First slide

Magnetism

Question

The magnetic moment of the assumed dipole at the earth’s centre is $7 .0 \times 10^{22} A m^{2}$ . The magnetic field B at the geomagnetic poles of the earth is $\frac{P \times 10^{14}}{R^{3}}$ (The radius of earth is ‘R’), then the value of ‘P ’ is__

Moderate

Solution

The geomagnetic pole is at the end on position of the earth (axial position) $B = \frac{μ_{0}}{4 π} . \frac{2 M}{d^{3}}$
$g i v e n : d = R \Rightarrow B = \frac{10^{7} \times 2 \times 7 \times 10^{22}}{R^{3}} = \frac{14 \times 10^{15}}{R^{3}}$
Hence,
$\frac{P \times 10^{14}}{R^{3}} = \frac{14 \times 10^{15}}{R^{3}} P = 140.00$

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