Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity

4.19 An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field  is transverse to its initial velocity

Answers (1)

best_answer

 (a) The electron has been accelerated through a potential difference of 2.0 kV. 

Therefore K.E of electron = 1.6\times10^{-19}\times2000=3.2\times10^{-16} J

\\\frac{1}{2}mv^{2}=3.2\times 10^{-16}\\ \\v=\sqrt{\frac{2\times 3.2\times 10^{-16}}{9.1\times 10^{-31}}}\\ v=2.67\times 10^{7}\ ms^{-1}

 Since the electron initially has velocity perpendicular to the magnetic field it will move in a circular path.

The magnetic field acts as a centripetal force. Therefore,

\\\frac{mv^{2}}{r}=evB\\ r=\frac{mv}{eB}\\ r=\frac{9.1\times 10^{-31}\times 2.67\times 10^{7}}{1.6\times 10^{-19}\times 0.15}=1.01mm 

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board Exams 2025: Guidelines issued for Class 10, 12 practicals
anam-khan

Board Exam Date Sheet 2025 Live: ICSE, ISC, CBSE, MP time table out; BSEB routine soon
anam-khan

Board exam dates 2025: CBSE, CISCE, UP, MP, Maharashtra, TN time table out; state-wise updates

Latest Question