Question

The magnetic dipole moment of Earth has magnitude 8.00 1022 J/T.Assume that this is produced by charges flowing in Earth’s molten outer core. If the radius of their circular path is 3500 km, calculate the current they produce.

Answer 1

Answer:

$2.08\cdot 10^9 A$

Explanation:

The magnetic dipole moment of a circular coil with a current is given by

$\mu = IA$

where

I is the current in the coil

$A=\pi r^2$ is the area enclosed by the coil, where

$r$ is the radius of the coil

So the magnetic dipole moment can be rewritten as

$\mu = I\pi r^2$ (1)

Here we can assume that the magnetic dipole moment of Earth is produced by charges flowing in Earth’s molten outer core, so by a current flowing in a circular path of radius

$r=3500 km = 3.5\cdot 10^6 m$

Here we also know that the Earth's magnetic dipole moment is

$\mu = 8.0\cdot 10^{22} J/T$

Therefore, we can re-arrange eq (1) to find the current that the charges produced:

$I=\frac{\mu}{\pi r^2}=\frac{8.00\cdot 10^{22}}{\pi (3.5\cdot 10^6)^2}=2.08\cdot 10^9 A$