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Magnetism

Blazing goIITian

Joined:	28 Sep 2007
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18 Dec 2007 11:20:07 IST

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fiitjee aits pt2 EMI

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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)

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Comments (2)

Lokesh $ardana

Blazing goIITian

Joined: 28 Sep 2007

Posts: 637

18 Dec 2007 11:22:15 IST

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click image and save it for clearity.........

pardesi .svk

Scorching goIITian

Joined: 7 Dec 2007

Posts: 219

18 Dec 2007 11:58:46 IST

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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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