Answer to Question #212484 in Electricity and Magnetism for John Mark

Question

A magnetic field exerts a force of 8.0 x 10–14 N toward the west on a proton moving vertically upward at a speed of 5.0 x 106 m/s. When moving straight in a northerly direction, the force on the proton is zero. Determine the magnitude and direction of the magnetic field B in this region. (The charge of a proton = 1.6 x 10-19 C).

Answer 1

The direction of the magnetic field is northward according to the left hand rule.

The magnitude:

$$F=qvB,\\\space\\ B=\frac{F}{qv}=0.1\text{ T}.$$