Get Answers to all your Questions

header-bg qa

A beam of cathode rays is subjected to crossed electric (E) and magnetic field (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)

Option: 1

\frac{\text{E}^{2}}{2\text{VB}^{2}}


Option: 2

\frac{\text{B}^{2}}{2\text{VE}^{2}}


Option: 3

\frac{2\text{VB}^{2}}{\text{E}^{2}}


Option: 4

\frac{2\text{VE}^{2}}{\text{B}^{2}}


Answers (1)

Its the electron beam is not deflected then,
\begin{aligned} & \mathrm{F}_{\mathrm{m}}=\mathrm{Ee} \\ & \mathrm{Bev}=\mathrm{Ee} \\ & \mathrm{v}=\frac{\mathrm{E}}{\mathrm{B}} \end{aligned}

According to the law of conservation of energy
\begin{aligned} & \frac{1}{2} \mathrm{mv}^2=\mathrm{eV} \\ & \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \\ & \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=\frac{\mathrm{E}}{\mathrm{B}} \\ & \frac{\mathrm{e}}{\mathrm{m}}=\frac{\mathrm{E}^2}{2 \mathrm{VB}^2} \end{aligned}

Posted by

Sumit Saini

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks