A particle having charge 10⁻⁹ C moving in x - y plane in fields of 0.4 j N/C and 4 10⁻³ k T experiences a force of (4 i + 2 j ) 10⁻¹⁰ N. The velocity of the particle at that instant is _______ m/s.

Step-by-step solution

The Lorentz force experienced by a charged particle is given by F = q( E + v B ) . Given that the particle moves 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 = 10⁻⁹ [0.4 j + (v_x i + v_y j ) (4 10⁻³ k )] Using the cross product rules i k = - j and j k = i : F = 10⁻⁹ [0.4 j - 4 10⁻³ v_x j + 4 10⁻³ v_y i ] F = 10⁻⁹ [4 10⁻³ v_y i + (0.4 - 4 10⁻³ v_x) j ] The given force is F = (4 i + 2 j ) 10⁻¹⁰ N, which can be rewritten as: F = 10⁻⁹ (0