Question

Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extending around Earth at its geomagnetic equator if we wished to set up such a dipole? Could such an arrangement be used to cancel out Earth's magnetism (b) at points in space well above Earth's surface or (c) on Earth's surface?

Answer 1

a

The current that would be produced is
`I = 6.26 *10 ^8 A`

b

Yes this arrangement can be used to cancel out earths magnetic field at points well above the Earth's surface.This is because this current loop acts as a magnetic dipole for point above the earth surface

c

No this arrangement can not be used to cancel out earths magnetic field at points on the Earth's surface .this because on the earth surface it shifts from its behavior as a magnetic dipole

Step-by-step explanation:

From the question we are told that

The magnetic moment of earth is
`M = 8.0*10 ^(22) J/T`

The radius of earth generally has a value of
`R = 6378 *10^3 m`

Magnetic moment is mathematically given as

`M = IA`

A is the area of the of the earth(assumption that the earth is circular ) and this evaluated as

`A = \pi R^2`

Now making
`I` the subject in the above formula

`I = (M)/(A)`

`= (M)/(\pi R^2)`

`= (8.0^10^(22))/(\pi (6378 *10^(3))^2)`

`= 6.26 *10^8 A`