Recall how we find the Electric field of spherically symmetric charge distribution, cylindrically symmetric charge distributions and also charge distributions with planar symmetry (infinite flat sheet), by Gauss's Law and arguements based on symmetry alone. This make me think that we can do the same in magnetism by using Ampere's law and symmetry alone. But I failed. So I would like to know if I have used the wrong method of arguments.

So take an infinite straight long current carrying wire as the simplest example. WLOG, let it be lying on the $z$ axis with current flowing in the positive $z$ direction. So using cylindrical coordinate system $$\vec{B}(\rho,\phi,z)=B_\rho(\rho,\phi,z)\hat{\rho}+B_{\phi}(\rho,\phi,z)\hat{\phi}+B_z(\rho,\phi,z)\hat{z}$$

Now by translational symmetry along the wire, we conclude that nothing can have $z$-dependence. And by rotational symmetry about $z$ axis, nothing can have $\phi$ dependence. Hence $$\vec{B}(\rho)=B_\rho(\rho)\hat{\rho}+B_{\phi}(\rho)\hat{\phi}+B_z(\rho)\hat{z}$$

Now if you flip the current upside down, $B_{\rho}(\rho)\hat{\rho}$ is unchanged. While it should have become $-B_{\rho}(\rho)\hat{\rho}$ because superimposing the flipped current and the original current gives zero current and therefore zero field.

Hence now we have $\vec{B}(\rho)=B_{\phi}(\rho)\hat{\phi}+B_z(\rho)\hat{z}$$

Now here comes the problem: How to show that $$B_z(\rho)\hat{z}=\vec{0}$$

I saw some people using Biot-Savart Law at this point, appealing to the fact that a current in the $z$ direction produce no $B_z\hat{z}$.

So is it possible to reach the same conclusion by using symmetry and superposition principle and Ampere's law only, without asking for help from Biot-Savart Law?

Similar issues arises in deriving the field due to an infinitely long solenoid. It seems that in addition to symmetry, superposition, and Ampere's law, we also need Biot-Savart law and Gauss's law for magnetism as well. I will post a separate question soon if necessary.


2 Answers 2


Here is one argument:

Draw a rectangle with normal pointing in the $\phi$ direction and not enclosing the line of current. Suppose the rectangle has height $h$ and ranges from $\rho = a$ to $\rho = b$. Since $B_\rho = 0$, the line integral of $\mathbf{B}$ around this rectangle is simply $h(B_z(b)-B_z(a))$, which must be zero by Ampère's law. So $B_z$ doesn't depend on $\rho$, since $a$ and $b$ were arbitrary. If we want $\mathbf{B}$ to vanish as we move infinitely far from the wire, then $B_z$ is zero everywhere.

Incidentally, I'm not sure I understand your argument that $B_\rho$ must be zero. To me the natural approach would be to use the divergenceless of $\mathbf{B}$.

  • $\begingroup$ The radial component could depend on $r^{-1}$ and remain divergence free. You do need another argument. If the current changes direction it is just like turning the wire around, so the radial B-field direction would not change. Hence the radial component must be zero, as the OP argues. $\endgroup$
    – ProfRob
    Jun 11, 2020 at 16:53
  • $\begingroup$ @ProfRob Except when $r = 0$. Such a field configuration gives a delta function in the divergence of $\mathbf{B}$. You may say that $\mathbf{B}$ is not defined at $r = 0$, but I say that this configuration corresponds to a line of magnetic charges, and that the equation $\mathrm{div}\, \mathbf{B} = 0$ precludes this possibility (if not by mathematical necessity, then by the understood subtext). $\endgroup$
    – gj255
    Nov 16, 2021 at 14:08

For static fields this can be done using Amperes circuit law.

$$ \nabla\times\vec{H}=\nabla\times(H_\phi(r)\hat{\phi}+H_z(r)\hat{z})=-\frac{\partial H_z}{\partial r}\hat{\phi}+r^{-1}\frac{\partial (r H_\phi)}{\partial r}\hat{z}\overset{!}{=}\vec{j}=j_0\hat{z} $$ And therefore by orthogonality $$ \frac{\partial H_z}{\partial r}=0 \rightarrow H_z(r)=C \quad \forall r $$ Enforcing a vanishing magnetic field infinitely far away from the current, a condition which every physically sensible solution should satisfy, one finds $$H_z(r)=0 \quad \forall r$$

It shall be stated, that this analysis can be extended for a time varying current (and therefore fields) in the region of the current carrying media (in the cable), assuming ohms law applies there and the conductivtiy is non-vanishing. For this convert the circuit law to frequency space and replace $\vec{E}(\vec{r},\omega)=\frac{\vec{j}(\vec{r}, \omega)}{\sigma(\omega)}$. This does not work outside the cable where $\sigma=0$.


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.