A particle of charge 1 C moves in a magnetic field B = (2 i + 3 j + k ) T . It experiences a magnetic force F = (4 i - 3 j + k ) N . If the velocity of the particle is known to lie entirely in the x-y plane, what is the velocity vector v of the particle?

Step-by-step solution

The magnetic Lorentz force is given by F = q( v B ) . Since the velocity lies in the x-y plane, we can write v = v_x i + v_y j . Substituting the given values: F = 1 ((v_x i + v_y j ) (2 i + 3 j + k )) Evaluating the cross product using a determinant or term-by-term expansion: v B = vmatrix i & j & k v_x & v_y & 0 2 & 3 & 1 vmatrix v B = i (v_y 1 - 0) - j (v_x 1 - 0) + k (v_x 3 - v_y 2) v B = v_y i - v_x j + (3v_x - 2v_y) k Equating this to the given force F = 4 i - 3 j + k : Comparing the i components: v_y = 4 Com