Question

The couple acting on a magnet of length $10\, cm$ and pole strength $15\, Am$, kept in a field of $B=2 \times 10^{-5} T$, at an angle of $30^{\circ}$ is

• A. $ 1.5\times 10^{-5}Nm $
• B. $ 1.5\times10^{-3}Nm $
• C. $ 1.5\times10^{-2}Nm $
• D. $ 1.5\times10^{-6}Nm $

Answer 1

Answer: (A)

$C = MB \sin \theta$
$=( m \times 21) \times 2 \times 10^{-5} \sin 30^{\circ}$
$=150 \times 10^{-2} \times 2 \times 10^{-5} \times \frac{1}{2}=1.5 \times 10^{-5} N - m$