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Magnetic field around an infinitely long wire

  1. Aug 17, 2007 #1
    Hello, I am trying to integrate

    [tex] \int \mu_0 I (dl X r) /4 \pi r^2 [/tex]
    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. jcsd
  3. Aug 17, 2007 #2


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    Your integral is wrong... in the numerator it should be ([tex]\vec{dl}X\hat{r}[/tex])
    [tex]\hat{r}[/tex] is different from [tex]\vec{r}[/tex]
    [tex]\hat{r}[/tex] denotes a unit vector in the direction of [tex]\vec{r}[/tex]
  4. Aug 18, 2007 #3
    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:

    [tex] \mu_0 I/4\pi \int_0^{2\pi} \sin \theta d\theta r/r^2 [/tex]

    Is that correct?
    Last edited: Aug 18, 2007
  5. Aug 18, 2007 #4
    no why do u take the angle to be [tex]0[/tex] to [tex]2 \pi [/tex]
    the angles should be [tex]\theta_{1}[/tex] to [tex]\theta_{2}[/tex] where tehse are the angles made by the wire's end points with point
    in case of infinite wire they turn out to be [tex]\frac{\pi}{2},\frac{\pi}{2}[/tex] or
    [tex]\frac{- \pi}{2},\frac{\pi}{2}[/tex] depending on how u integrate
  6. Aug 18, 2007 #5


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    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 [tex] \frac{\mu_0 I}{4\pi R}\int_0^{\pi} \sin \theta d\theta [/tex]

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

    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
  7. Aug 18, 2007 #6

    OK. So my integral should be

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

    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!
  8. Aug 18, 2007 #7


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    There shouldn't be an r in the numerator for the fraction on the right. I think that's what's causing the trouble.
  9. Aug 18, 2007 #8
    YES! Because r-hat is the unit vector. I see. Thanks.
  10. Aug 18, 2007 #9


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    You're welcome. :smile:
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