Q 12-04-245JEE MainJEE Main 2025 (3 Apr, Shift 1)Easy
A $4.0\ \text{cm}$ long straight wire carrying a current of $8\ \text{A}$ is placed perpendicular to a uniform magnetic field of strength $0.15\ \text{T}$. The magnetic force on the wire is ______ mN.
Numerical value type. Enter your answer.
Answer: 48
$$F = IlB\sin90^\circ = 8\times0.04\times0.15 = 0.048\ \text{N} = 48\ \text{mN}$$
Solution by Sreeraj P, M.Sc Physics