Calculate the couple (in $\mu \mathrm{N} \mathrm{m}$) acting on a magnet of length $10 \mathrm{cm}$ and pole strength $15 \mathrm{A} \mathrm{m}$, kept in a field of $B=2\times {10}^{-5} T$ at an angle of $30°$.

Solution

$C=MB\sin \theta$

$=\left(m\times 2l\right)\times 2\times {10}^{-5}\sin 30°$

$=15\times 10\times {10}^{-2}\times 2\times {10}^{-5}\times \frac{1}{2}$




$=15\times {10}^{-6} \mathrm{N} \mathrm{m}$

Hence, the correct answer is 15.