Question

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 (Where $V$ is the potential difference between cathode and anode)

• A. $\frac {B^2}{2VE^2}$
• B. $\frac {2VB^2}{E^2}$
• C. $\frac {2VE^2}{B^2}$
• D. $\frac {E^2}{2VB^2}$

Answer 1

Answer: (D)

Its the electron beam is not deflected then,
$ \begin{array}{l} F _{ m }= Ee \\ Bev = Ee \\ v =\frac{ E }{ B } \end{array} $
According to the law of conservation of energy
$ \begin{array}{l} \frac{1}{2} mv ^{2}= eV \\ v =\sqrt{\frac{2 eV }{ m }} \\ \sqrt{\frac{2 eV }{ m }}=\frac{E}{ B } \\ \frac{ e }{ m }=\frac{ E ^{2}}{2 VB ^{2}} \end{array} $