# Equation for magnetic field line of dipole

1. Sep 19, 2011

### erogard

Hi, given the equation for a dipole magnetic field in spherical coordinates:

$\vec{B} = \frac{\mu_0 M}{4 \pi} \frac{1}{r^3} \left[ \hat{r} 2 \cos \theta + \bf{\hat{\theta}} \sin \theta \right]$

I need to show that the equation for a magnetic field line is $r = R \sin^2 \theta$
where R is the radius of the magnetic field at the equator (theta = pi/2)

Not sure where to start. I know that the gradient of B would give me a vector that is perpendicular to a given field line...

I also know that a vector potential for a dipole magnetic field in spherical coordinate is given by

$A_\theta = \frac{\mu_0 M}{4 \pi} \frac{sin\theta}{r^2}$

Last edited: Sep 19, 2011
2. Sep 19, 2011

### clem

The equation for a field line is $$\frac{dr}{d\theta}=B_r/B_\theta$$.
I dont think this gives $$sin^2\theta$$.

3. Sep 21, 2011

### Philip Wood

Clem has omitted an r.

Clem meant: $\frac{B_r}{B_\theta} = \frac{dr}{r d\theta}$.

dr is the radial increment in the line corresponding to a tangential increment r d$\theta$.

The resulting DE is solved by separating variables, and yield logs on each side. You use the condition that r = R when $\theta$ = $\pi$/2 to re-express the arbitrary constant. You do get just what you said.

4. Oct 17, 2012

### JMeck

But why is the equation for the field lines:

$\frac{B_r}{B_\theta} = \frac {dr}{rd\theta}$ ??

I can see how solve this to give the equation:

$r = R sin^2 \theta$

where R is r when θ is ∏/2. Any help would be greatly appreciated.

5. Oct 17, 2012

### Philip Wood

You need to recall what is meant by a field line: a line whose direction at every point along it is the direction of the field at that point. So the ratio of radial to tangential field components must be the same as the ratio of tangential to radial components of the line increment.

6. Oct 17, 2012

### JMeck

Thanks that is great I get it now.

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