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ashgirl (1429)

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a bar magnet  falls through a metal ring.will its acceleration be equal to "g"? if  the ring is cut somewhere, what would be the answer?

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Greatdreams (3083)

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The value of acceleration will be < g because according to Lenz's law the direction of any magnetic induction is such as to oppose the the cause of the effect.Here the cause is the free fall of magnet.So the induced current will oppose it and accn will be less than g.

Ring is cut means ring is broken somewhere???

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ashgirl (1429)

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well...actually thats where i'm having doubt....its not said where it has been cut....but yeah it means that it is broken somewhere.

Remember Cedric. Remember, if the time should come when you have to make a choice between what is right and what is easy, remember what happened to a boy who was good, and kind, and brave, because he strayed across the path of Lord Voldemort. Remember Cedric Diggory.


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feynmann (2083)

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( Editing a little bit )
 
Then it is g as no current is present.
 
Because when the ring is cut i= 0 ; and   dF = i dL X B = 0
 
Hence no  force on the ring . So there is no opposing force on the magnet ( 3 rd law ) .
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iitian_aakash (41)

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YAAH WHEN RING IS BROKEN THAT MEANS CIRCUIT IS NOT COMPLETE THAT MEANS THERE IS NO CURRENT IN THE LOOP HENCE NO ITS MAGNETIC FLUX THAT WOULD OPPOSE THE MOTION OF MAGNET. HENCE A=G
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Aatish (2293)

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very correct.......well done guys.....

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iitian_aakash (41)

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WELL IF YOUR DOUBT IS OVER CAN YOU PLEASE EXPLAIN ME IF THIS LOOP IS NT BROKEN ACCC IS LESS THAN G. WHY?
SEE MAGNETIC FIELD LINES ARE UPWARD AND VELOCITY OF MAGNET DOWNWARD THEN HOW THIS FORCE IS ACTING
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Aatish (2293)

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Hey aakash you will have to take Lenz's Law into consideration........

I hope you know that........

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astatine19 (1108)

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when the magnet is falling through the loop, flux is increasing through the loop. By lenzs law, the loop will oppose the magnet and decrease its accn. Hence another question-in the case of the broken loop, will there be a momentary current or will it 'always' be zero?

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chimanshu_007 (11344)

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it will remain g according to me.....

wen the ring is cut....induced emf will indeed be there.....but as the ring is cut or broken , no induced current can flow , so no opposing force...(i mean no lenz law here) , therefore g will remain = to g


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Nithy (400)

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yes i think so too like himanshu..

open circuit so no induced current..

BUT what abt eddy currents???? will it b valid? won't that continue opposing....
metal ring equivalent to bulk solid conductor...

???? rite?

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ashgirl (1429)

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but change in magnetic flux takes place in open circuit too..........

Remember Cedric. Remember, if the time should come when you have to make a choice between what is right and what is easy, remember what happened to a boy who was good, and kind, and brave, because he strayed across the path of Lord Voldemort. Remember Cedric Diggory.


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Goblet of Fire, Chapter 37, Page 724
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astatine19 (1108)

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wht im trying to say is that this case is similar to that of a capacitor connected across a battery. In that case, a momentary current flows, charging up the capacitor. In this case, wont the broken ends of the ring facing each other act as a capacitor?
i feel tht they will, nd during that moment when current flows, a will be less than g

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Nithy (400)

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can't that b explained by eddy currents like i said earlier?

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