Question

A square loop of side $l$ is kept in a uniform magnetic field $B$ such that its plane makes an angle $\alpha$ with $\vec{B}$. The loop carries a current $i$. The torque experienced by the loop in this position is

• A. $B i l^2$
• B. $B i l^2 \sin \alpha$
• C. $B i l^2 \cos \alpha$
• D. Zero

Answer 1

Answer: (C)

$\tau=\vec{M} \times \vec{B}$
$\vec{M}=i A=i l^{2}$
$\Rightarrow \tau=i l^{2} \sin (90-\alpha)$
$=B i l^2 \cos \alpha$