An alpha particle, which is a helium nucleus, has a mass of 6.64 x 10-27 kg. It is traveling horizontally at 42.3 km/s when it enters a uniform, vertical 1.60 T magnetic field. What is the magnitude of the acceleration of the alpha particle while it is in the magnetic field?

7.70 x 1012 m/s2

3.26 x 1012 m/s2

5.49 x 1012 m/s2

1.06 x 1012 m/s2

3.26 x 1012m/s2

The answer is found using Newton's second law, F = ma. The force experienced on the charge in the magnetic field, that is entering perpendicular to the field, the Lorentz force is given by F = qvB. Equating this with Newton's second law, we have ma = qvB. Rearranging, a = qvB/m. Plug in the values for v, b and m given in the question. Q, the coulombic electric charge of the helium nucleus is 2 x (1.6 x 10^-19) as thats the charge of a proton, and a helium nucleus has two protons.