Question

A triangular loop of side $a$ carries a current $I$. It is placed in a magnetic filed $B$ such that the plane of the loop makes an angle ${30}^{\circ }$ with the direction of $B$. The torque on the loop is

• A. $\frac{BI{a}^2}{8}$
• B. $\frac{3BI{a}^2}{8}$
• C. $\frac{\sqrt{3}BI{a}^2}{8}$
• D. $\frac{\sqrt{3}BI{a}^2}{4}$

Answer 1

The correct option is B $\frac{3BI{a}^2}{8}$

When current $I$ is passed through the loop having area $A$, then the magnetic dipole moment is given by, $$\mu =IA$$
Where,

$A=\frac{1}{2}\times a\times \left(a\sin {60}^{\circ }\right)=\frac{\sqrt{3}{a}^2}{4}$

Therefore, magnetic dipole moment is,

$\mu =IA=\frac{I{a}^2\sqrt{3}}{4}$


The angle between the normal of the plane and the applied magnetic field is ${60}^{\circ }$

Now, the torque acting on the loop is,

$\tau =\mu B\sin \theta =\frac{I{a}^2\sqrt{3}}{4}\times B\times \sin {60}^{\circ }$

$\therefore \tau =\frac{3BI{a}^2}{8}$

Hence, option (b) is the correct answer.