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Magnetic field on t:quatur of Earth is 4 xx 10^(-5) T. Radius of Earth is 6400 km, then magnetic dipole moment of Earth is about ...... Am^(-2). |
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Answer» `10^(23)` `B_(E) = (mu_0 m)/( 4pi R^3)` `THEREFORE m= (4pi B_(E) R^(3) )/( mu_0)` `therefore m= (4 xx 10^(-5) xx (6.4 xx 10^(6) )^(3) )/(10^(-7) ) [ because (mu_0)/( 4pi ) = 10^(-7)]` `therefore m= 1048 . 576 xx 10^(20) ` `therefore m ~~ 1.0xx 10^(23) ""therefore m~~ 10^(23) "Am"^(2)` |
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