Magnetic field outside a conducting hollow cylinder

[FranzDiCoccio]

Homework Statement

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

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

B=\frac{\mu_0}{2 \pi} \frac{I}{r}

Homework Equations

Ampere's law:
\Gamma_\gamma(\vec B)=\mu_0 \sum_k I_k

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
\mu_0 \sum_k I_k= \mu_0 I

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 \gamma that is a circle with radius r, 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

\Gamma_\gamma(\vec B)=\sum_k \vec{B}_k \cdot \vec{s}_k=\sum_k B_k s_k \cos \alpha_k where the \vec{s}_k 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 B_k=B, \alpha_k=\alpha.
The remaining sum on the displacements adds up to the circumference
\Gamma_\gamma(\vec B)=B \cos \alpha \sum_k s_k =2\pi r B \cos \alpha.

Therefore, since 2\pi r B \cos \alpha=\mu_0 I,
B =\frac{\mu_0}{2\pi} \frac{I}{r \cos \alpha}

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 \alpha_k=\alpha, but also \alpha=0.

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.

 

[Charles Link]

I think with the $B_k \cdot s_k$, you might be asking if the $B$ could also have a radially outward or radially inward component. The answer to that is it doesn't, because of $\int B \cdot dA=0$ , which is a result of Maxwell's equation: $\nabla \cdot B=0$. This basically says the lines of flux of $B$ are conserved and don't originate from a point like the electric field from a point charge can.

 

[TSny]

Yes. Very good. Your reasoning looks sound to me (including the comments in the edit for showing that B is tangent to the path of integration.)

[Oops, I see that Charles replied before me.]

 

[FranzDiCoccio]

Thanks Charles,

I'm sure that the angle is zero and the cosine is 1.
Perhaps I did not explain myself clearly. I am looking for a simple explanation of that, where simple means that it does not require advanced knowledge, such as integrals or the differential formulation of Maxwell's equations.

The problem is intended for students that do not have such knowledge yet. They (or should I say some of them) understand the circulation in Ampere's law as a sum of tiny contributions obtained from seeing the path as the result of many tiny displacements.

I was wondering if there's an argument that could convince them that B is tangent to the circle.
Rotational symmetry of the field per se is not enough, since it is compatible with a nonzero (constant) angle between B and the tangent. The only explanation I thought of is the one I sketched above, but it's kind of cumbersome.

 

[Charles Link]

Thanks Charles,

I'm sure that the angle is zero and the cosine is 1.
Perhaps I did not explain myself clearly. I am looking for a simple explanation of that, where simple means that it does not require advanced knowledge, such as integrals or the differential formulation of Maxwell's equations.

The problem is intended for students that do not have such knowledge yet. They (or should I say some of them) understand the circulation in Ampere's law as a sum of tiny contributions obtained from seeing the path as the result of many tiny displacements.

I was wondering if there's an argument that could convince them that B is tangent to the circle.
Rotational symmetry of the field per se is not enough, since it is compatible with a nonzero (constant) angle between B and the tangent. The only explanation I thought of is the one I sketched above, but it's kind of cumbersome.

Even Ampere's law in its calculus form does not give any information about a radially inward or outward component. I had to think about this one for a minute or two when you first posted it, because Ampere's law simply gives information of the $B$ field component that is tangent to the circle. It is a combination of $\nabla \cdot B=0$ along with Gauss' law that shows there is no radial component. I don't see any simple argument from symmetry that would exclude a radial component of $B$.

 

[TSny]

You want an argument for why there can't be a radial component to the B field. Suppose there were. Then reversing the direction of the current should do what to the direction of the radial component? Compare that to flipping the cylinder over (end for end).

 

[FranzDiCoccio]

@Charles,

of course, Ampere's law is not enough. That was kind of the point of my question. Apologies if I was not clear, I'm not a native English speaker.

@TSny

that's smart. I did not think of using mirror symmetry (in such an elegant way).
I have to think that through, but I'd say that if you reverse the current every B vector in the field should flip wrt a plane perpendicular to it. Therefore the outwards component become inwards. On the other hand changing the direction of the current should definitely not do that. After all it's like turning the cylinder upside down, and hence the outwards component stays outwards.
The only way avoid an absurd is that the vector is tangent. Cool!

 

[Delta2]

You want an argument for why there can't be a radial component to the B field. Suppose there were. Then reversing the direction of the current should do what to the direction of the radial component? Compare that to flipping the cylinder over (end for end).

I don't understand this argument, flipping the cylinder will do nothing (unless we also reverse the polarity of the electrodes that supply current to the cylinder, or unless the cylinder has permanent magnetization).

 

[Charles Link]

I don't understand this argument, flipping the cylinder will do nothing (unless we also reverse the polarity of the electrodes that supply current to the cylinder, or unless the cylinder has permanent magnetization).

Having the current go in the opposite direction will put a "minus sign" on the magnetic field $B$ at each location, (by Biot-Savart's law). This will make any outward component of $B$ go inward, and visa versa , which thereby means the radial component must be zero. (Otherwise, you would have two different results for the same geometry, because you can rotate yourself 180 degrees, and then you see the current going in the forward direction).$\\$ I don't agree with a non-algebraic explanation either, because the argument e.g. would not work if you tried it on the electric field from electrical charges, which has an outward radial component.