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?

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aristocles aristocles · Answer 1 · 2019-07-27T12:52:22+02:00

Answer:

[tex]v = 4.51 \times 10^3 m/s[/tex]

Explanation:

electric field = 2170 N/C

now the speed of the charge particle is given as

[tex]v = 5.45 \times 10^3 m/s[/tex]

here we know that charge particle moves without any deviation

so we will have

[tex]qvB = qE[/tex]

now magnetic field in this region is given as

[tex]B = \frac{E}{v}[/tex]

[tex]B = \frac{2170}{5.45 \times 10^3}[/tex]

[tex]B = 0.398 T[/tex]

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

[tex]F = qE - qvB[/tex]

[tex]1.54 \times 10^{-9} = (4.10\times 10^{-12})[2170 - v(0.398)][/tex]

[tex]375.6 = 2170 - v(0.398)[/tex]

[tex]v = 4.51 \times 10^3 m/s[/tex]