Calculating current to produce magnetic field of Earth

  • #1
Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius 2988 km through the earth’s molten core. The strength of the Earth’s magnetic field on the surface near a magnetic pole is about 6.00E-5 T. About how large a current would be required to produce such a field?

I have to use the equation B=((mu_naught*i)/2)* (x^2/(x^2+R^2)^(3/2))
and then for i get:
i=(2*(x^2+R^2)^(3/2))/(mu_naught*R^2)

How do I get x?
 

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  • #2
Redbelly98
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Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius 2988 km through the earth’s molten core. The strength of the Earth’s magnetic field on the surface near a magnetic pole is about 6.00E-5 T. About how large a current would be required to produce such a field?

I have to use the equation B=((mu_naught*i)/2)* (x^2/(x^2+R^2)^(3/2))
and then for i get:
i=(2*(x^2+R^2)^(3/2))/(mu_naught*R^2)

How do I get x?
Is x is the distance from the center of the circle to the point where the magnetic field is measured? The center of the circle is at the center of the Earth, and the magnetic field is on the surface of the Earth. Your textbook probably has what that distance is, perhaps in the chapter that discusses gravitation and Keppler's laws.

p.s. are you sure the equations you wrote are correct? The last one you wrote,
i=(2*(x^2+R^2)^(3/2))/(mu_naught*R^2)​
does not include B, the magnetic field. It seems that B should be involved in the calculation somehow.
 

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