Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An electron is accelerated to a high speed down the axis of a cathode ray tube by the application of a potential difference of <img alt="\mathrm{~V}" src="https://entrancecorner.oncodecogs.com/gif.

An electron is accelerated to a high speed down the axis of a cathode ray tube by the application of a potential difference of \mathrm{~V} volts between the cathode and the anode. The particle then passes through a uniform transverse magnetic field in which it experiences a force \mathrm{F}. If the potential difference between the anode and the cathode is increased to 2 \mathrm{~V}, the electron will now experience a force
 

Option: 1

\mathrm{F / \sqrt{2}}


Option: 2

\mathrm{F / 2}


Option: 3

\mathrm{\sqrt{2} F}


Option: 4

\mathrm{2 F}


Answers (1)

best_answer

The velocity when the potential difference is V is

\mathrm{v=\sqrt{\frac{2 e V}{m}}}

and force \mathrm{F=e v B}
When the potential difference is doubled, i.e.\mathrm{ V^{\prime}= 2 \mathrm{~V}} the velocity is
\mathrm{ v^{\prime}=\sqrt{\frac{2 e V^{\prime}}{m}}=\sqrt{\frac{2 e \times 2 V}{m}}=\sqrt{2} v }
\mathrm{ \therefore Force ~F^{\prime}=e v^{\prime} B=\sqrt{2} e v B=\sqrt{2} F } . Hence the correct choice is (c).

Posted by

avinash.dongre

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 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question