# Homework Help: Magnetic field of hollow cylinder

1. Feb 7, 2015

### whatisreality

1. The problem statement, all variables and given/known data
A hollow cylinder with thin walls has radius R and current I, which is uniformly distributed over the curved walls of the cylinder. Determine the magnetic field just inside the wall and just outside, and the pressure on the wall.

2. Relevant equations
F = IL×B

3. The attempt at a solution
The field just inside is zero by ampere's law, and μ0I/2πr just outside. I can't find the pressure though! If I can find an expression for dI, then I can find an expression for dF, and dF/dA would be the pressure.

dI = dr/2πr * I where I is total current, which would make dF=(dr/2πr)L(μI/2πr)sinθ,
the angle between them is 90° so sinθ=1
dF = μ I dr / (2πr)2
I'm pretty sure this is wrong though! And I don't know how to express dA.

Thanks for any help :)

2. Feb 7, 2015

### TSny

If dr represents a small increment in the radial direction, then I don't understand why you have dr in this equation.

A tricky point in calculating the force this way is to decide what value of magnetic field to use. The current is located right where the B field is changing from 0 on the inside to Bo on the outside. So, there is the question: what value of B actually acts on the current? You will need to think about that.

Another approach to the problem is to consider energy stored in the magnetic field, but I don't know if you are allowed to approach the problem that way.

3. Feb 8, 2015

### whatisreality

dr was supposed to represent an infinitesimal part of the circumference. I wasn't sure if it should be dI = dr/2πrL * I, where L is the length of the cylinder.
I think B0 would act on the current, wouldn't it?

4. Feb 8, 2015

### TSny

If the symbol r represents the radius, then dr would represent a small increment in radius. For an element of arc length, we could use ds.

If I do that, then your expression for the current in the arc length ds is dI = ds/2πrL * I. I interpret this as dI = ds/(2πrL) * I. That is, I assume that you meant the denominator to be 2πrL. If you check your dimensions (or units) you can see that this expression can't be correct. In order for the right hand side to represent a current, the quantity ds/(2πrL) would need to be dimensionless. But you can see that it has the dimensions of 1/length. However, your expression is fairly close to the correct expression.

The way to think about getting the correct expression for dI is to realize that the entire circumference of the cylinder contains the current I. The length L of the cylinder is not important here. So the current dI is the current contained in the fraction of a circumference subtended by ds.

The magnetic field that acts on the current dI is not the magnetic field, Bo, at the outside surface of the cylinder. This is kind of tricky. In reality, the current on the surface of the cylinder does not have zero thickness. So, imagine that the current occupies a very thin layer of thickness δr as shown in blue in the attached figure. The B field is not really discontinuous at the surface of the cylinder. It is zero at the inside surface of the layer of current and Bo at the outside surface of the layer. The B field changes continuously from 0 to Bo as you go through the layer of thickness δr. Different parts of the current in the layer experience different magnetic field strengths. If you imagine that the B field increases linearly from 0 to Bo, what is the average value of B in the layer of current?

File size:
2.2 KB
Views:
474
5. Feb 8, 2015

### whatisreality

Oh, would it be B0/2? And my expression for dI should be
dI = ds/(2πr) * I, where 2πr is the denominator.
Sub this into F=IL×B
dF=dI LB sinθ where sinθ is 1 by the reasoning in the first post
dF= (I ds L μI) / (4πr)
dF/dA = pressure dP

Last edited: Feb 8, 2015
6. Feb 8, 2015

### TSny

Yes.
Yes.
Of course, r has a certain specific value in this expression.

7. Feb 8, 2015

### whatisreality

r has been called R! I forgot about that.
Edit: I edited the previous post while you were posting! Sorry!

8. Feb 8, 2015

### whatisreality

And dA = Lds, I think. So
dF/dA = μI2/8(πr)2
Presumably I then have to integrate, but I don't actually know what to integrate with respect to.

9. Feb 8, 2015

### TSny

You are asked to find the pressure. What does dF/dA represent?

10. Feb 8, 2015

### whatisreality

Pressure on the area element dA?

11. Feb 8, 2015

### whatisreality

Pressure is force per unit area, so actually I don't have to integrate!

12. Feb 8, 2015

### TSny

Right!

13. Feb 8, 2015

### whatisreality

Thank you for helping :)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted