoidally-charging capacitor and verified the formula in post #8; i.e., $\vec{B}(r)\propto r\,\hat{\theta}$. In 1985 Bartlett and Corle immersed a capacitor in liquid helium within a superconducting sphere and used a SQUID probe to quantify the field. More recently, Scheler and Paulus in 2015 employed a different apparatus to make the measurement:

For a good overview of this subject see: John A. Milsom, Untold secrets of the slowly charging capacitor.

 

[Baluncore]

@Baluncore, that's simply wrong.

I agree that if you measure the radial current at a point on the disc, there will be a local magnetic field due to that current, but at the same instant, on the other side of the disc axis, is an equal and opposite current and magnetic field. The distance between those two points is so small, in wavelengths, that the two local fields, will in effect cancel, within the volume of the capacitor.

The displacement current flows between the two plates, with an equal current density across the plate and dielectric. Two equal neighbouring small areas of the disc have circular magnetic lines about them, with the same sense, but where the areas meet, the circular fields run counter, and so cancel within the dielectric, between the plates.

Sticking a ferrite rod into a capacitor, to sense the internal magnetic field, is ridiculous. There will be so much peripheral field coupled into the side of the rod, that the signal detected will not be a sample of a magnetic field at that end of the rod. Surely, the sense coil will be more closely coupled to the closer outer end of the rod, than the distant inner.

 

[renormalize]

...The distance between those two points is so small, in wavelengths, that the two local fields, will in effect cancel, within the volume of the capacitor. ...
...Two equal neighbouring small areas of the disc have circular magnetic lines about them, with the same sense, but where the areas meet, the circular fields run counter, and so cancel within the dielectric, between the plates.

@Baluncore, the two independent experiments I cited both agree and clearly demonstrate that:
a) there is a nonzero magnetic field in the dielectric (liquid helium for the first and air for the second) between the capacitor plates and
b) said magnetic field agrees well with the theoretical prediction:

Please clarify your position: do you reject these peer-reviewed and published experiments that contradict your statements above? Can you offer an alternative explanation for the graphs?

 

[A.T.]

Two equal neighbouring small areas of the disc have circular magnetic lines about them, with the same sense, but where the areas meet, the circular fields run counter, and so cancel within the dielectric, between the plates.

Isn't the local current on one plate opposite to the local current at the closest position on the other plate? But then the magnetic fields between the plates would add up, not cancel.

 

[Baluncore]

Isn't the local current on one plate opposite to the local current at the closest position on the other plate? But then the magnetic fields between the plates would add up, not cancel.

Yes. You are correct. The displacement current does not generate a magnetic field, but an equivalent field to the displacement current is generated by the currents on the inside faces of the plates. I have not yet reconciled the two different approaches. It seems to depend on the ratio of separation to radius.

 

[A.T.]

but an equivalent field to the displacement current is generated by the currents on the inside faces of the plates.

Aren't the currents concentrated on the inside faces of the plates, due to the attraction by the opposite mobile charges on the other plate?