A particle carrying a charge of 1.0 10⁻⁹ C travels in the x – y plane. A magnetic field of 4.0 10⁻³ k T exerts a force of (4.0 i + 3.0 j ) 10⁻¹⁰ N on it. Determine the velocity of the particle.

Step-by-step solution

The magnetic force on a moving charge is given by F = q( v B ) . Since the particle travels in the x - y plane, its velocity can be written as v = v_x i + v_y j . Substituting the given values into the force equation: F = 1.0 10⁻⁹ [ (v_x i + v_y j ) (4.0 10⁻³ k ) ] F = 4.0 10⁻¹² [ v_x ( i k ) + v_y ( j k ) ] Using the cross product rules i k = - j and j k = i : F = 4.0 10⁻¹² (v_y i - v_x j ) N Equating this to the given force F = (4.0 i + 3.0 j ) 10⁻¹⁰ N : 4.0 10⁻¹² v_y = 4.0 10⁻¹⁰ v_y = 100 m s ⁻¹ -4.0 10⁻¹² v_x =