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

Answer:

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

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

                  $I = \frac{M}{A}$

                     $= \frac{M}{\pi R^2}$

                     $= \frac{8.0^10^{22}}{\pi (6378 *10^{3})^2}$

                     $= 6.26 *10^8 A$