Answer:
[tex]r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}[/tex]
Explanation:
Let m and e are the mass and charge of an electron. It is accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. Let v is the velocity of the electron. It can be calculated as :
[tex]\dfrac{1}{2}mv^2=eV[/tex]
[tex]v=\sqrt{\dfrac{2eV}{m}}[/tex]
When the electron enters the magnetic field, the centripetal force is balanced by the magnetic force as :
[tex]\dfrac{mv^2}{r}=evB[/tex]
[tex]r=\dfrac{mv}{eB}[/tex]
or
[tex]r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}[/tex]
So, the radius of the resulting electron trajectory is [tex]\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}[/tex]. Hence, this is the required solution.
