What is the approximate earth's dipole moment, if the earth's magnetic field at the Equator is approximately 0.5 G and diameter is \( 15.0 \times 10^{6}\,\mathrm{m} \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.1 \times 10^{23}\,\mathrm{Am^2} \)
The earth's field is treated as that of a magnetic dipole at its centre. The magnetic equator lies on the dipole's equatorial line (perpendicular bisector), where
\( \displaystyle B_{E} = \frac{\mu_0}{4\pi}\,\frac{m}{R^{3}} \)
\( B_{E} = 0.5\,\mathrm{G} = 0.5\times10^{-4}\,\mathrm{T} \)
\( R = 15.0\times10^{6}/2 =7.5\times10^{6}\,\mathrm{m} \)
\( \mu_0/4\pi = 10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle R^{3} = (7.5\times10^{6})^{3} = 4.22\times10^{20}\,\mathrm{m^{3}} \)
\( \displaystyle m = \frac{B_{E}\,R^{3}}{\mu_0/4\pi} = \frac{0.5\times10^{-4} \times 4.22\times10^{20}}{10^{-7}} = 2.1\times10^{23}\,\mathrm{A\,m^{2}} \)