Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need clarity, kindly explain! - Magnetic Effects of Current and Magnetism - NEET-2

A bar magnet having a magnetic moment of 2\times 10^{4}JT^{-1} is free to rotate in a horizontal plane. A horizontal magnetic field B=6\times 10^{-4}T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60^{o} from the field is

  • Option 1)

    12J

  • Option 2)

    6J

  • Option 3)

    2J

  • Option 4)

    0.6J

 

Answers (1)

best_answer

 

Work done by current carrying coil -

W=MB(1-cos heta )

-

 

 work done = u_{f}-u_{i}

= -MB .\cos 60^{\circ}-\left ( -MB \cos 0^{\circ} \right )

= \frac{MB}{2}= \frac{2\times 10^{4}\times 6\times 10^{-4}}{2}J

= 6J

 


Option 1)

12J

Incorrect Option

Option 2)

6J

Correct Option

Option 3)

2J

Incorrect Option

Option 4)

0.6J

Incorrect Option

Posted by

Plabita

View full answer

Similar Questions