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[Post New]30 Jan 2008 23:51:34 IST
    Subject: Re:Magnetism challenge :
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rhd92781 (686)

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Olaaa!! Perrrfect answer. 118  [166 rates]
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Consider a ring of radius r n thickness dr in d disc.
dA = 2r.dr
charge in dA, dq = 2rdr
f=w/2pi
hence current, di = dq*f = 2rdr * w/2 = wrdr
 
Magnetic field due to small ring at P is,
dB = /4 * 2r*(di)/d2
dB = /4 * 2r*(wrdr)/d2
dB = /2 * w/d2 * r2dr
Hence, net magnetic field (assuming d<<<R),
B = [ ][ dB
  
 = w/2d2           [ 0][ R] r2dr
=wR3/6d2
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