Q 12-04-173JEE MainJEE Main 2021 (26 Aug, Shift 2)Easy
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 $\sqrt{x}\times10^{-5}$ N m. The value of $x$ is
Numerical value type. Enter your answer.
Answer: 3
Area $= \dfrac{\sqrt{3}}{4}(0.1)^2 = \dfrac{\sqrt{3}}{4}\times10^{-2}$ m$^2$.
With the plane parallel to $B$, the magnetic moment is perpendicular to $B$:
$$\tau = IAB = 0.2\times\frac{\sqrt{3}}{4}\times10^{-2}\times20\times10^{-3} = \sqrt{3}\times10^{-5}\ \text{N m}$$
So $x = 3$.
Solution by Sreeraj P, M.Sc Physics