If a proton were released in the electric field above, what direction would it move?
1 answer:
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Answer:
here
Explanation:
Answer:
The beam used is a negatively charged electron beam with a velocity of
v = E / B
Explanation:
After reading this long statement we can extract the data to work on the problem.
* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.
* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
q E = qv B
v = E / B
this configuration is called speed selector
They ask us what type of beam was used.
The beam used is a negatively charged electron beam with a velocity of v = E / B