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.

enter image description here

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?


3 Answers 3


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


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.


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.

  • $\begingroup$ why does the current distribution remain the same? If the plane is horizontal and perpendicular to the computer's screen and the current is going inside the screen, flipping the system, will make the current invert its direction and flow out of the screen. $\endgroup$ Commented Jan 5, 2020 at 19:19
  • $\begingroup$ Just to clarify, I don't mean total inversion. I mean rotating the system about x-axis by 180 degree. Then the current is still pointing in the same direction. $\endgroup$
    – Leo L.
    Commented Jan 5, 2020 at 23:02

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].
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.

  • $\begingroup$ Are you associated with lcwangpress (which seems likely)? You should be very clear about that. $\endgroup$
    – Jon Custer
    Commented Oct 27, 2022 at 17:53
  • $\begingroup$ All the papers in lcwangpress are written by Li-Chung Wang. $\endgroup$ Commented Oct 27, 2022 at 21:33
  • $\begingroup$ I own lcwangpress. lcwangpress.com is my website. $\endgroup$ Commented Oct 29, 2022 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.