Aproton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5 × 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction. the mass of a proton is 1.673 × 10−27 kg. find the magnitude of the field. answer in units of t.

The magnetic force experienced by the proton is given by

where q is the proton charge, v its velocity, B the magnitude of the magnetic field and  the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so  and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:


So we have

from which we can find the magnitude of the field: