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The couple acting on a magnet of length 10cm and pole strength 15A-m, kept in a field of `B=2xx10^(-5)`, at an anlge of `30^(@)` isA. `1.5xx10^(-5)N-m`B. `1.5xx10^(-3)N-m`C. `1.5xx10^(-2)N-m`D. `1.5xx10^(-6)N-m` |
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Answer» As, `tau=MB sin theta=(mxx2l)xx2xx10^(-5)sin30^(@)` `=15xx10xx10^(-2)xx(1)/(2)=1.5xx10^(-5)N-m` |
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