Question

A coil in the shape of an equilateral triangle of side $0.02\text{m}$ is suspended from a vertex such that it is hanging in a vertical plane. A magnetic field of magnetic strength $5\times {10}^{-2}\text{T}$ is applied along the plane of coil. When a current of $0.1\text{A}$ is passed through the coil, the torque experienced by its will be:

• A. $5\sqrt{3}\times {10}^{-7}\text{Nm}$
• B. $2\sqrt{2}\times {10}^{-7}\text{Nm}$
• C. $10\sqrt{5}\times {10}^{-7}\text{Nm}$
• D. $5\times {10}^{-5}\text{Nm}$

Answer 1

The correct option is A $5\sqrt{3}\times {10}^{-7}\text{Nm}$


The diagram shows the arrangement of coil & magnetic field.

As the coil is in the form of equilateral triangle, its area is,

$A=\frac{1}{2}\times L\times L\sin {60}^0=\frac{1}{2}\times (0.02{)}^2\times \frac{\sqrt{3}}{2}$

$\Rightarrow A=\sqrt{3}\times {10}^{-4}{\text{m}}^2$

Now, the magnetic moment of the coil:

$\mu =iA=0.1\times \sqrt{3}\times {10}^{-4}=\sqrt{3}\times {10}^{-5}{\text{Am}}^2$

The torque acting on the coil will be,

$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$

$\Rightarrow \tau =\mu B\sin \theta$

Since, the plane of the coil is parallel to the magnetic field, the vector($\overrightarrow{A}$) will be perpendicular to the magnetic field.
Thus, $\overrightarrow{\mu }\perp \overrightarrow{B}$ and $\theta ={90}^0$

$\Rightarrow \tau =\mu B=\sqrt{3}\times {10}^{-5}\times 5\times {10}^{-2}$

$\therefore \tau =5\sqrt{3}\times {10}^{-7}\text{Nm}$

Hence, option (a) is the correct answer.