Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A conducting rod of length l = 0.5 m is moving with velocity of 

A conducting rod of length l = 0.5 m is moving with velocity of 3 ms^{-1}  in a constant magnetic field f strength  b = 5T .Resistance of rod is 10 \Omega  . find the power dissipited in rod 

Option: 1

5.625 W 


Option: 2

5.625 mW 


Option: 3

1.125 W 


Option: 4

1.125 mW 


Answers (1)

best_answer

As we have learned

Power dissipated in moving the conductor -

P=F\cdot v=\left ( \frac{B^{2}vl^{2}}{R} \right )\cdot v

P= \frac{B^{2}l^{2}v^{2}}{R}

-

 

 Force experienced by rod in magnetic field is 

F_m= \frac{B^2vl^2}{R}

Power dissipitated P = F_m \cdot v

P = \left ( \frac{B^2vl^2}{R} \right )\cdot v

P = \frac{B^2v^2l^2}{R}

P = \frac{(5)^2(3)^2(0.5)^2}{10}

P = 5.625 W

 

 

 

 

 

Posted by

qnaprep

View full answer

NEET 2024 Most scoring concepts

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

Similar Questions

Trending Articles/News

anam-khan

Punjab’s new 2-year MBBS bond, fee hike ‘exploitation’, ‘unfair’; interns, doctors shut OPDs in protest
anam-khan

NEET UG 2025 counselling likely to begin on July 1; AIQ seat allocation, court hearing updates
anam-khan

NEET UG: Delhi HC seeks NTA's response on plea alleging defective biometric system at centre

Latest Question