# Magnetic Field due to a Long Straight Wire (Proof)

1. Apr 29, 2012

### FatPhysicsBoy

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

Whilst revising the Biot-Savart law, I came across the result:

for the magnetic field due to a long straight wire. I understand that this can be found by integrating the Biot-Savart equation over the length of the wire, however after attempting the integral myself and looking at the integration in my notes I find myself unable to understand how to come to the result.

2. Relevant equations

3. The attempt at a solution

Outlined above. Any help will be appreciated, thank you.

2. Apr 29, 2012

### Steely Dan

Could you clarify what confuses you about the integral? The way to set it up would be as an integral over $z$, say, along the wire, with each wire element $dz$ contributing to the field at P based on the distance to point P and the angle between the radius vector and the z-axis.

3. Apr 29, 2012

### FatPhysicsBoy

This is the proof from my lecture notes:

I understand where the cross product has been replaced with dssinθ up to the integral of sinθ/r2.

I also understand the first relationship r2 = s2 + R2 as just being Pythagoras, however I do not understand it's necessity, also I do not understand the second relationship involving sinθ either.

So I suppose you could say I don't understand the relationships, and also why and how they are used.

4. Apr 29, 2012

### Staff: Mentor

In order to do the integral, you have to write it in terms of a single variable. Here, the writer chose s as the variable of integration, so he has to eliminate $\theta$ and r in favor or s. R is a constant as far as the integral is concerned.

This is not the only way to do it. You can also do it by using $\theta$ as the integration variable, in which case you have to eliminate s and r. You should of course end up with the same final result.

You can also (at least in principle) use r as the integration variable. I've never done it that way in this particular example, but I've done something similar in other situations.

5. Apr 29, 2012

### FatPhysicsBoy

Oh ok yeah I see, however I still do not understand this relationship then:

sin(∏-θ) = R/(s2 + R2)0.5

6. Apr 29, 2012

### Villyer

Theta is the outside angle in that diagram, pi-theta makes it the inside angle so that you can use the simple definition (sin theta = opp / hyp) to remove the theta from the equation.

7. May 1, 2012

### FatPhysicsBoy

Thank you, I fully understand the derivation now!