- #1

FranzDiCoccio

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- 41

## Homework Statement

A current [itex]I[/itex] flows along the surface of a hollow conducting cylinder. The radius of the cylinder section is [itex]r[/itex].

By using Ampere's law, show that the magnetic field [itex]B[/itex] outside the cylinder is

[tex]B=\frac{\mu_0}{2 \pi} \frac{I}{r} [/tex]

## Homework Equations

Ampere's law:

[tex]\Gamma_\gamma(\vec B)=\mu_0 \sum_k I_k[/tex]

## The Attempt at a Solution

It is expected that this problem is solved without resorting to integrals.

I think I'm pretty much there.

The right-hand side of Ampere's law above is simply

[tex]\mu_0 \sum_k I_k= \mu_0 I[/tex]

As to the lhs one can observe that the field should have a cylindrical symmetry. Also, it should be everywhere orthogonal to the cylinder axis.

Now the trick is to use a path [itex]\gamma[/itex] that is a circle with radius [itex]r[/itex], whose center is on the cylinder axis and lies in a plane orthogonal to such axis.

The lhs of Ampere's law should then be

[tex]\Gamma_\gamma(\vec B)=\sum_k \vec{B}_k \cdot \vec{s}_k=\sum_k B_k s_k \cos \alpha_k[/tex] where the [itex]\vec{s}_k[/itex] are very small displacement vectors that go around the circle.

On account of the symmetry of the problem, the magnitude of the field and the angle between the field and the displacement vectors should not depend on the point on the circle [itex]B_k=B[/itex], [itex]\alpha_k=\alpha[/itex].

The remaining sum on the displacements adds up to the circumference

[tex]\Gamma_\gamma(\vec B)=B \cos \alpha \sum_k s_k =2\pi r B \cos \alpha[/tex].

Therefore, since [itex]2\pi r B \cos \alpha=\mu_0 I[/itex],

[tex] B =\frac{\mu_0}{2\pi} \frac{I}{r \cos \alpha}[/tex]

Now, I know that the cosine should not be there, and I am pretty sure that it is possible to show this with a "real" integral (as opposed to the naive "sum" I have performed).

I cannot find a "simple" argument that allows me to conclude that not only [itex]\alpha_k=\alpha[/itex], but also [itex]\alpha=0[/itex].

Thanks a lot for any insight.

Franz

[EDIT]

Now that I'm ready to post I maybe have an idea. It's kind of cumbersome so, any more elegant solution is welcome.

What I thought is that I can see the surface of the cylinder as composed by a bunch of wires.

Biot & Savart's law applies to each wire.

It should be possible to prove that at any point on the circular path the field is perpendicular to the diameter going through that point.

The field is the sum of all the fields of the "imaginary" wires forming the surface of the cylinder.

Surely the wires on the diameter of the circle produce fields that are perpendicular to the diameter.

Now I can consider a wire that is on one half circle, and its field is not perpendicular to the diameter.

However, when combined with the field of the wire in the symmetric position wrt the diameter, it is.

Since for each point on one half circle there is a symmetric one on the other half, everything works, although, as I mention, this is a bit cumbersome.