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gOlU g3n|[0]uS Grade: Upto college level
        




A non conducting ring of radius r has a charge Q uniformily distriubuted over it. a magnetic field perpendicular to the plane of the ring changes at the rate dB/dT. the torque experienced by the ring = ?? . will there be any difference if the ring is conducting?whyor why not?



 

6 years ago

Answers : (1)

AJIT AskiitiansExpert-IITD
68 Points
										

Dear golu ,


Case I : Apply following equation -   ∫line E.dl = - dØ/dt   i.e line integral of electric field over a surface is the rate change of flux through that surface. note that bold letters are vectors.


  now , varying magnetic field produces electric field , direction given by right hand thumb rule. we can see that electric field is in circular direction along the ring , assume it to be E (magnitude) . it will be constant as ring is symmetrical.


Therefore , E.2∏r  = ∏r2  * dB/dt  ;  E =  (1/2)*dB/dt.  now we can easily see torque = EQr  = (1/2)*dB/dt*Qr.


 Case II: in first case there was no conduction so ring moved with charges but in this case charges shall move inside the ring because ring allows conduction . so a current will flow and ther will not b any torque on ring.


Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.


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6 years ago
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