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