Question

A electron passes through a magnetic field at 60° to the field at a velocity of 3.5×10⁶m/s. What is the magnitude of the force acting on the electron having a 0.45T magnetic field?

Answer 1

$$F=qvB\sin\theta\\ =1.6*10^{-19}*3.5*10^6*0.45*\sin 60^\circ\\ =2.2*10^{-13}\:\rm N$$