Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field B. The work done by the fiel

A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field B. The work done by the field when the particle completes one full circle is.

Option: 1

\mathrm{\left(\frac{m v^2}{R}\right) 2 \pi R}


Option: 2

zero
 


Option: 3

\mathrm{2 \pi R B Q}


Option: 4

\mathrm{2 \pi R B v Q}


Answers (1)

best_answer

When a charged particle describes a circular path, the necessary centripetal force is provided by the magnetic force F. Since the velocity v is always tangential, vectors F and v are perpendicular to each other \left(\theta=90^{\circ}\right).


\mathrm{Power =\mathbf{F} \cdot \mathbf{v}=F v \cos \theta=F v \cos 90^{\circ}=0}. Since power consumed is zero, the work done is zero.

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: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 city slip soon; admit card, exam dates, here's how to prepare
anam-khan

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

Latest Question