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    Hi Everyone,

    I am a bit stuck on this question from a past AQA paper, unfortunately I can't find a mark scheme to help!The question is about the coil and magnets of a simple motor. The coil has 24 turns and carries a current of 4.3 A. The coil is made from copper wire of diameter 0.36 mm. The uniform magnetic field produced by the magnets has a flux density of 0.17 T.The work done to rotate a coil 90 degrees is 0.084 J.

    5 (b) (ii) The conversion of work done to the kinetic energy of the coil is 83% efficient. Assume that only the long sides of the coil share the kinetic energy.The coil starts from rest. Show that the speed of a side of the coil after it has rotated through 90 degrees is approximately 5.2 m s-1.

    density of copper = 8900 kg m-3

    My attempt so far:

    0.084 * 0.83=0.69 J

    KE=1/2 mv2
    =0.69 J

    v=sqroot(2KE/m)

    m=p x A x L

    A=(pi) x 0.000182 =1.02 x 10-7 m2

    I have tried various values for m because I'm not sure if I need to consider m as the mass of 24 lots of wire on both sides of the coil or just one side? but nothing gives the answer as 5....Anybody able to help? I have attached the whole question paper as I couldn't do the question on its own, but it is question 5 if anybody needs to look at it. Thanks!
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  1. File Type: pdf AQA-PHB5-W-QP-JUN09.PDF (302.5 KB, 63 views)
 
 
 
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