## 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 fur

Question

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?

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5 days 2021-07-22T16:04:21+00:00 1 Answers 5 views 0

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