Q 12-04-098JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
The horizontal component of earth's magnetic field at a place is $3.5\times10^{-5}\ \text{T}$. A very long straight conductor carrying a current of $\sqrt2\ \text{A}$ in the direction from South-East to North-West is placed. The force per unit length experienced by the conductor is ______ $\times10^{-6}\ \text{N m}^{-1}$.
Numerical value type. Enter your answer.
Answer: 35
The horizontal field points from south to north. A wire running from SE to NW makes $45^\circ$ with the north direction.
$$\frac FL = IB_H\sin45^\circ = \sqrt2\times3.5\times10^{-5}\times\frac1{\sqrt2} = 35\times10^{-6}\ \text{N m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics