Question

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=?

Answer 1

@Mashy

Answer 2

do u know the expression for magnetic moment? :P

Answer 3

\[\Huge M=N I A\]

Answer 4

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 5

=IA

Answer 6

ok.. good boy :P.. Well.. here.. its gonna be a lil complicated ..u need to do an integral!

Answer 7

integration wagera karna hai shayad jo nahi aata

Answer 8

^^

Answer 9

lemme thinkaa!

Answer 10

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)

Answer 11

ahh!.. that ain't gonna work!

Answer 12

:/
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?

Answer 13

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!..

Answer 14

do u have the solution?!

is the answer \[M = \frac{wR^2q}{8}\]

Answer 15

umm no..

Answer 16

is it

\[M = \frac{\pi^4wR^2q}{}\]

Answer 17

divided by 2 i mean!

Answer 18

umm no

Answer 19

whats the answer?! :P

Answer 20

\[\Huge M=\frac{q \omega R^2 }{5}\]

Answer 21

lemme try again!

Answer 22

okay :|

Answer 23

i dunno.. this would involve a double integral i think.. seriously which book are you referring?!

Answer 24

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!

Answer 25

what on earth is this? :|
and dunno..and DC pande's question I guess..similar question in HC verma

Answer 26

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
\[ T = \frac{2 \pi }{\omega } \]

Answer 27

ans \( \sigma = Q/4\pi R^2 \)

Answer 28

*and

Answer 29

i already posted the answer!

Answer 30

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.