DLS
  • DLS
A solid conducting sphere of Radius R and total charge q rotates about its diametric axis with constant Angular speed omega,Magnetic moment of the sphere=?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
@Mashy
anonymous
  • anonymous
do u know the expression for magnetic moment? :P
DLS
  • DLS
\[\Huge M=N I A\]

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More answers

DLS
  • DLS
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
DLS
  • DLS
=IA
anonymous
  • anonymous
ok.. good boy :P.. Well.. here.. its gonna be a lil complicated ..u need to do an integral!
DLS
  • DLS
integration wagera karna hai shayad jo nahi aata
DLS
  • DLS
^^
anonymous
  • anonymous
lemme thinkaa!
anonymous
  • anonymous
u know how to find the current right?! imagine.. a small cross sectional ring on the sphere.. let dq be the charge on that ring.. then the current di = dq/(2*3.14/omega)
anonymous
  • anonymous
ahh!.. that ain't gonna work!
DLS
  • DLS
:/ Since the question says "Conducting sphere" what does that mean?charge will only reside on the outer surface?If yes,why so?what will be the difference in solving the question if it was not conducting?
anonymous
  • anonymous
the conducting spehre makes sure the charge is uniformly distributed over the outer surface.. supposedly making it simpler.. but i seriously don't understand how to!..
anonymous
  • anonymous
do u have the solution?! is the answer \[M = \frac{wR^2q}{8}\]
DLS
  • DLS
umm no..
anonymous
  • anonymous
is it \[M = \frac{\pi^4wR^2q}{}\]
anonymous
  • anonymous
divided by 2 i mean!
DLS
  • DLS
umm no
anonymous
  • anonymous
whats the answer?! :P
DLS
  • DLS
\[\Huge M=\frac{q \omega R^2 }{5}\]
anonymous
  • anonymous
lemme try again!
DLS
  • DLS
okay :|
anonymous
  • anonymous
i dunno.. this would involve a double integral i think.. seriously which book are you referring?!
anonymous
  • anonymous
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&ved=0CCoQFjAA&url=http%3A%2F%2Fwww.lmnoeng.com%2FTank%2FTankTime.htm&ei=GXIjUrOEHsbJrAeB8oFQ&usg=AFQjCNGXnWn9OkLvJ2hWN52xjmjYG9LI1g&sig2=So_-vSkus1E9TtX8-yIOGA&bvm=bv.51495398,d.bmk here's ur answer!
DLS
  • DLS
what on earth is this? :| and dunno..and DC pande's question I guess..similar question in HC verma
experimentX
  • experimentX
or \[ \int_0^{\pi } \pi (R \sin \theta)^2 \cdot \sigma ( R d\theta \cdot 2 \pi R \sin \theta ) \cdot \frac{1}{T } \] where \[ T = \frac{2 \pi }{\omega } \]
experimentX
  • experimentX
ans \( \sigma = Q/4\pi R^2 \)
experimentX
  • experimentX
*and
DLS
  • DLS
i already posted the answer!
experimentX
  • experimentX
the integral evaluates to \[ \frac 1 3 Q \omega R^2\] http://www.wolframalpha.com/input/?i=Integrate%5B%5C%5BPi%5D+%28R+Sin%5Bx%5D%29%5E2+Q%2F%284+%5C%5BPi%5D+R%5E2%29+%28R%5E2+Sin%5B++++++x%5D+%5C%5BOmega%5D%29%2C+%7Bx%2C+0%2C+Pi%7D%5D check it out here. http://lynx.uio.no/trine/fys3510/oppg_3_20_eng.pdf \[ \Huge M=\frac{Q \omega R^2 }{5} \] is magnetic moment of a uniformly charged (non conducting) sphere. For conducting sphere, the charges only lie on the surface. possibly you made error on your question or your answer sheet is wrong. or I misunderstood the above literature on that paper.

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