Answer to Question #329724 in Physics for Kys

Question

A proton moving horizontally at 3.5×10⁴ m/s enters a uniform magnetic field of 0.075 T directed upward at 30° with the horizontal. Find the magnitude of the magnetic force acting on the proton

Answer 1

$$F=qvB\sin\theta\\ =1.6*10^{-19}*3.5*10^4*0.075*\sin30^\circ\\ =2.1*10^{-16}\:\rm N$$

The force is directed out of the page.