Chapter 29: Q. 19 (page 831)

The earth’s magnetic dipole moment is $8.0 \times 10^{22} A . m^{2}$ .
a. What is the magnetic field strength on the surface of the earth at the earth’s north magnetic pole? How does this compare to the value in Table 29.1? You can assume that the current loop is deep inside the earth.
b. Astronauts discover an earth-size planet without a magnetic field. To create a magnetic field at the north pole with the same strength as earth’s, they propose running a current through a wire around the equator. What size current would be needed?

Expert verified

a. The magnetic field is $61.8 μT .$

b. current is $6.3 \times 10^{8} A .$

Step by step solution

We have given,

magnetic moment = $8 \times 10^{22} A . m^{2}$

we have to find the magnetic field at north pole.

The magnetic field due to dipole

$B = \frac{μ_{0} \vec{μ}}{2 {πr}^{3}} B = \frac{(4 π \times 10^{- 7}) (8 \times 10^{22})}{2 π {(6.4 \times 10^{6})}^{3}} B = 61.8 μT$

We have given,

magnetic dipole moment = $8 \times 10^{22} A . m^{2}$

We have to find the current needed.

The magnetic dipole moment due to loop is

$\vec{μ} = A . I I = \frac{\vec{μ}}{π r^{2}} I = \frac{8 \times 10^{22}}{π {(6.4 \times 10^{6})}^{2}} I = 6.3 \times 10^{8} A$

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