The earth's magnetic field at the equator is | vedclass

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(D) The equatorial magnetic field of the Earth is given by the formula:$B_{E} = \frac{\mu_{0} m}{4 \pi r^{3}}$We are given that $B_{E} \approx 0.4 \;

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$1.05 	imes 10^{23} \; A m^{2}$

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(D) The equatorial magnetic field of the Earth is given by the formula:$B_{E} = \frac{\mu_{0} m}{4 \pi r^{3}}$We are given that $B_{E} \approx 0.4 \; G = 0.4 	imes 10^{-4} \; T = 4 	imes 10^{-5} \; T$.For $r$,we take the radius of the Earth,$r = 6.4 	imes 10^{6} \; m$.Rearranging the formula to solve for the magnetic dipole moment $m$:$m = \frac{B_{E} \cdot 4 \pi r^{3}}{\mu_{0}} = B_{E} \cdot r^{3} \cdot \left( \frac{4 \pi}{\mu_{0}} ight)$Since $\frac{\mu_{0}}{4 \pi} = 10^{-7} \; T m/A$,we have $\frac{4 \pi}{\mu_{0}} = 10^{7} \; A/T m$.Substituting the values:$m = (4 	imes 10^{-5}) 	imes (6.4 	imes 10^{6})^{3} 	imes 10^{7}$$m = 4 	imes 10^{2} 	imes (6.4)^{3} 	imes 10^{18}$$m = 4 	imes 10^{2} 	imes 262.144 	imes 10^{18}$$m = 1048.576 	imes 10^{20} \; A m^{2}$$m \approx 1.05 	imes 10^{23} \; A m^{2}$.

The earth's magnetic field at the equator is approximately $0.4 \; G$. Estimate the earth's dipole moment.

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