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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Jan 2008 22:37:58 IST
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Consider a uniform disc rotating with an angular velocity w. Consider a point P at a distance d from the center of the disc, such that d>>>radius of the disc. The disc has a charge density of  . Find the magnetic field at the point P.
Rates assured for the correct method. I will post the solution in case no one answers this.
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Will nip in at times to solve problems :)
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30 Jan 2008 22:47:25 IST
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similar type of qn. is in DC pandey...here the trick is
dq = (sigma).(2.pie.d.dr)
dq/dt = i = sigma.omega.d.dr
now use similar integration....
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30 Jan 2008 23:51:34 IST
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Consider a ring of radius r n thickness dr in d disc. dA = 2  r.dr charge in dA, dq = 2 r  dr f=w/2pi hence current, di = dq*f = 2 r dr * w/2 = wr dr Magnetic field due to small ring at P is, dB =  /2 * w /d 2 * r 2dr Hence, net magnetic field (assuming d<<<R), B = [ ] [ dB = w R 3/6d 2 |
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30 Jan 2008 23:52:54 IST
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Re:Magnetism challenge :
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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31 Jan 2008 10:32:24 IST
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consider a small elementary ring or radius r and thickness dr from the centre of the disc as shown in the fig...
the magnetic field due to this ring is given as
dB = u0wr2p 2 pi r dr / 4pi (d2+r2)3/2 where p = sigma
integrating for the limits 0 to R
B = u0wp/2 0 R r3dr/(d2+r2)3/2
since d>>r
B = u0wp/2d3 0R r3dr
u'll get
B = u0wpR4/8d3
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~ANSHUMAN
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31 Jan 2008 10:39:47 IST
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take an elament a concentric ring of width dx area 2pie x dx.apply biot savarts law for dB and iontegrate
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31 Jan 2008 19:07:15 IST
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One thing - the radius of the disc is not mentioned anywhere in the problem
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Will nip in at times to solve problems :)
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31 Jan 2008 21:22:59 IST
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edited...
post truncated...
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~ANSHUMAN
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