Question

A proton (charge 1.61 x 10^-19 C) is moving at the speed of 3.42 x 10^+4 m/s parallel to the magnetic field equal to 3.5T. What is the magnitude of the force that will act on the proton? _______N

Answer 1

$$F=qvB=1.61\cdot10^{-19}\ C\cdot3.42\cdot10^4\ \dfrac{m}{s}\cdot3.5\ T=1.93\cdot10^{-14}\ N.$$