A coil in the shape of an equilateral triangle of side 0.02 m is suspended from its vertex such that

Question

A coil in the shape of an equilateral triangle of side \(0.02 \mathrm{~m}\) 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 \(5 \times 10^{-2} \mathrm{~T}\). If a current of \(0.1 \mathrm{~A}\) is passed through the coil, what is the couple acting?

(a) \(5 \sqrt{3} \times 10^{-7} \mathrm{Nm} \quad\)

(b) \(5 \sqrt{3} \times 10^{-10} \mathrm{Nm}\)

(c) \(\frac{\sqrt{3}}{5} \times 10^{-7} \mathrm{Nm}\)

(d) None of these

Answer 1

The correct option of this question will be (a).

Solution —

Torque, $\tau=I A B \sin \theta, I=0.1 \mathrm{~A}, \theta=90^{\circ}$

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

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

$=\frac{\sqrt{3} a^{2}}{4}=\frac{\sqrt{3} \times(0.02)^{2}}{4}=\sqrt{3} \times 10^{-4} \mathrm{~m}^{2}$

$\tau=0.1 \times \sqrt{3} \times 10^{-4} \times 5 \times 10^{-2} \sin 90^{\circ}$

$=5 \sqrt{3} \times 10^{-7} \mathrm{~N} \mathrm{~m}$