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A short bar magnet is placed horizontally along the magnetic N-S direction with its axis along the magnetic E-W direction.The resultant horizontal magnetic induction on its equator at a distance of 20 cm from centre is (B_(H)=4xx10^(-5)T) |
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Answer» `4.5 Am^(2)` `4xx10^(-5)=(10^(-7)xx2xxM)/(27xx10^(-3))` `therefore M=5.4Am^(2)` |
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