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)

$B=4\times10^{-5}T$
$B=\frac{\mu_{0}}{4\pi}\times\frac{M}{r^{3}}$
$=10^{-7}\times\frac{M}{\left(6.4\times10^{6}\right)^{3}}=4\times10^{-5}$
$M=\frac{4\times 10^{-5}\times 10^{18}\times6.4^{3}}{10^{-7}}$
$=1.048\times10^{3+18+7-5}$
$=10^{23}$