Homework Help: Magnetic field due to two wires

1. Mar 16, 2013

Siune

1. The problem statement, all variables and given/known data
We have wires like in the picture and we want to know the value of magnetic field as
function of x:

NOTE: We want to know the magnetic field at point P ( P is on +x axis ) ( Which is totally arbitary!) The picture is there for just to give the idea.

R is the distance from each wire to point P.

2. Relevant equations

$H = \frac{I}{2 \pi R }$

3. The attempt at a solution

Okey, so the currents are going in opposite ways. By the right hand rule we get the direction of the magnetic field due to each wire!

We notice that the y-component cancels out ( due to symmetry ), and that the magnetic field is only the x component due to each wire at point P.

So by superposition principle we get the magnetic field at point P on x -axis:

$H = sin(\theta) \frac{ I }{ \pi R }$

Now $sin(\theta)$ is the angle between x axis and the distance R from each wire.

So we get from trigonometry:

$H = \frac{d \cdot I }{ \pi R^2}$

$H = \frac{d \cdot I }{ (x ^2 + d ^2 ) \pi }$

Not sure if this is right? At least the units match ( A / m ), but... problem is:

My assigment gives us only the current I and the x -axis distance to point P? So I assume the right answer of the magnetic field at point P should be independent of d? But that doesn't make sense to me.

Sincerely yours,
Siune

2. Mar 16, 2013

TSny

If θ is the angle between the x-axis and R, make sure you are using the correct trig function for getting the x-component.

3. Mar 16, 2013

Siune

Hmm, as the distance vector R from one wire and magnetic field from same wire are perpendicular I get that the x component of magnetic field by that wire

so $sin(\theta) = \frac{d}{R}$

and

$R^2 = x^2 + d^2$
?

4. Mar 16, 2013

TSny

Sorry, my mistake. You are entirely correct. The answer does depend on both x and d.

5. Mar 16, 2013

Siune

No problem at all, wouldn't have been the first time I make the mistake on the easiest part of the problem. ^^