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Question: A large parallel plate capacitor with uniform surface charge $\sigma$ on the upper plate and $-\sigma$ on the lower plate is lower with a constant speed V as in the figure. Use Ampere's law with the appropriate Amperian loop to find the magnetic field between the plates and also above and below them.

By Ampere's law:

$\vec{\bigtriangledown }\times \vec{B}=\mu_{0}\vec{J}+\mu_{0}\epsilon\frac{\partial \vec{E}}{\partial t}$

The current sheet due to the moving surface charge plate is $\vec{K}=\left \langle 0,\pm\sigma,0 \right \rangle$. Since the charges are constant across the surface, it can be expected that the current is steady/ constant so this is a case of magnetostatic. Additionally, the fact that the current is on the x-y plane suggests there is no x and y dependence.

By the Biot-Savart law for surface current:

$\vec{B}\left ( \vec{r} \right )=\frac{\mu_{0}}{4 \pi}\int \frac{\vec{K}\left ( \vec{r'} \right )\times \hat{\eta}}{\left \| \vec{\eta} \right \|^{2}}da'$

where $\vec{\eta}=\vec{r}-\vec{r'}$ where $\vec{r}$ is the vector distance from the origin to the field point and $\vec{r'}$ is the vector distance from origin to the source charge.

We expect the magnetic field not to be in the $\hat{y}$ direction due to the cross product in the integrand.

Fig

However, I do understand why the magnetic field does not have a $\hat{z}$ direction. In fact, I am unfamiliar with the right hand rule for 'surface' current.

Edit:I figured that if I rotate the plate 180 degrees about the z-axis in the CCW direction, the direction of the surface current changes-opposite to the direction of the surface current before the rotation-but the magnetic field continues to point in the positive z-axis which is a contradiction.

Would someone kindly clear my doubts? Thanks in advance.

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  • $\begingroup$ I did not quite understand what exactly your question was. can you state it once again ? just the part where you're confused . $\endgroup$ – Lelouch Aug 16 '16 at 13:32
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    $\begingroup$ Yes, are you interested in the magnitude of the B-field in space or the direction? $\endgroup$ – GeeJay Aug 16 '16 at 13:51
  • $\begingroup$ @JayJay I am interested in the direction of the B-field in space. I understand the B-field has no y component direction. This comes from the mathematics of the cross product in the integrand. But what about the B-field having no z component direction? I suggested that the B-field cannot have a z-component direction because if it did a contradiction arises. Am I correct? $\endgroup$ – Physkid Aug 16 '16 at 14:01
  • $\begingroup$ Yes, you are correct. $\endgroup$ – GeeJay Aug 16 '16 at 16:10
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I wasn't sure how to proceed with this question, since the figure makes it seem like a finite capacitor and does not provide the dimensions. I will attack the problem with the assumption that the capacitor is infinite in the x-y plane.

Now, there is one way to figure out the direction of the net B-field: logical reasoning as Griffiths does (Introduction to Electrodynamics, 3rd Edition, p.226, example 5.8) for the case of one plate with surface current $K$. Note that the current is in the +x direction in this example.

enter image description here

First of all, what is the direction of B? Could it have any x -component? No: A glance at the Biot-Savart law (5.39) reveals that B is perpendicular to K. Could it have a z -component? No again. You could confirm this by noting that any vertical contribution from a filament at +y is canceled by the corresponding filament at —y. But there is a nicer argument: Suppose the field pointed away from the plane. By reversing the direction of the current, I could make it point toward the plane (in the Biot-Savart law, changing the sign of the current switches the sign of the field). But the z -component of B cannot possibly depend on the direction of the current in the xy plane. (Think about it!) So B can only have a y -component, and a quick check with your right hand should convince you that it points to the left above the plane and to the right below it.

Griffiths then proceeds to make an Amperian loop, and find the B-field:

enter image description here

So, in your case, where the current in in +y direction you are correct that there cannot be a z component. There will only be an x component.

Now for the real fun. Since you put that Biot-Savart law for surface current in your question, I thought I might as well use it and go ahead and show that the net B-field is in the x-direction the integration way. So here goes

We know,

$\vec{B}\left ( \vec{r} \right )=\frac{\mu_{0}}{4 \pi}\int \frac{\vec{K}\left ( \vec{r'} \right )\times \vec{\eta}}{\left \| \vec{\eta} \right \|^{3}}da'$

where $\vec{\eta}=\vec{r}-\vec{r'}$ where $\vec{r}$ is the vector distance from the origin to the field point and $\vec{r'}$ is the vector distance from origin to the source charge.

K=$\sigma$v$\hat{y}$

Let us take any point ($\ x_1,y_1,z_1$) that we want to find the net B-field at. We find the field here due to the surface current at the general point ($\ x,y,0 $)

$\vec{\eta}=(x_1-x)\hat{x}+(y_1-y)\hat{y}+z_1\hat{z}$

${\vec{K} ( \vec{r'} )\times \vec{\eta}}$=$\sigma$v$\hat{y} \times ((x_1-x)\hat{x}+(y_1-y)\hat{y}+z_1\hat{z}) $ =$z_1\hat{x}-(x_1-x)\hat{z} $

Now for the integral, which I take over the entire xy plane:

$\vec{B}\left ( \vec{r} \right )=\frac{\mu_{0}\sigma v}{4 \pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{\vec{K}\left ( \vec{r'} \right )\times \vec{\eta}}{( {(x_1-x)^{2}+(y_1-y)^{2}+{z_1}^{2}} )^{3/2}}dxdy$= $\frac{\mu_{0}\sigma v}{4 \pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{z_1\hat{x}-(x_1-x)\hat{z}}{( {(x_1-x)^{2}+(y_1-y)^{2}+{z_1}^{2}} )^{3/2}}dxdy$

Separating this integral into two separate ones, one that finds the field in the x-direction and one in the z-direction: the latter comes out to be zero and the former:

$\frac{\mu_{0}\sigma v}{4 \pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{z_1\hat{x}}{( {(x_1-x)^{2}+(y_1-y)^{2}+{z_1}^{2}} )^{3/2}}dxdy =\frac{\mu_{0}\sigma v}{4 \pi}*\frac{2 \pi z}{|z|} $

So, you get:

$ \vec{B}=\frac{\mu_0\sigma v\hat{x}}{2} (z>0 $, ie-above the positive plate)

$ \vec{B}=-\frac{\mu_0\sigma v\hat{x}}{2} (z<0 $, ie-below the positive plate)

Extending these results to both plates (adding the contributions from both $\sigma $ and -$\sigma $ you will find that

$ \vec {B}=-\mu_0\sigma$v$\hat{x}$ (between the plates)

$ \vec {B}=0 $ (above and below)

I hope I addressed all your doubts!

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    $\begingroup$ Thanks for your answer! I think it addresses all the issues @Physkid had very well. $\endgroup$ – Gauri Aug 17 '16 at 9:02

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