Question

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

• A. $\frac{3}{4}IB{a}^2$
• B. $\frac{3}{2}IB{a}^2$
• C. $\frac{3}{8}IB{a}^2$
• D. $\frac{3}{16}IB{a}^2$

Answer 1

The correct option is C $\frac{3}{8}IB{a}^2$


Area of the loop,

$A=\frac{1}{2}\times \text{base}\times \text{height}$

$\Rightarrow A=\frac{1}{2}\times a\times \frac{a\sqrt{3}}{2}=\frac{a^2\sqrt{3}}{4}$

Torque on the loop,

$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}=i(\overrightarrow{A}\times \overrightarrow{B})$

$\therefore \tau =iAB\sin {60}^{\circ }$

$=I\times \frac{a^2\sqrt{3}}{4}\times B\times \frac{\sqrt{3}}{2}$

$=\frac{3}{8}IB{a}^2$