Magnetic field around an infinitely long wire

1. Aug 17, 2007

ehrenfest

Hello, I am trying to integrate

$$\int \mu_0 I (dl X r) /4 \pi r^2$$
in order to get the magnetic field at a point a distance R from a wire with current I. r is the distance between the differential length and the point. I integrate over the entire wire (which becomes an angle from 0 to 2 pi).

I am off by a factor of 1/pi from the correct answer of of mu_0 I/2 pi R. Is my integral or my integration wrong?

2. Aug 17, 2007

learningphysics

Your integral is wrong... in the numerator it should be ($$\vec{dl}X\hat{r}$$)
$$\hat{r}$$ is different from $$\vec{r}$$
$$\hat{r}$$ denotes a unit vector in the direction of $$\vec{r}$$

3. Aug 18, 2007

ehrenfest

You're right. That is what I meant. I still am off by the factor of 1/pi.

I tried to change to change it to an integral with respect to an angle and got:

$$\mu_0 I/4\pi \int_0^{2\pi} \sin \theta d\theta r/r^2$$

Is that correct?

Last edited: Aug 18, 2007
4. Aug 18, 2007

pardesi

no why do u take the angle to be $$0$$ to $$2 \pi$$
the angles should be $$\theta_{1}$$ to $$\theta_{2}$$ where tehse are the angles made by the wire's end points with point
in case of infinite wire they turn out to be $$\frac{\pi}{2},\frac{\pi}{2}$$ or
$$\frac{- \pi}{2},\frac{\pi}{2}$$ depending on how u integrate

5. Aug 18, 2007

learningphysics

Are you mixing up the big R with the little r? If that is supposed to be r (which is varying) then the integral is wrong. And the limits should be 0 and pi.

I get $$\frac{\mu_0 I}{4\pi R}\int_0^{\pi} \sin \theta d\theta$$

write everything in terms of R (which is constant) and $$\theta$$ and after taking all the constants outside, inside the integral you should just get $$sin(\theta)$$

The way I did was setting:
x = -R/tan(theta)

calculate dx (which is dl) in terms of d(theta)

also set r = R/sin(theta)

For some reason latex is messing up for me.

Last edited: Aug 18, 2007
6. Aug 18, 2007

ehrenfest

OK. So my integral should be

$$\int \mu_0 I (dl X r) /4 \pi r^2$$ = $$\int \mu_0 I sin(\theta)dl r /4 \pi r^2$$

I just had a dtheta instead of a dl.

I am confused about your solution. I am not sure why you introduced x. Anyway I get with your trig substitution,

dx = R csc^2(x) dtheta

then all the sines cancel!

7. Aug 18, 2007

learningphysics

There shouldn't be an r in the numerator for the fraction on the right. I think that's what's causing the trouble.

8. Aug 18, 2007

ehrenfest

YES! Because r-hat is the unit vector. I see. Thanks.

9. Aug 18, 2007

learningphysics

You're welcome.

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