Q. A coil in the shape of an equilateral triangle of side $0.02\ m$ is suspended from its vertex such that it is hanging in a vertical plane between the pole pieces of permanent magnet producing a uniform field of $5 \times 10^{- 2}T$. If a current of $0.1\ A$ is passed through the coil, what is the couple acting
A) $5\sqrt{3} \times 10^{- 7}N - m$
B) $5\sqrt{3} \times 10^{- 10}N - m$
C) $\frac{\sqrt{3}}{5} \times 10^{- 7}N - m$
D) None of these
Explanation
(a) Torque $\tau = iAB\sin\theta,\ i = 0.1A,\ \theta = 90{^\circ}$
$A = \frac{1}{2} \times base \times height$
or $A = \frac{1}{2}a \times \frac{a\sqrt{3}}{2}$
$= \frac{\sqrt{3}a^{2}}{4} = \frac{\sqrt{3} \times (0.02)^{2}}{4}$
$= \sqrt{3} \times 10^{- 4}m^{2};\ \theta = 90{^\circ}$
$\tau = 0.1 \times \sqrt{3} \times 10^{- 4} \times 5 \times 10^{- 2}\sin{90{^\circ}}$
$= 5\sqrt{3} \times 10^{- 7}N - m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.