Question

A velocity selector has an electric field of magnitude 2170 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 5.45 × 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 +4.10 × 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.54 × 10-9 N, pointing directly upward. What is the speed of this particle?

Answer 1

`v = 4.51 * 10^3 m/s`

Step-by-step explanation:

electric field = 2170 N/C

now the speed of the charge particle is given as

`v = 5.45 * 10^3 m/s`

here we know that charge particle moves without any deviation

so we will have

`qvB = qE`

now magnetic field in this region is given as

`B = (E)/(v)`

`B = (2170)/(5.45 * 10^3)`

`B = 0.398 T`

Now another charge particle enters the region with different speed and experience the force upwards

`F = qE - qvB`

`1.54 * 10^(-9) = (4.10* 10^(-12))[2170 - v(0.398)]`

`375.6 = 2170 - v(0.398)`

`v = 4.51 * 10^3 m/s`