A straight wire of length 4.0 cm carrying current 8 A is placed perpendicular to a uniform magnetic field of 0.15 T. Find the magnetic force on the wire in mN.

Question

G1 Physics · Accepted Answer

Magnetic force on a current-carrying straight wire is F = BIl sinθ. Given, B = 0.15 T, I = 8 A, l = 4.0 cm = 4.0×10⁻² m, θ = 90°. Therefore, F = 0.15×8×4.0×10⁻²×1 = 0.048 N. Convert to millinewton: 0.048 N = 48 mN. Answer: 48.