Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two parallel metallic plates are separated by a distance , and a magnetic field of induction <img alt=

Two parallel metallic plates are separated by a distance \mathrm{d}, and a magnetic field of induction \mathrm{B} acts between the plates as shown in the figure. The work function of the emitter is \phi and light of frequency \nu is incident on the plate. For what value of \mathrm{B} will the photo current between the plates be equal to zero? (where V is the velocity)

Option: 1

\frac{m V}{e d}


Option: 2

\frac{2 m V}{e d}


Option: 3

\frac{2 m V}{2 e d}


Option: 4

\frac{4 m V}{e d}


Answers (1)

best_answer

Suppose that the incident radiation causes photo-emission in all directions, then those electrons that are emitted
almost parallel to the plate, just manage to reach the other plate when the magnetic field B is at its minimum value so as to cut off the current. We can write,

\frac{1}{2} {mV}^2+\mathrm{e} \phi=\mathrm{h\nu}                                    ...(1)

\text { and, } \quad B e V=\frac{m^2}{d / 2}                               ...(2)

The value of B satisfies equation (2) and therefore,

\mathrm{B}=\frac{2 m V}{e d}, \text { where} \ V=\sqrt{\frac{2}{m}\{h\nu-e \phi\}}

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question