# The Earth's magnetic field is essentially that of a magnetic dipole...

1. May 12, 2015

### hitemup

1. The problem statement, all variables and given/known data

The Earth's magnetic field is essentially that of a magnetic dipole. If the field near the North Pole is about 10^-4 T, what will it be (approximately) 13,000 km above the surface at the North Pole?

2. Relevant equations

$$B = \frac{\mu _0}{2\pi}\frac{\mu}{(R^2+x^2)^{3/2}}$$

for x >> R

$$B = \frac{\mu _0}{2\pi}\frac{\mu}{x^3}$$

3. The attempt at a solution

I know how to find the magnetic field of a dipole on its axis for a circle. My textbook uses the second formula to solve this problem. But I cannot understand what "x" and "R" are for the earth, and how we assumed x>>R.

2. May 12, 2015

### haruspex

Wherever you got that equation from (link?), it should state what the variables mean and what constraints there are on its validity. If you can't provide a link, please quote them.
The form of it ($R^2+x^2$), suggests R and x are orthogonal, which doesn't sound right for the given question.

3. May 13, 2015

### hitemup

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html#c3

This is the same formula except pi*r^2*I is changed with mu(magnetic dipole moment) in mine.

4. May 13, 2015

### BvU

"Dipole field" is the name for the field from a current loop in the limit $R\rightarrow 0$, with m or ${\bf \mu} = \pi R^2 \;I$ constant. So Haru is right and your book is right too.

That eliminates the R question. Now you should think about where this dipole could possibly be located, and therefore what to take for the x.