Asked by picm
Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude 2607 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.81 103 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 +3.72 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.80 10-9 N, pointing directly upward. What is the speed of this particle?
Answers
Answered by
Elena
Motion of the particle without deflection =>
F(el) = F mag)
qE =qvB
B=E/v=2607/6.81•10³=0.38 T
For the second particle
F=qE-qv₁B
v₁=(F+qE)/qB=
=1.8•10⁻⁹•3.72•10⁻¹²•2607}/
3.72•10⁻¹²•0.38 =
=8.13•10³ m/s
F(el) = F mag)
qE =qvB
B=E/v=2607/6.81•10³=0.38 T
For the second particle
F=qE-qv₁B
v₁=(F+qE)/qB=
=1.8•10⁻⁹•3.72•10⁻¹²•2607}/
3.72•10⁻¹²•0.38 =
=8.13•10³ m/s
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