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)

    10^{23}A\; m^{2}

  • Option 2)

    10^{20}A\; m^{2}

  • Option 3)

    10^{16}A\; m^{2}

  • Option 4)

    10^{10}A\; m^{2}

 

Answers (1)

As we discussed in

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

-

 

 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)

10^{23}A\; m^{2}

This solution is correct 

Option 2)

10^{20}A\; m^{2}

This solution is incorrect 

Option 3)

10^{16}A\; m^{2}

This solution is incorrect 

Option 4)

10^{10}A\; m^{2}

This solution is incorrect 

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