Earth's magnetic field may be supposed to be due to a small bar magnet located at the centre of the earth. If the magnetic field at a point on the magnetic equator is $0.3 \times 10^{- 4} T$ . Magnet moment of bar magnet is: (Radius of earth= 64000 Km)

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$6.4 \times 10^{22} a m p \times m^{2}$

$7.8 \times 10^{22} a m p \times m^{2}$

$6.4 \times 10^{8} a m p \times m^{2}$

$\begin{aligned} 7.8 \times 10^{8} a m p \times m^{2} \end{aligned}$

answer is B.

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Detailed Solution

When a magnet is freely suspended in earth’s magnetic field, it's north pole points north, so the magnetic field of the earth may be suppose to be due to a magnetic dipole with it's south pole towards north and as equatorial point is on the broad side on position of the dipole.

$B_{e} = \frac{μ_{0}}{4 π} \cdot \frac{M}{r^{3}}$

$\begin{array}{l} \Rightarrow 0.3 \times 10^{- 4} = 10^{- 7} \times \frac{M}{{(6.4 \times 10^{6})}^{3}} \\ \Rightarrow M = 7.8 \times 10^{22} A - m^{2} \end{array}$