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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: fiitjee aits pt2 EMI
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[Post New]18 Dec 2007 11:20:07 IST
    Subject: fiitjee aits pt2 EMI
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lokeshsardana (685)

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ABCD, a wire frame of negligible resistance, is placed in uniform magnetic field (B) acting perpendicular to the plane of wire frame.
It is rotated about point O in the same plane. The direction of angular velocity w is perpendicular and into the plane. If a resistance R is connected between A and D. Then find the current flowing through resistance.
 
Answer:   (3Bwl2 )/(2R)
 

LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.


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Harry Potter and the Prisoner of Azkaban movie


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[Post New]18 Dec 2007 11:22:15 IST
    Subject: fiitjee aits pt2 EMI
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lokeshsardana (685)

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LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.


Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie


There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein

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[Post New]18 Dec 2007 11:58:46 IST
    Subject: Re:fiitjee aits pt2 EMI
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pardesi (541)

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according to me the solution provided was useless.
well here is one
the line integration of the force per unit charge through the wire loop is the current times resistance.
clearly at any momen the line integration along a path BC is 0.
also for AB.
the force is df=Bwxdx(just take the component of spped along the wire since we need line integration along wire)
so f=2Bwl^{2} as length is 2;
similarly  for frame DC the line integral is Bwl^{2}/2
but oppositely directed so
net 3Bwl^{2}/2
so current I=3Bwl^{2}/2R
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