Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A conducting rod of length of l = 2 m slides at constant velocity 'v = 5 m/s' on two parallel conducting rails, placed in a uniform and constant magnetic field B = 0.5T perpendicu

A conducting rod of length of l = 2 m slides at constant velocity 'v = 5 m/s' on two parallel conducting rails, placed in a uniform and constant magnetic field B = 0.5T perpendicular to the plane of the rails as shown in the figure. Find resistance R which is connected between the two ends of the rail. If the electric power dissipated in the resistor is 20 W

 

 

Option: 1

1 \Omega


Option: 2

1.25 \ \Omega


Option: 3

1.5 \ \Omega


Option: 4

1.75 \ \Omega


Answers (1)

best_answer

 

 

\begin{array}{l}{P=\frac{B^{2} l^{2} v^{2}}{R}}\\ \\ {R=\frac{B^{2} l^{2} v^{2}}{P}} \\ \\ {R=\frac{(0.5)^{2} *(2)^{2} *(5)^{2}}{20}}\\ \\ {R=1.25 \Omega}\end{array}

Posted by

Ritika Jonwal

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 2025, NEET, CUET UG Dates Live: NTA exam calendar awaited at nta.ac.in
anam-khan

JEE Main 2025 session 1 exam in the last week of January; registration details
anam-khan

Over 656 UPSC, JEE, NEET, other coaching aspirants refunded more than Rs 1 crore: Govt

Latest Question