Answer:
a
The current that would be produced is [tex]I = 6.26 *10 ^8 A[/tex]
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 [tex]M = 8.0*10 ^{22} J/T[/tex]
The radius of earth generally has a value of [tex]R = 6378 *10^3 m[/tex]
Magnetic moment is mathematically given as
[tex]M = IA[/tex]
A is the area of the of the earth(assumption that the earth is circular ) and this evaluated as
[tex]A = \pi R^2[/tex]
Now making [tex]I[/tex] the subject in the above formula
[tex]I = \frac{M}{A}[/tex]
[tex]= \frac{M}{\pi R^2}[/tex]
[tex]= \frac{8.0^10^{22}}{\pi (6378 *10^{3})^2}[/tex]
[tex]= 6.26 *10^8 A[/tex]
