A gyromagentic ratio for a cylinder

[TheSodesa]

Homework Statement

In classical physics, a system's magnetic moment can be written like so: [tex]\mu = g\frac{Q}{2M}L,[/tex] where ##Q## is the total charge, ##M## is the total mass of the system and ##L## the angular momentum.

a) Show, that for a cylinder (##I = \frac{1}{2}MR^2##) spinning around its axis of symmetry, and that has an even charge distribution on its surface, the value of ##g = 2##. (Think! Don't integrate.)
correct answer: 2

b) Calculate ##g## for a sphere of radius ##R## whose charge is concentrated on it's equator.
(Again, think. Do not integrate.)

correct answer: ##\frac{5}{2}##

Homework Equations

Moment of inertia for a cylinder:
\begin{equation}
I_c = \frac{1}{2} MR^2
\end{equation}
Moment of inertia for a sphere:
\begin{equation}
I_s = \frac{2}{5} MR^2
\end{equation}
The magnectic moment:
\begin{equation}
\mu = g\frac{Q}{2M}L
\end{equation}

The Attempt at a Solution

[/B]
Alright, let's start out by solving for ##g## in ##(3)##:
\begin{equation}
g = \frac{2M\mu}{QL}
\end{equation}

For a cylinder this becomes:
\begin{equation}
g = \frac{2M\mu}{Q \frac{1}{2}MR^2} = \frac{4\mu}{QR^2}
= 2 \frac{2\mu}{QR^2},
\end{equation}
and for the sphere:
\begin{equation}
g = \frac{2M\mu}{Q \frac{2}{5}MR^2} = \frac{5}{2} \frac{2\mu}{QR^2}
\end{equation}

Now for a cylinder the surface charge can be written:
\begin{equation}
Q = \sigma A = \sigma (2\pi R h + 2\pi R^2),
\end{equation}
where ##\sigma## is the charge density, ##A## the surface area, ##R## the cylinder radius and ##h## its height.

For a sphere, whose charge is on its equator this is:
\begin{equation}
Q = \sigma C = \sigma (2\pi R).
\end{equation}

Plugging ##(7)## into ##(5)##:
\begin{equation}
g = \frac{2M\mu}{Q \frac{1}{2}MR^2} = 2\frac{2\mu}{ \sigma (2\pi R h + 2\pi R^2) R^2}
\end{equation}
Again, for the sphere ##(6)## this is:
\begin{equation}
g = \frac{5}{2} \frac{2\mu}{ \sigma (2 \pi R) R^2}
\end{equation}

This is where I'm stuck. It looks like, looking at the correct answers, that once we solve for ##g##, we should end up with the inverses of the coefficients in ##(1)## and ##(2)##, meaning
\begin{equation}
\frac{2\mu}{QR^2} = 1
\end{equation}
in both cases.

For a single particle rotating in a circle, ##\mu = \frac{q}{2m}L = \frac{q}{2m} mR^2 \omega = \frac{q \omega}{2} R^2 = \frac{qmRv}{2}##, but surely I can't use that to cancel out any ##R##s, since in ##(3)##, ##\mu## is the magnetic moment of the entire system. Also, ##\sigma## isn't going anywhere even if I do manage to cancel out the areas/circumferences.

In short, what am I missing here?

 

[haruspex]

In going from eqn 4 to eqn 5, you substituted the moment of inertia for L, but L is angular momentum. What equation relates the two?

 

[TheSodesa]

In going from eqn 4 to eqn 5, you substituted the moment of inertia for L, but L is angular momentum. What equation relates the two?

Whoopsie. ##L = I\omega##.

Then ##(5)## becomes:
\begin{equation}
g = \frac{2 M \mu}{Q \frac{1}{2}MR^2 \omega} = 2 \frac{2 \mu}{Q R^2 \omega} = 2 \frac{2 \mu}{Q R^2 \frac{v}{R}} = 2 \frac{2 \mu}{Q R v}
\end{equation}
Now if we are allowed to use ##\mu = \frac{1}{2} qmrv## (the ##R## below ##(11)## should not have been a capital one, for general purposes), then things do cancel out quite nicely at a distance ##R## from the axis of symmetry, except ##q \neq Q##, is it? I mean [tex] \mu = \frac{1}{2} qmrv [/tex] for single particles only, isn't it?

In any case, if we do plug it into the above expression: [tex]g = 2 \frac{2qmRv}{2QRv} = 2\frac{qm}{Q}.[/tex] Not quite what I was looking for, but close.

 

[haruspex]

##\mu = \frac{1}{2} qmrv## for single particles only, isn't it?

You can consider the charge as made of of single particles all with the same velocity and radius.
But what is m there? Isn't it just ##\mu = \frac{1}{2} qrv##?

 

[TheSodesa]

You can consider the charge as made of of single particles all with the same velocity and radius.
But what is m there? Isn't it just ##\mu = \frac{1}{2} qrv##?

Yeah, forgot to cancel out the ##m## from the numerator (or more like put it back in for some reason).

Ok, so now we have [tex]g = 2 \frac{2 \mu}{QRv} = 2 \frac{2 \sum_{i} \frac{q_i R v}{2} }{QRv} = 2 \frac{2 QRv}{2QRv} = 2[/tex]

And there we have it (I think).

Applying the same procedure to case b) (which we can do since all of the charge is on the equator):
[tex]g = \frac{5}{2} \frac{2 \mu}{QRv} = \frac{5}{2} \frac{2 \sum_{i} \frac{q_i R v}{2} }{QRv} = \frac{5}{2} \frac{2 QRv}{wQRv} = \frac{5}{2}.[/tex]

Boom shaka laka?

 

[haruspex]

Yeah, forgot to cancel out the ##m## from the numerator (or more like put it back in for some reason).

Ok, so now we have [tex]g = 2 \frac{2 \mu}{QRv} = 2 \frac{2 \sum_{i} \frac{q_i R v}{2} }{QRv} = 2 \frac{2 QRv}{2QRv} = 2[/tex]

And there we have it (I think).

Applying the same procedure to case b) (which we can do since all of the charge is on the equator):
[tex]g = \frac{5}{2} \frac{2 \mu}{QRv} = \frac{5}{2} \frac{2 \sum_{i} \frac{q_i R v}{2} }{QRv} = \frac{5}{2} \frac{2 QRv}{wQRv} = \frac{5}{2}.[/tex]

Boom shaka laka?

looks right.

 

[TheSodesa]

looks right.

Boom shaka laka it is, then.

Thank you very much.