A proton moving at 8.9 * 106 m/s through a magnetic field of 0.96 T experiences a magnetic force of magnitude 3.8 * 10-13 N. What is the angle between the proton's velocity and the field? The charge on a proton is 1.60218 * 10-19 C and its mass is 1.67262 * 10-27 kg.

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v*B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F / (q*v*B) = sin α

3.8 * 10^-13 / (1.6*10^-19*8.9*10^6 * 0.96) = 0.27

then α = 15.66°