# Homework Help: Magnetic Field of a Straight Current Carrying Conductor

1. Apr 9, 2012

### forestmine

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

The figure shows an end view of two long, parallel wires perpendicular to the xy-plane, each carrying a current I but in opposite directions.

Derive the expression for the magnitude of B at any point on the x-axis in terms of the x-coordinate.

2. Relevant equations

B=(mu*I/4*pi)∫(dl x $\hat{r}$)/r^2

3. The attempt at a solution

Ok, so here's my issue with this problem..I understand that the magnetic field of a straight current carrying conductor is

B = (mu*I)/2pi*r

I completely understand how to derive this equation, say for example we have a current running along the y-axis, and we want to know the magnetic field at a point along the x axis. This is easy enough and sure enough, I wind up with the above equation. But I'm completely confusing myself in this situation...this time, the current (and dl) run along the direction of z.

That being said, I'm completely confused about dl x $\hat{r}$. $\hat{r}$ is point from the y axis to the x, but dl is along the z axis, correct?

Dealing with this in 3-dimensions is really beginning to confuse me.

I know that for this problem, essentially, we want to look at the x-component for the fields produced from each conductor, so it will be

Bcos(0), where B is (mu*I)/2pi*r

Does this require integration by polar coordinates? Should I just use the standard equation or B from a straight current carrying conductor and forego all this integration nonsense? I'm beginning to think I ought to, heh...

Any help would be greatly appreciated! I've spent too long trying to rotate coordinate systems in thin air.

2. Apr 10, 2012

### tiny-tim

hi forestmine!
but where are they?
you're confusing me

dl is along the wire, r is from the particular element of the line (from l to l+dl) that you're integrating over, to the point

anyway … yes, either use the standard equation, or find it by applying the ∫ B.dl formula around a circle enclosing the wire

3. Apr 11, 2012

### forestmine

Whoops, guess I forgot to include the image. Sorry about that!

Anyone, I've got this one figured out.

Thanks for the help!