# Homework Help: Components of a magnetic field in a uniform wire

1. Mar 11, 2012

### ItsFootballNo

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

A straight wire, of current I, radius a is centred at (α,β). What are the x and y components of the magnetic field B inside one of the wires?

2. Relevant equations

∮B.dl = μ I_enc
∫∫J.dS = I

3. The attempt at a solution

Any point (x,y) in the wire has a constant current density J.

Hence:
∫∫J.dS = J pi r^2 = J pi ((x-α)^2 + (y-β)^2)

The wire has total current I and the current density J is uniform, hence:
J = I / (pi a^2)

Therefore:
I_enc = ∫∫J.dS = I ((x-α)^2 + (y-β)^2)/a^2

Therefore:
∮B.dl = μ I ((x-α)^2 + (y-β)^2)/a^2

It is from here that I get stuck, mostly how to evalutate the integral without it becoming one big equation without staying in its components. If I was just looking at magnitude of the magnetic field, I know we could show:
∮B.dl = B (2 pi r)
=> B = μ I r / (2 pi a^2)

But looking at the answers, just the y component comes out as:
B_y = μ I (x-α) / (2 pi a^2) - μ I x / (2 pi [(x+α)^2 + (y+β)^2])

Am I going about this the wrong way or are there any tips on how to get to the next step? Any help is greatly appreciated!

2. Mar 11, 2012

### SammyS

Staff Emeritus
Hello ItsFootballNo. Welcome to PF !   (I see that you posted in PF once before without getting any reply.)

Your first sentence indicates to me that there is more than one wire. What can you tell us about that?

3. Mar 11, 2012

### ItsFootballNo

Apologies, the other wire is centred at (-α,-β), also with radius a.

4. Mar 11, 2012

### ItsFootballNo

The other wire also has current -I (that is, it is parallel but running in the opposite direction)

5. Mar 11, 2012

### ItsFootballNo

I have been working on it for the past hour or so, and seem to have come up with some sort of solution, but it isn't the most elegant one.

I looked at the case of the magnetic field inside just one of the wires (say, in this case, the one to the right, centred at (α,β) and solved for the magnitude of B. I assumed that:
vec(B) = (B_x, B_y, B_z) = (-B sin(t), B cos(t), 0) = (-B(y-β)/r, B(x-α)/r, 0) = (-μI(y-β)/(2 pi a^2), μI(x-α)/(2 pi a^2), 0)

I then looked at the contribution of the other wire which, as the point we're inspecting is outside of the wire, is:
B=-μI/(2 pi r) [The negative sign arising from the current being negative in this wire]

Again looking at the effect at a point (x,y)
vec(B) = (B_x, B_y, B_z) = (-B sin(t), B cos(t), 0) = (-B(y+β)/r, B(x+α)/r, 0) = (μI(y+β)/(2 pi ((y+β)^2 + (x+α)^2)), (μI(x+α)/(2 pi ((y+β)^2 + (x+α)^2)), 0) [y+β and x+α arising from the new centre of (-α,-β)]

Finding the sum, using the superposition principle, of these gives a correct looking solution.

I just have a problem in that I've just used: vec(B) = |B| theta-hat and subbed in from there. Does this represent a satisfactory means to solve the solution or is there a better method when doing the integral in Ampere's Law?