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The magnetic dipole moment of earth is 6...

The magnetic dipole moment of earth is `6*4xx10^(21)Am^2`. If we consider it to be due to a current loop wound around the magnetic equator of the earth, then what should be the magnitude of the current? Take earth to be a sphere of radius `6400km`.

A

amper `metre^(2) (Am^(2))`

B

newton metre / tesla

C

newton `"metre"^(3) //"weber"`

D

joule tesla

Text Solution

Verified by Experts

The correct Answer is:
A
The magnetic dipole moment of the earth M=IA
`therefore M=I.piR^(2)`
`=(10^(9))/(6.4xx3.14)=(100xx10^(7))/(20.096)`
`=5xx10^(7) A`
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