Q 12-04-118JEE MainJEE Main 2023 (25 Jan, Shift 2)Medium
For a moving coil galvanometer, the deflection in the coil is $0.05$ rad when a current of $10\ \text{mA}$ is passed through it. If the torsional constant of suspension wire is $4.0\times10^{-5}\ \text{N m rad}^{-1}$, the magnetic field is $0.01\ \text{T}$ and the number of turns in the coil is $200$, the area of each turn (in $\text{cm}^2$) is
Answer: (B) 1.0
$NIAB=k\theta$:
$$A=\frac{k\theta}{NIB}=\frac{4\times10^{-5}\times0.05}{200\times0.01\times0.01}=\frac{2\times10^{-6}}{0.02}=10^{-4}\ \text{m}^2=1\ \text{cm}^2$$
Solution by Sreeraj P, M.Sc Physics