- #1
greg_rack
Gold Member
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Considering the device above, which uses electric and magnetic fields placed properly to avoid charged moving particles with velocities different from the ratio ##\frac{E}{B}## to exit, getting deflected upwards or downwards. All that is easily demonstrable by equalling the forces acting on the particle: ##qE=qvB##.
Now, my question is: what happens to particles with ##v\neq \frac{E}{B}##?
What I found online if ##v>\frac{E}{B}## Lorentz's force will be greater(since it's proportional to speed) than the electric one, so the particle will get deflected upwards; otherwise, if ##v<\frac{E}{B}##, the electric force will "win" so the particle will move downwards... but this explanation doesn't satisfy me:
what if the electric field is much greater than the magnetic one, and even if ##v>\frac{E}{B}##, ##E>vB## and vice-versa?
Shouldn't we know the magnitude of both fields to predict the behavior of particles with ##v\neq \frac{E}{B}##?