Suppose that the magnetic field of the Earth were due to a single current moving

Question

Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius 2.00 · 10^3 km through the Earth's molten core. The strength of the Earth's magnetic field on the surface near a magnetic pole is about 6.00 · 10^5 T. About how large a current would be required to produce such a field? I understand the formula setup, which I found in the textbook solutions on this website. However, I'm confused as to why the value in the denominator, which is usually R (radius of the loop of current), is instead R^2/(R^2+x^2)^(3/2), with x being the radius of earth from its center to the surface of the earth. Why is the value for what the loop of the current is so complex?

Explanation / Answer

B=((uo*I)/2)* (x^2/(x^2+R^2)^(3/2))

Then I=(2B*(x^2+R^2)^(3/2))/(uo*R^2)

At a magnetic pole at the surface, x is the radius of the earth.

I = (2B*(R^2+R^2)^(3/2))/(uo*R^2) = 5.66*B*R/uo = (5.66*6*10^-5*2*10^6)/(4pi*10^-7) = 5.4*10^8 A

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