A proton moves perpendicularly to a magnetic field that has a magnitude of 4.64×10^−2 T.
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A proton moves perpendicularly to a magnetic field that has a magnitude of 4.64×10^−2 T.

[From: ] [author: ] [Date: 12-04-10] [Hit: ]
B is magnetic field in teslas, q is charge in coulombs,v = 5.34 x 10^-14N / ((.0464T)(1.v = 7,......
The charge on a proton is 1.60×10^−19C.
What is the speed of the particle if the magnitude of the magnetic force on it is 5.34×10^−14N?
Answer in units of m/s

-
F = Bqv sin theta (no angle is mentioned here so you can ignore the sin theta part)
where F is force in newtons, B is magnetic field in teslas, q is charge in coulombs, and v is velocity

Rearrange solving for v:
v = F/Bq
v = 5.34 x 10^-14N / ((.0464T)(1.60 x 10^-19 C))

v = 7,192,887.93 m/s, or 7.19 x 10^6 m/s.

Just as a side note, make sure your answer for a velocity doesn't exceed 3 x 10^8, that would be impossible.
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