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Consider the infinity current carrying wire wire drawn below along with the three possible components of the magnetic field at a point a distance $r$ from the centre. enter image description here I know how to use symmetry to explain why the $x$ component of the field is 0. But I cannot think of one to explain why the $z$ component is 0. I know that this is trivially 0 with use of Biot-Savart's law, but I don't want to use that as it defeats the whole point of using the shorter Ampere's law. So my question basically is:

Is there a symmetry argument that I can use to explain why the $z$ component is 0? (I don't mind if it contains rotations, reflections, translations etc.

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  • $\begingroup$ if it's not zero, then should it be up or down? $\endgroup$ – Azad May 10 '15 at 8:35
  • $\begingroup$ @Azad You make a good point, but simply not been able to decide which way it should be does not imply that it is 0. $\endgroup$ – Quantum spaghettification May 10 '15 at 8:40
  • $\begingroup$ think why you can't decide $\endgroup$ – Azad May 10 '15 at 8:42
  • $\begingroup$ You want prove the Biot-Savart law or Ampere's formula? $\endgroup$ – Constantine Black May 10 '15 at 9:10
  • $\begingroup$ @ConstantineBlack Neither only that the $z$ component of the field is 0 without using Biot-Savart law and using symmetry $\endgroup$ – Quantum spaghettification May 10 '15 at 9:12
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There's no real symmetry argument that can explain why the $z$ component is zero. This is because it doesn't have to be: there can always be a uniform magnetic field added to the existing field, without affecting Maxwell's equations. They are all differential equations in the fields, like the following: $$\nabla\times\mathbf{B} = 0$$ which is Ampere's (magnetostatic) law in vacuum. This is where boundary conditions play a role.

In my opinion, a slightly better argument is this: you can show that if there is a $z$-component, then it must be constant in space - there is no current through by any loop in a vertical ($z$-direction containing) plane. The solution you get is then the usual circulating magnetic field due to a wire plus a constant field, say $\mathbf{b}$: $$\mathbf{B} = \frac{\mu_0 I}{2\pi r}\hat{\mathbf{e}}_\phi + \mathbf{b}$$ so that when $I = 0$, $\mathbf{B} = \mathbf{b}$.

Now, you have to enforce the following by hand: that when there is no current, you expect no magnetic field. You can also require the magnetic field to be zero infinitely far away from the wire ("boundary condition"). Either of this is possible only when $\mathbf{b} = 0$.

It is possible to have (at least theoretical) cases where $\mathbf{b} \ne 0$, such as when you have the wire inside an infinitely large solenoid (having an infinitely large product of turns per unit length and current). The principle of linear superposition then dictates that the contribution to the magnetic field from the wire would then add up to the existing uniform magnetic field. To deduce from the (still-existing) symmetry along the $z$ axis that there is no 'vertical' magnetic field would then be incorrect.

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    $\begingroup$ As a practical example, if I make a very long current-carrying wire in a lab and run a current through it, the field I measure near the center of the wire will be the circulating magnetic field plus $b$ from the Earth's (and the Sun's, etc) magnetic field. $\endgroup$ – Neal May 10 '15 at 12:03
  • $\begingroup$ @Neal true, but the presence of a symmetry is questionable in these cases. $\endgroup$ – AV23 May 10 '15 at 12:06

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