Answer to Question #172060 in Physics for Ashley

Question

Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude 2780 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.27 × 10³ m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +3.10 × 10^-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.52 × 10^-9 N, pointing directly upward. What is the speed of this particle?

Answer 1

The electric and magnetic forces are equal for the particle that does not experience deflection:

$Eq=qvB,\\\space\\ B=\frac{E}{v}.$

For another particle, the electric force Eq acts upward, the magnetic (Lorentz's) force acts downward according to the left-hand rule, and the difference between these two particles is

$F=EQ-QuB,\\\space\\ u=\frac{EQ-F}{QB}=\frac{v(EQ-F)}{QE}=v\bigg(1-\frac{F}{QE}\bigg),\\\space\\ u=6.27·10^3\bigg(1-\frac{1.52·10^{-9}}{2780·3.1·10^{-12}}\bigg)=5164\text{ m/s}.$