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Homework Help: Closed integral problem: field in the plane of a magnetic dipole

  1. Jul 2, 2013 #1
    I have read about Biot-Savarts law but I have no idea how to solve it when the curve is dependent of some variable. My books in mathematics dont help, nor my books in physics. You are welcome to give me a link where I can read about this, but first let me know if I have got it right this far. Now, to the problem!

    1. The problem statement, all variables and given/known data
    The magnetic field of the Earth is found to fall off according to an inverse cube law at several Earth radii. The field can thus be roughly described in terms of a magnetic dipole at the centre of the Earth. If the field strength is 10-4 G at 15 Earth radii in the equatorial plane what is the magentic dipole moment of the Earth? (Radius of Earth≈6000 km.)

    2. Relevant equations
    [tex]\textbf{B}(\textbf{r})= \frac{\mu_0 I}{4 \pi} \oint_s \frac{d\textbf{l´}\times(\textbf{r-r´)}}{|\textbf{r-r´}|^3}[/tex]

    3. The attempt at a solution
    So I want to find out the magnetic field of the dipole at a large distance. So first I made a sketch of the problem in the thumbnail below. Then I dig in to the Biot-Savart formula and realise that according to the law of cosines
    [tex]d^2=(\textbf{r-r´})^2=x^2+R^2-2xR\cos \theta \simeq x^2-2xR\cos \theta[/tex]
    [tex]d\textbf{l´}\times(\textbf{r-r´})=\sin (\pi/2-\varphi)dl d \hat{y}=\cos (\varphi)Rdd\theta \hat{y}[/tex]
    But now since the angle ##\phi \simeq 0 ## I see that
    [tex]\cos \varphi \simeq \cos (\pi-\theta ) = -\cos \theta[/tex]
    So putting it all together
    [tex]\textbf{B}(x)= \frac{\mu_0 I}{4 \pi} \oint_s \frac{Rd \theta \cos \varphi}{d^2}\hat{y}\simeq \frac{\mu_0 I}{4 \pi} \oint_s \frac{-R \cos \theta}{x^2-2Rx\cos \theta}\hat{y}d \theta=-\frac{\mu_0 I}{4 \pi} \frac{R}{x^2} \oint_s \frac{\cos \theta}{1-2R/x \cos \theta}\hat{y} d \theta [/tex]
    And since I don't know how to integrate this I don't even know if I'm right. That's my dilemma.

    Attached Files:

  2. jcsd
  3. Jul 2, 2013 #2


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    Hello, Order.

    I think you went too far in this approximation when you approximated ##\phi## by zero. I made the same approximation at first, but kept getting the wrong answer by a factor of 2. Only when I went back and kept the first order approximation in ##\phi## did it all work out.

    You can use the approximation ##1/(1-\epsilon) \approx 1+\epsilon## for small ##\epsilon## to simplify the integral.
  4. Jul 3, 2013 #3

    Yes, obviously the cosine of the angle is dependent of ##\phi## to first order. I should have understood that. I got an estimation of ##\cos \varphi ## without making this assumption (##\phi=0##) but it was half a page of trigonometric calculations and approximations! (There must be an easier way.) I got it all right in the end (after 3 Days).

    Anyway, the real reason I posted was because I did not understand the closed integral. I thought I would have to use Green's theorem or some other gadget to solve it :redface:. But now I am no longer afraid of them.
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