Question

The magnetic field of earth at the equator is approximately $4 \times 10^{-5} \, T$. The radius of earth is $6.4 \times 10^6 \, m$. Then the dipole moment of the earth will be nearly of the order of:

• A. $10^{23} \, A \, m^2$
• B. $10^{20} \, A \, m^2$
• C. $10^{16} \, A \, m^2$
• D. $10^{10} \, A \, m^2$

Answer 1

Answer: (A)

Given, $B = 4 \times10^{-5} T$
$ R_{E } = 6.4 \times10^{6}m $
Dipole moment of the earth M = ?
$B = \frac{\mu_{0}}{4\pi} \frac{M}{d^{3}}$
$ 4 \times10^{-5} = \frac{4 \pi\times10^{-7} \times M}{4 \pi\times\left(6.4 \times10^{6}\right)^{3}} $
$\therefore M \cong 10^{23} Am^{2}$