1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A gyromagentic ratio for a cylinder

  1. Nov 27, 2016 #1
    1. The problem statement, all variables and given/known data
    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}##

    2. Relevant 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}

    3. The attempt at a solution

    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 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. jcsd
  3. Nov 27, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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?
     
  4. Nov 27, 2016 #3
    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.
     
  5. Nov 27, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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##?
     
  6. Nov 27, 2016 #5
    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?
     
  7. Nov 27, 2016 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    looks right.
     
  8. Nov 27, 2016 #7
    Boom shaka laka it is, then. :cool:

    Thank you very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A gyromagentic ratio for a cylinder
  1. Velocity Ratio (Replies: 19)

  2. Ratio of power (Replies: 1)

  3. Ratio of momentums (Replies: 8)

  4. Ratio Question (Replies: 1)

Loading...