Answer to Question #300309 in Electricity and Magnetism for Zack

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 10^6 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, the north direction is into the page.).

Answer 1

The magnetic force

$$F=qvB$$

Hence, the magnitude of the magnetic field

$$B=\frac{F}{qv}\\ =\frac{8.0*10^{–14}}{1.6*10^{–19}*5.0*10^{6}}=0.1\:\rm T$$

The field is directed into the page.