Question

A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be $\sqrt{x}\times {10}^{-5}Nm$. The value of x is .

• A. 3
• B. 3.0
• C. 3.00

Answer 1

Answer: (A), (B), (C)

The situation is shown in the figure below:

Let us assume the direction of current in the loop to be clockwise.

Clearly, the magnetic moment $\overrightarrow{\mu }$ would point in a direction perpendicular to the plane of the coil.

The angle between $\overrightarrow{\mu }$ and $\overrightarrow{B}$ is ${90}^{\circ }$

Now, torque acting on the loop is given by,

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

$|\overrightarrow{\tau }|=i|\overrightarrow{A}||\overrightarrow{B}|\sin {90}^{\circ }$

$|\overrightarrow{\tau }|=i\times \frac{\sqrt{3}}{4}{a}^2\times 20\times {10}^{-3}$

$|\overrightarrow{\tau }|=\sqrt{3}\times {10}^{-5}$

$\Rightarrow x=3$