Q 12-04-215JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
A moving coil galvanometer has a coil with 175 turns and area $1\ \text{cm}^2$. It uses a torsion band of torsion constant $10^{-6}\ \text{N m/rad}$. The coil is placed in a magnetic field $B$ parallel to its plane. The coil deflects by $1^\circ$ for a current of $1\ \text{mA}$. The value of $B$ (in tesla) is approximately
Answer: (C) $10^{-3}$
At equilibrium the magnetic torque balances the torsional torque: $NIAB = c\theta$.
$$B = \frac{c\theta}{NIA} = \frac{10^{-6}\times\frac{\pi}{180}}{175\times10^{-3}\times10^{-4}} = \frac{1.75\times10^{-8}}{1.75\times10^{-5}} = 10^{-3}\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics