# Homework Help: Magnetic field from conducting pipe

1. Jun 18, 2010

### sarahgarden

1. The problem statement, all variables and given/known data

A long, straight copper pipe of inner diameter 16mm and outer diameter 24mm carries a current of 100A. Calculate the magnetic flux density at distances i)5mm, ii)10mm and iii)15mm from its axis?

Answer : i)0, ii)9*10^-4T iii) 13.3*10^-4T

2. Relevant equations

I thought it would be B=mu_0*I/(2pi)r
It does in fact say this in my notes.
Also I get i) resulting in 0, if it's taken from line through middle of pipe (it irked me then that ii) had a value) but axis is used here and in my textbook differently - to mean off the surface of the pipe.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 18, 2010

### RoyalCat

EDIT:
Your textbook is a bit confusing, do they want the magnetic field at distances of 5, 10 and 15 mm from the axis of the wire, or from its outer surface, or inner surface?

If they mean from the axis, then the answer is 0 across the board, see the reasoning below.

Two assumptions are necessary to solve this problem, the first is that of symmetry, which is pretty straightforward, and the other is that the current density in the wire is uniform.

Use Ampere's Law and the uniformity of how the current is spread out across the cross section of the wire to find the magnetic field there.

The radial distance you see in your equation refers to a circle arranged around the axis of the cylinder, not the distance from the surface of the pipe.

Ampere's Law:

$$\oint \vec B \cdot \vec d\ell = \mu_0 I_{penetrating}$$

What Ampere's Law means is that the closed line integral of the B field around any path, is proportional to the current that penetrates through any surface attached to that path.

For instance, for a single straight wire, we can assume radial symmetry. Taking our closed path as a circle around the axis of the wire, we can take B to be a constant, and always in the direction of the path, so the closed path integral is just the value of B times the length of the path, the circumference of the circle.
$$\oint \vec B \cdot d\ell = B\cdot2\pi r$$

Taking the simplest surface attached to this loop, we see that our wire penetrates it. So the current, I, in the wire penetrates the surface.

Therefore $$B\cdot 2\pi r = \mu_0 I$$

$$B=\frac{\mu_0 I}{2\pi} \frac{1}{r}$$

Now apply Ampere's Law and see what current penetrates through a surface attached to a loop through the volume of the pipe!

3. Jun 18, 2010

### stevenb

Actually, their answers are correct. Note that the pipe specifications are for diameter and not radius. The radii of 5 mm, 10 mm and 15 mm place the points of interest inside the pipe, within the pipe and outside the pipe, respectively.

Good explanation though.

In a nutshell, due to symmetry and the static version of Ampere's law, the current outside the radius of interest contributes nothing to the magnetic field. The current inside the radius of interest behaves like an infinitely long and infinitely thin wire, on the central axis with the same current flow (as within the radius).

4. Jun 18, 2010

### sarahgarden

Thankyou both very much for your help.

Yikes! I'm embarassed about that. It must seem like I didn't spend long thinking about it but really I never think to re-read a question. I'm just surprised I didn't notice when I was typing it out.

For 10mm from centre, I got the answer but not the way I expected to. The following is how I expected to get my answer;

a is outer radius (12mm), b is inner (8mm).

$$(a)\gg(r)\gg(b)$$?

$$cross-sectional area=\pi(a^2-b^2)$$

$$J=\frac{I}{\pi(a^2-b^2)}$$

$$I_{enc}=J\times\pi(r^2-b^2)$$

so

$$B=\frac{\mu_{0}I(r^2-b^2)}{(a^2-b^2)}$$

$$=5.7\times10^-5T$$

I got the correct answer for parts ii) and iii) using simply

$$B=\frac{\mu_{0}I}{2\pi\cdot(r)}$$

Why isn't it the way I showed above for part ii), when r is a distance inbetween b and a?

Thankyou.

P.S. How is my Latex looking?

Last edited: Jun 18, 2010
5. Jun 18, 2010

### stevenb

I think your method is correct, but double check your derivation. Just looking quickly, there may be a mistake in there.