Q. A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\overrightarrow{B}$ is in plane of the coil. If due to a current $i$ in the triangle a torque $\tau$ acts on it, the side $l$ of the triangle is
A) $\frac{2}{\sqrt{3}}\left( \frac{\tau}{Bi} \right)^{\frac{1}{2}}$
B) $\frac{2}{3}\left( \frac{\tau}{Bi} \right)$
C) $2\left( \frac{\tau}{\sqrt{3}Bi} \right)^{\frac{1}{2}}$
D) $\frac{1}{\sqrt{3}}\frac{\tau}{Bi}$
Explanation
(c) Torque acting on equilateral triangle in a magnetic field $\overrightarrow{B}$ is
$\tau = iAB\sin\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
[image:4_image126.emf]
Area of triangle $LMN$
$A = \frac{\sqrt{3}}{4}l^{2}$
And $\theta = 90{^\circ}$
$\therefore\ \tau \times \frac{\sqrt{3}}{4}l^{2}B\sin{90{^\circ}}$
$= \frac{\sqrt{3}}{4}il^{2}B\ \ \ \ \ \ \ \ \ \ \left\lbrack \because\sin{90{^\circ}} = 1 \right\rbrack$
Hence, $l = 2\left( \frac{\tau}{\sqrt{3}Bi} \right)^{1/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.