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A bar magnet of `5 cm` long having a pole strenght of `20 A.m`. Is deflected through `30^(@)` from the magnetic meridian. If `H=(320)/(4pi)A//m`, the deflecting couple isA. `1.6xx10^(-4)Nm`B. `3.2xx10^(-5)Nm`C. `1.6xx10^(-5)Nm`D. `1.6xx10^(-2)Nm` |
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Answer» Correct Answer - C `C=M mu_(0)H sin theta` |
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