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.