3
$\begingroup$

Imagine a circular toroid coiled with a wire through which some current $I$ is flowing. Everywhere it is stated that the magnetic field inside this toroid can be calculated as $$\vec{B} = \mu \frac{NI}{2\pi r} \ \hat{\phi}$$

where $\mu$ is the magnetic permeability of the toroid, $N$ the number of loops the coil presents, $r$ the distance to the center of the toroid and $\hat{\phi}$ the typical versor in cylindrical coordinates.

My cuestion is: why doesn't the magnetic field depend on $z$, the vertical position?

As I see it, there is no symmetry in $z$ that allows us to automatically discard this coordinate. Namely, as we move radially (varying $r$), the field changes because the situation differs from one radius to another: we get closer to (or further away from) the wires, and that makes the field vary. If we moved along the $z$ direction, the case would be analogous. If we center the coordinate system such that the plane $z=0$ slices the toroid in two halves, we can see that at $z=0$ the current has just a component in the $\hat{z}$ direction, but if we analyze this for any other value of $z$ the current acquires other components as well. So I don't see why the magnetic field would not depend on $z$.

Does it depend on $z$ or not? If yes, how can one then calculate the actual magnetic field (the technique that would be used if the section was squared would no longer apply, I guess)?

$\endgroup$
4
  • $\begingroup$ The formula that you give for the magnetic field holds only in the $z=0$ plane. It can be obtained easily using Ampère theorem. $\endgroup$
    – Christophe
    Commented Jun 25, 2020 at 20:03
  • $\begingroup$ @Christophe That's what I wondered, but there are lots of places where it appears to be valid for any $z$ (like here or here, but there are lots of additional places). Is there a way to calculate the actual magnetic field for any $z$? $\endgroup$
    – Tendero
    Commented Jun 25, 2020 at 20:10
  • 2
    $\begingroup$ I was wrong and I learnt something! The current is symmetric with respect to all the planes containing the $(Oz)$ axis. In any point M, the magnetic field $\vec B(M)$ is therefore perpendicular to the plan containing both M and $(Oz)$, i.e. $\vec B(r,\phi,z)\sim\hat\phi$ for any $z$! The Ampère theorem can by applied to a circular path in any plane z=Cst. When the path is inside the torus, it gives the expression that you give, independently of $z$! $\endgroup$
    – Christophe
    Commented Jun 25, 2020 at 20:21
  • $\begingroup$ Related : Magnetic Induction at the centre of a Toroid. $\endgroup$
    – Frobenius
    Commented Jun 25, 2020 at 20:21

1 Answer 1

2
$\begingroup$

Yes, it does depend on $z$.

If you think about it, it must depend on $z$ to satisfy the boundary conditions when you are close to the wires. You expect a constant value with $z$ only in the middle of the torus, where you can ignore the edge effects.

Ampère's law only reduces to that simple formula if the path element $\mathrm{d}\boldsymbol{\ell}$ is parallel to magnetic field $\mathbf{B}$.

But anyway, I did the maths.

I have 20 current loops azimuthally distributed, so that the magnetic field magnitude in the $xy$ plance, at $z=0$, looks like this:

enter image description here

The radius of each loop is $3$, and the torus is "centred" at $10$, so that its inner and outer radii are $7$ and $13$.

Now let's look at the three components of the magnetic field, at $x=10, y=0$:

enter image description here

And then I plotted the only $B_{\phi}$, still at $y=0$ but now varying $x$: enter image description here

You can see that actually the $B_\phi$ value is quite constant with $x$ provided that you are not too close to the edge (this time in the $x$ direction). This is wrong though, as you’d expect the field to go down $\propto 1/r$ — I suspect this is an artifact of a finite number of current loops.

Conclusion

  • Away from the edges, the field is essentially independent of $z$.

  • The field is always in the azimuthal $\phi$ direction, as suggested by @Christophe in the comment).

$\endgroup$

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.