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When deriving the magnetic field inside of an infinitely long solenoid carrying a stationary current, it's useful to take into consideration the symmetries of the problem, in order to understand which components the field actually has and on which variables it depends.

Considering cylindrical coordinates, and placing the $z$ axis so that it passes through the solenoid's center and it is perpendicular to the loops, we can say that the field doesn't depend on $\theta$ because there is a rotation symmetry around the z axis, and it doesn't depend on $z$ either, because there is a translation symmetry along the $z$ axis, so it depends on $r$ only.

Since there isn't a parity symmetry, if we put the solenoid upside down, a potential $r$ component of the field would remain the same, although the system isn't equal to the first one, because the current would flow in the opposite verse. Therefore, there can't be an $r$ component.

What I don't understand is why we can assume that there isn't a $\theta$ component either, and therefore compute the field assuming that it only has a $z$ component.

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    $\begingroup$ If there is a $\theta$ component, wouldn't a line-integral along a circle (say the circle's center is at the center of the solenoid) in the xy plane and inside the solenoid result in a nonzero value? This would mean there should be nonzero current going through this circle. But there isn't such current for a solenoid. $\endgroup$
    – aystack
    Sep 18, 2023 at 9:16
  • $\begingroup$ Oh, I see, I didn't think about that aspect. Thank you $\endgroup$
    – Fede
    Sep 18, 2023 at 9:34
  • $\begingroup$ Cylindrical coordinates use $\rho$ for the distance from the axis, so $r$ is a bit confusing since usually means the distance to an origin. $\endgroup$
    – JEB
    Sep 18, 2023 at 11:32
  • $\begingroup$ @JEB sorry, in my course we do the opposite and use $\rho$ for spherical coordinates and $r$ for cylindrical coordinates $\endgroup$
    – Fede
    Sep 18, 2023 at 11:43
  • $\begingroup$ to add to @LPZ-s answer, here are some more on the pseudo (axial) vector nature of the B-field physics.stackexchange.com/questions/410714/…, physics.stackexchange.com/questions/160993/… $\endgroup$
    – hyportnex
    Sep 18, 2023 at 14:34

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In general, for magnetic fields it is faster to look for planes of symmetry. It usually gives more information than parity. These are planes that leave the current distribution invariant by reflection, assuming that the current density is a vector. Since the magnetic field is a pseudo vector, it is a plane of antisymmetry of the field. This specifies its direction since $B$ is normal to this plane.

In your case, any plane parallel to the $xy$ plane is a plane of symmetry, so the magnetic field is necessarily along $z$.

Now that you know the only non zero component, you can use the usual symmetry arguments to see that it has only a radial dependence (invariant by translation along $z$ and rotation about the $z$ axis).

Hope this helps.

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  • $\begingroup$ Thank you, I had never heard the concept of "pseudovector" before: it actually seems much easier to help understand the direction of the magnetic filed. Maybe our professor will introduce it later on in the course. For now though, if I have to justify the assumption that the magnetic field only has a certain component during an exam, I don't think I can use these concepts, since we've never discussed them or proved that the magnetic field is normal to such planes. Can I ask if this fact is equivalent to saying that the magnetic field is never parallel to the current density? $\endgroup$
    – Fede
    Sep 18, 2023 at 13:24
  • $\begingroup$ It's a bit strange because you mentioned the effect of parity on the magnetic field. To know it is affected, you need to know that it is a pseudo vector. $\endgroup$
    – LPZ
    Sep 25, 2023 at 9:23
  • $\begingroup$ Not quite, it's just how the magnetic field is transformed under odd transformations, you need to add an extra minus sign. Strictly speaking, I don't think that the two are never parallel, but it is somewhat related. Your property comes from Ampère's law, which also shows that the magnetic field is a pseudo vector (due to the rotational). $\endgroup$
    – LPZ
    Sep 25, 2023 at 9:25

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