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Ghella [55]
4 years ago
13

The magnetic dipole moment of Earth has magnitude 8.00 1022 J/T.Assume that this is produced by charges flowing in Earth’s molte

n outer core. If the radius of their circular path is 3500 km, calculate the current they produce.
Physics
1 answer:
ziro4ka [17]4 years ago
3 0

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

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