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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 don't 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!
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.)
[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]
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 into the Biot-Savart formula and realize 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]
And
[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.
Homework Statement
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.)
Homework 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]
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 into the Biot-Savart formula and realize 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]
And
[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.