Question

The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the earth's outer core. Suppose we model the core current as a 3000-km-diameter current loop made from a 1000-km-diameter “wire.” The loop diameter is measured from the centers of this very fat wire. A)What is the current in the current loop? B)What is the current density J in the current loop? C)To decide whether this is a large or a small current density, compare it to the current density of a 8.0 A current in a 1.8-mm-diameter wire.

Answer 1

magnetic dipole moment = I*A

A = area of loop = pi*R^2

R = radius of loop = 1500 km = 1500000 m

current I = 8*10^22/(pi*1500000^2)

I = 1.13*10^10 A

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B)

current density J = I/a

a = area of crosssection = pi*r^2

r = radius of wire = 1000/2 = 500 km = 500000 m

J = 1.13*10^10/(pi*500000^2)

J = 0.0144 A/m^2

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C)

for wire

radius r = 1.8/2 = 0.9 mm = 0.9*10^-3 m

current density in wire = 8/(pi*(0.9*10^-3)^2) = 3.14*10^6 A/m^2