Question

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

Answer 1

$C=MBsin \theta $
$=\left(m \times 2 l\right)\times 2\times \left(10\right)^{- 5}sin 30^\circ $
$=15\times 10\times 10^{- 2}\times 2\times 10^{- 5}\times \frac{1}{2}$


$=15\times 10^{- 6} \, Nm$