Griffiths' symmetry argument for magnetic field direction of an infinite uniform surface current $K = K \hat x$, flowing over the xy plane

Question

This example is from Griffiths' Introduction to electrodynamics, 4th edition,

Example 5.8. Find the magnetic field of an infinite uniform surface current $K = K \hat x$, flowing over the xy plane First of all, what is the direction of B? Could it have any x component? No: A glance at the Biot-Savart law (Eq. 5.42) 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.

I can't figure out the second argument, why can't the z component of B depend on the direction of the current in the xy plane?

Answer 1

Let's say you're at the origin and your up/down/left/right directions point respectively towards +z/-z/+y/-y axes. Let's also assume that the surface current $K$ is flowing in the +x direction (flowing into the plane of your vision). Let's call the z-component of the magnetic field $B_z(x,y,z)$.

Refinement #1

$B_z(x,y,z)$ is only a function of z.

Reasoning: Since the plane of the surface current is infinite, the magnetic field $\mathbf{B}$ at two points $(x_1,y_1,z)$ and $(x_2,y_2,z)$ cannot be distinguished, and hence are exactly the same.

Refinement #2

The z-component of the magnetic field should either point toward the x-y plane or away from the x-y plane when viewed from both sides of the x-y plane, at all points that are the same distance away from the x-y plane. What I mean to say, is that $\mathbf{B} (z)\cdot \hat{z}=-\mathbf{B} (-z)\cdot \hat{z}$.

Reasoning: Let's say the z-component of the magnetic field at $(5,5,5)$ points away from the x-y plane. If you turned 180$^o$ with respect to the x-axis, then your up/down/left/right directions are changed to -z/+z/-y/+y, and as a result, you observe the point $(5,-5,-5)$ in the place of $(5,5,5)$. To you, the problem setup hasn't changed: The current is still flowing into the plane away from you. Therefore, the z-component of the magnetic field at $(5,-5,-5)$ should also point away from the x-y plane and have the same magnitude as the magnetic field at $(5,5,5)$. Combining the knowledge of Refinement #1 ($\mathbf{B}(x,y,z)=\mathbf{B}(z)$), we see that Refinement #2 follows.

Refinement #3

$B_z(z)=0$

Reasoning: Let's say the z-component of the magnetic field at $(5,5,5)$ points away from the x-y plane, for a given value of $K$, irrespective of the direction of the surface current in the x-y plane (rotational symmetry about the z-axis). Alright, no contradictions so far. Let's now reverse the direction of the current. Due to the rotational symmetry about z-axis, the problem hasn't changed and the z-component of the magnetic field at $(5,5,5)$ must still point away from the x-y plane. However, one can see from the Biot Savart law that the direction of $\mathbf{B}$ gets flipped at all points if the direction of current is reversed. According to this new information, the z-component of the magnetic field at $(5,5,5)$ should instead point towards the x-y plane for the same value of $K$, which is a contradiction. The only possible solution that Nature can take to resolve this problem is to set $B_z(5)=0$. And similarly it applies to all z.

Keep in mind that Refinement #3 involves the application of a key information from the Biot Savart law.

Answer 2

It is a symmetry argument, basing on the symmetry of the current distribution if you invert the coordinate system along z direction.

Suppose the B field indeed has a z component, let's say along the positive z axis. If we now flip the entire system upside down, then the B field will now have a negative z-component, yet the current distribution is still the same!. It is not possible that the B field's direction is dependent on the coordinate axis you define (as it is arbitrary), so B field cannot have any z-component.

Answer 3

Griffiths = Griffiths, D. J.: Introduction to Electrodynamics, 4th ed., New York: Pearson, 2013.
The $z$-component of $\pmb{B}$ cannot possibly depend on the direction of the current in the $xy$-plane [Griffiths, p.235, l.11-l.12].
Proof.
I. In order to facilitate finding a key to solving a practical problem, we should build a simple model to see the big picture for the problem.
Simple model: springhead $=$ the origin; surface current $=$ a water flow in a channel
$=$ a ray from the origin [we regard the wide water flow as narrow as a ray].
II. On the $xy$-plane, if we increase the angle of the ray (i.e. the channel $=$ surface current) by angle $\theta$, the situation will obviously be the same as that after we rotate the ray by angle $\theta$ around the z-axis.
III. Fix a point $\pmb{r}_0=(x_0,y_0,z_0)$ not on the $xy$-plane. After the rotation, the point $\pmb{r}_0=(x_0,y_0,z_0)$ moves to a new point $\pmb{r}_0'=(x_0',y_0',z_0)$. Let $B_z^{(\theta)}(\pmb{r}_0')$ be the $B_z$ value at $\pmb{r}_0'$ after rotation. By II, $B_z(\pmb{r}_0)=B_z(z_0)=B_z^{(\theta)}(z_0)=B_z^{(\theta)}(\pmb{r}_0')$.
IV. Let $B_z^{Rev}(\pmb{r}_0)$ be the $B_z$ value at the point $\pmb{r}_0$ after the current is reversed. Then
$B_z^{Rev}(\pmb{r}_0)=-B_z(\pmb{r}_0)$ [by the Biot-Savart law]
$B_z^{(\theta)}(z_0)=B_z(\pmb{r}_0)(*)$ (by III).
By II, $B_z^{Rev}(\pmb{r}_0)=B_z^{(\theta)}(z_0)(**)$. By $(*)$ and $(**)$, $-B_z(\pmb{r}_0)=B_z(\pmb{r}_0)$. Thus, $B_z(\pmb{r}_0)=0$.QED.
This answer is excerpted from §1.10.(A), Remark 34 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/quantum-mechanics.