A beam of cathode rays is subjected to crossed electric (E ) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by

Options

A. `(B^(2))/(2VE^(2))`
B. `(2VB^(2))/(E^(2))`
C. `(2VE^(2))/(B^(2))`
D. `(E^(2))/(2VB^(2))`

Answer

D

Explanation

As the electron bea is not deflected, then
`F_(m)=F_(e) implies Bev=Ee`
or `v=(E)/(B)` . . . (i)
As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode =eV. also kinetic energy of the electron at anode `=(1)/(2)mv^(2)`. according to law of conservation of energy , `(1)/(2)mv^(2)=eV`
or `vsqrt((2eV)/(m))` . . (ii)
From eqs. (i) and (ii), we have
`sqrt((2eV)/(m))=(E)/(B)implies (e)/(m)=(E^(2))/(2VB^(2))`.