11
rkrish
·2009-02-09 22:51:14
Angular Momentum = L (let) = (2/3) mr2. ω
Magnetic Monent = M (let)
For any regular body with uniform mass & charge distribution rotating with ω :
M/L = Q/(2m) m : mass of body.
So,for this case,
M = Q((2/3)mr2. ω) / (2m) = (Q/3) . r2 ω
33
Abhishek Priyam
·2009-02-09 22:52:51
u can do this by integrations... by assuming rings...
or M/L=q/2m
M is mag mom... L is ang momentum about COM... q is charge.. m is mass...
1
Philip Calvert
·2009-02-09 22:54:46
@ rkrish
arre that's why i am asking the question
and u are underlining i don't know what
i asked for proof and this is not it
give me a proof for hollow sphere
@ abhi can u give me the proof
i have been tru all this b4 and i am asking bcozt i had problem with integration of those "rings" why don't u help me with that part
1
Philip Calvert
·2009-02-11 03:07:50
anyone ....
i have problems in integrating this...
ok can someone prove the thing for any rigid body in general ??(if it is true for all that is)
33
Abhishek Priyam
·2009-02-11 05:19:54

Charge on ring.. is Qsinθdθ
2
i=Qsinθdθω/2π
2
idA=Qsinθdθ ωπ(Rsinθ)2
2 2Ï€
M=∫idA
=QR2ω 2 0∫π/2sin3θdθ
4
QR2ω/3
1
Philip Calvert
·2009-02-11 08:30:22
ok thanks i tried it three times but with no result somehow i was getting sin^4 there..
it must be ok .. anyway it was just curiosity u know..