Question

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

• A. $Bi{l}^2$
• B. $Bi{l}^2\sin \alpha$
• C. $Bi{l}^2\cos \alpha$
• D. $Zero$

Answer 1

The correct option is C $Bi{l}^2\cos \alpha$

Plane makes angle $=\alpha$

$\Rightarrow$ Normal to the plane will make angle $\theta =90-\alpha$

terque is $\overline{C}$, given as

$\overline{C}=\overrightarrow{m}\times \overrightarrow{B}=i\overrightarrow{A}\times \overrightarrow{B}=i|\overrightarrow{A}||\overrightarrow{B}|\sin \theta$

$\Rightarrow \overline{C}=i\times i^2\times B\times \sin (90-\alpha )$

$\Rightarrow \overline{C}=i{l}^{2}B\cos \alpha$