A charge of 1 μC moves with velocity v = (î-2ĵ+3k̂) m·s⁻¹ in a magnetic field B = (2î+3ĵ-5k̂) T. If the magnitude of the magnetic force is √α×10⁻⁶ N, find α.

Question

G1 Physics · Accepted Answer

Magnetic force is F = q(v×B). Given, v = î-2ĵ+3k̂, B = 2î+3ĵ-5k̂. Compute the cross product: v×B = | î ĵ k̂ | | 1 -2 3 | | 2 3 -5 |. Therefore, v×B = [10-9]î -[-5-6]ĵ +[3+4]k̂ = î+11ĵ+7k̂. Its magnitude is |v×B| = √(1²+11²+7²) = √(1+121+49) = √171. Since q = 1 μC = 1×10⁻⁶ C, the force magnitude is F = 1×10⁻⁶√171 = √171×10⁻⁶ N. Comparing with √α×10⁻⁶ N, we obtain α = 171.