as shown in figure, potential difference V exists between two parallel plates (a,b) that have a small hole. Uniform electric field `vecE` and magnetic flux density `vecB` exist in the region above the plates, and are perpendicular to each other. The direction of `vecE` is parallel to this page and plate b. The direction of `vecB` is perpendicular to this page, from back to front. A positively charged particle of charge q and mass m, initially at rest in the hole of plate a, is accelerated by potential difference V so that it enter the region above the plate perpendicularly to `vecE` and `vecB` and travels straight through the region``
what is the ratio of the magnitude E of the electric field to the magnitude B of the magnetic flux density E/B?
1 Answer | Add Yours
The particle is accelerated from rest by the potential difference U until it reaches a speed v when it exits the space between plates (a,b).
Kinetic energy when particle leaves the plates = Electric potential energy given between plates
`v = sqrt((2*q*U)/m)`
Now, in the region where there exist both fields E and B the Lorentz force on the charge q is (v is perpendicular to both E and B):
`F = q*(E +vB)`
Condition of non-deflection is
`F =0` or equivalent `E = -vB` which means that in absolute
value the ratio E/B is
`E/B = v = sqrt((2*q*U)/m)`
In practice this experiment is used to determine the rapport (q/m) for the electron or any given particle of charge q and mass m, by measuring the fields E and B under zero deflection conditions.
We’ve answered 320,019 questions. We can answer yours, too.Ask a question