Q 12-04-082JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
A coil having 100 turns, area of $5\times10^{-3}\ \text{m}^2$, carrying a current of $1\ \text{mA}$ is placed in a uniform magnetic field of $0.20\ \text{T}$ in such a way that the plane of the coil is perpendicular to the magnetic field. The work done in turning the coil through $90^\circ$ is ______ $\mu\text{J}$.
Numerical value type. Enter your answer.
Answer: 100
With the plane perpendicular to $\vec B$, the magnetic moment is along $\vec B$ ($\theta = 0$).
$M = NIA = 100\times10^{-3}\times5\times10^{-3} = 5\times10^{-4}\ \text{A m}^2$.
$$W = MB(1 - \cos90^\circ) = 5\times10^{-4}\times0.2 = 10^{-4}\ \text{J} = 100\ \mu\text{J}$$
Solution by Sreeraj P, M.Sc Physics