In the given arrangement, the space between a pair of co-axial cylindrical conductors is evacuated. The outer cylinder, called anode, may be given a positive potential V relative to the inner cylinder.
A static homogeneous magnetic field `vec(B)` parallel to the cylinder axis, directed out of plane of figure is aslo present. induced charges in the conductors are neglected.
We study the dynamics of electrons with rest mass m and charge e. The electrons are released at the surface of inner cylinder.
Consider the following two cases

Case 1: Firstly the potential V is turned on, but `vec(B) = 0`. An electron with negligible velocity is ejected at the surface of inner cylinder. It is found to hit the anode.
Case 2: Now V = 0 but `vec(B)` is present. An electron starts out with an initial velocity `vec(v_(0))` in radial direction. For magnetic field larger than critical value `B_(c)` the electron will not reach the anode.
The critical magnetic field `B_(c)` is given by

Options

A. `(2mv_(0))/(eb)`
B. `(mv_(0))/(eb)`
C. `(2bmv_(0))/(e(b^(2)-a^(2)b)`
D. `(2amv_(0))/(e(b^(2)-a^(2)b)`

Answer

C

Explanation

From the fig., we see that in critical core
the radius R of the circle satisfies
`sqrt(a^(2)+R^(2))=b-R`
or `a^(2)+R^(2)=b^(2)+R^(2)-2bR`
or `R=((b^(2)-a^(2))/(2B))`...(i)
The radius R of the circular path is determined by equating centripetal force and Lorentz force
`eB_(c)v_(0)=(mv_(0)^(2))/(R) implies B_(c)=(mv_(0))/(eR)`
Using equating (i)
`B_(c)=(2mbv_(0))/(e(b^(2)-a^(2)))`
Hence choice (c) is correct.