# Does magnetic field depend on $z$ inside a toroidal coil?

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)?

• The formula that you give for the magnetic field holds only in the $z=0$ plane. It can be obtained easily using Ampère theorem. Commented Jun 25, 2020 at 20:03
• @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 $z$? Commented Jun 25, 2020 at 20:10
• 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 $z$! Commented Jun 25, 2020 at 20:21
• Commented Jun 25, 2020 at 20:21

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:

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 $$x=10, y=0$$:

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

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).