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