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

`2.08\cdot 10^9 A`

Step-by-step 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=(\mu)/(\pi r^2)=(8.00\cdot 10^(22))/(\pi (3.5\cdot 10^6)^2)=2.08\cdot 10^9 A`