# The magnetic field of earth at the equator is approximately 4x10-5 T. The radius of earth is 6.4x106 m. Then the dipole moment of the earth will be nearly of the order of : Option 1) Option 2) Option 3) Option 4)

P Plabita

As we discussed in

If a<$B_{e}= \frac{\mu _{o}}{4\pi }\frac{M}{r^{3}}$

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Given $B=4\times 10^{5}$

$R_{f}=6.4\times 10^{6}m$

Dipole moment of the earth M=?

$B=\frac{\mu _{0}m}{4\pi d^{3}}= 4\times 10^{-5}=\frac{4\times 10^{-7}}{4\pi\times \left ( 6.4\times 10^{6} \right )^{3}}$

$m\widetilde{=}10^{23} Am^{2}$

Option 1)

This solution is correct

Option 2)

This solution is incorrect

Option 3)

This solution is incorrect

Option 4)

This solution is incorrect

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