# A gyromagentic ratio for a cylinder

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1. Nov 27, 2016

### TheSodesa

1. The problem statement, all variables and given/known data
In classical physics, a system's magnetic moment can be written like so: $$\mu = g\frac{Q}{2M}L,$$ 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.)

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

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

2. Relevant equations
Moment of inertia for a cylinder:

I_c = \frac{1}{2} MR^2

Moment of inertia for a sphere:

I_s = \frac{2}{5} MR^2

The magnectic moment:

\mu = g\frac{Q}{2M}L

3. The attempt at a solution

Alright, let's start out by solving for $g$ in $(3)$:

g = \frac{2M\mu}{QL}

For a cylinder this becomes:

g = \frac{2M\mu}{Q \frac{1}{2}MR^2} = \frac{4\mu}{QR^2}
= 2 \frac{2\mu}{QR^2},

and for the sphere:

g = \frac{2M\mu}{Q \frac{2}{5}MR^2} = \frac{5}{2} \frac{2\mu}{QR^2}

Now for a cylinder the surface charge can be written:

Q = \sigma A = \sigma (2\pi R h + 2\pi R^2),

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:

Q = \sigma C = \sigma (2\pi R).

Plugging $(7)$ into $(5)$:

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}

Again, for the sphere $(6)$ this is:

g = \frac{5}{2} \frac{2\mu}{ \sigma (2 \pi R) R^2}

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

\frac{2\mu}{QR^2} = 1

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 cant 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?

Last edited: Nov 27, 2016
2. Nov 27, 2016

### 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?

3. Nov 27, 2016

### TheSodesa

Whoopsie. $L = I\omega$.

Then $(5)$ becomes:

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}

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

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

4. Nov 27, 2016

### haruspex

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 $\mu = \frac{1}{2} qrv$?

5. Nov 27, 2016

### TheSodesa

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 $$g = 2 \frac{2 \mu}{QRv} = 2 \frac{2 \sum_{i} \frac{q_i R v}{2} }{QRv} = 2 \frac{2 QRv}{2QRv} = 2$$

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):
$$g = \frac{5}{2} \frac{2 \mu}{QRv} = \frac{5}{2} \frac{2 \sum_{i} \frac{q_i R v}{2} }{QRv} = \frac{5}{2} \frac{2 QRv}{wQRv} = \frac{5}{2}.$$

Boom shaka laka?

6. Nov 27, 2016

### haruspex

looks right.

7. Nov 27, 2016

### TheSodesa

Boom shaka laka it is, then.

Thank you very much.