The magnetic field of the Earth at the equato | vedclass

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Question

(A) Given:Magnetic field at the equator,$B = 4 	imes 10^{-5} \, T$Radius of the Earth,$R_E = 6.4 	imes 10^6 \, m$The magnetic field at the equator d

Vedclass · Accepted Answer

$10^{23} \, A \cdot m^2$

Vedclass · Answer

(A) Given:Magnetic field at the equator,$B = 4 	imes 10^{-5} \, T$Radius of the Earth,$R_E = 6.4 	imes 10^6 \, m$The magnetic field at the equator due to a magnetic dipole is given by the formula:$B = \frac{\mu_0}{4\pi} \cdot \frac{M}{R_E^3}$Rearranging for the magnetic dipole moment $M$:$M = \frac{B \cdot 4\pi \cdot R_E^3}{\mu_0}$Substituting the values:$M = \frac{(4 	imes 10^{-5}) \cdot (6.4 	imes 10^6)^3}{10^{-7}}$$M = 4 	imes 10^{-5} 	imes 10^7 	imes (6.4)^3 	imes 10^{18}$$M = 4 	imes 10^2 	imes 262.144 	imes 10^{18}$$M \approx 1048 	imes 10^{20} \approx 1.048 	imes 10^{23} \, A \cdot m^2$Thus,the order of magnitude is $10^{23} \, A \cdot m^2$.

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

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