Question

a proton moving horizontally at 3.7 x 10 to the fifth power ms and thirst and a uniform magnetic field of 0.08 e60 directed upward at 35 degrees with the horizontal find the magnitude of the magnetic force acting on the proton

Answer 1

$$F=qvB\sin\theta\\ =1.6*10^{-19}*3.7*10^5*0.08*\sin 35^\circ\\ =2.7*10^{-15}\:\rm N$$