At the north magnetic pole the earth's magnetic field is vertical and has a strength of 0.62 gauss. The earth's field at the surface and further out is approximately that of a central dipole. (a) What is the magnitude of the dipole moment in joules/tesla? (b) Imagine that the source of the field is a current ring on the "equator" of the earth's metallic core, which has a radius of 3000 km, about half the earth's radius. How large would the current have to be?

A) Dipole moment; m = 8.02 x 10^ (22) J/T

B) I = 3.51 x 10^ (9) A

Explanation:

The components of a magnetic field of a dipole are;

B_r = (μ_o•m/2πr³). cosθ

B_θ = (μ_o•m/4πr³). sin θ

B_Φ = 0

Let's make m the subject in the B_r equation;

m = (2πr³•B_r) / (μ_o•cosθ)

Where;

B_r is magnetic field = 0.62 Gauss = 6.2 x 10^ (-5) T

μ_o is the magnetic constant and has a value of 4π * 10^ (-7) H/m

m is magnetic moment.

r is equal to radius of earth = 6.371 x 10^ (6) m

Thus, if we set θ = 0, we can solve for m as below;

m = (2π (6.371 x 10^ (6)) ³•6.2 x 10^ (-5)) / (4π * 10^ (-7) •cos0)

Thus, m = 8.02 x 10^ (22) J/T

B) Now, to find the current, let's use the expression for the magnetic field on the z-axis of the current ring.

B_z = (μ_o•Ib² / (2 (z² + b²/2) ^ (3/2)))

So, let's set z = R and b = R/2

Thus, we now have;

B_z = (μ_o•I) / (5^ (3/2) •R)

Making I the subject, we have;

I = [ (5^ (3/2) •R) •B_z]/μ_o

Plugging in the relevant values, we have;

I = [ (5^ (3/2) x 6.371 x 10^ (6)) x 6.2 x 10^ (-5) ] / (4π * 10^ (-7))

I = 3.51 x 10^ (9) A