electric-currentelectricityelectromagnetismelectrostaticsmagnetostatics
Steady currents generates a constant magnetic field-magnetostatic
Formally, this is $\frac{\partial \vec{J}}{\partial t}=\vec{0}$
Any currents steady or otherwise are due to the movement of point charges-electric charges.
But how does the a moving point charge ties in with the fact that moving point charges cannot constitute a steady current?
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.
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:
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!
Best Answer
The same way one defines a current density J even though current is caused by discrete moving charges and should not have a continuous density. We look at a macroscopic volumes over which the discrete nature of the current can be ignored but not so big so as to wash out any large scale variations. In microscopic volumes moving charges don't constitute a steady current ,but over volumes containing thousands of atoms we can regard it a continuous current. In these volumes the average charge density does not change with time and that's what defines a steady current.