Q 12-04-195JEE MainJEE Main 2020 (3 Sep, Shift 2)Easy
A galvanometer coil has $500$ turns and each turn has an average area of $3\times10^{-4}\ \text{m}^{2}$. If a torque of $1.5\ \text{N m}$ is required to keep this coil parallel to a magnetic field when a current of $0.5\ \text{A}$ is flowing through it, the strength of the field (in T) is ______.
Numerical value type. Enter your answer.
Answer: 20
With the plane of the coil parallel to $B$, the magnetic moment is perpendicular to $B$ and the torque is $\tau = NIAB$.
$$B = \frac{1.5}{500\times0.5\times3\times10^{-4}} = 20\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics