• # question_answer The couple acting on a magnet of length 10 cm and pole strength 125 A-m, kept in a field of $B=2\times {{10}^{-5}}T,$at an angle of ${{30}^{o}}$ is A) $1.5\times {{10}^{-5}}N-m$ B)  $1.5\times {{10}^{-3}}N-m$ C)  $1.5\times {{10}^{-2}}N-m$ D)  $1.5\times {{10}^{-6}}N-m$

$C=MB\sin \theta$ $=(m\times 2l)\times 2\times {{10}^{-5}}\sin {{30}^{o}}$ $=150\times {{10}^{-2}}\times 2\times {{10}^{-5}}\times \frac{1}{2}=1.5\times {{10}^{-5}}N-m$