Question

A coil in the shape of equilateral tringale of side $0.02m$ is suspended from the vertex such that it is hanging in a vertical place between the pole-pieces of a permanent magnet producing a horizontal magnetic field of $5\times {10}^{-2}T.$ When a current of $0.1A$ passed through it and the magnetic field is parallel to its plane then couple acting on the coil is :

• A. $8.65\times {10}^{-7}N-m$
• B. $6.65\times {10}^{-7}N-m$
• C. $3.35\times {10}^{-7}N-m$
• D. $3.91\times {10}^{-7}N-m$

Answer 1

The correct option is A $8.65\times {10}^{-7}N-m$

The torque acting on a coil is given by,

$\tau$ = N B I A sin$\theta$

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

$\begin{array}{c}=\frac{1}{2}\times 0.02\times \sqrt{{0.02}^2-{0.01}^2}\\ =1.732\times {10}^{-4}{m}^2\end{array}$

$\theta$ = Angle between direction of the magnetic field and normal to the plane of the coil. Also, as the magnetic field is parallel to the plane of the coil, so $\theta$ = 90

thus,

$\tau$ = $1\times 5\times {10}^{-2}\times 0.1\times 1.732\times {10}^{-4}\times 1$

= 8.6$\times {10}^{-7}$ Nm