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Help with Griffiths Electrodynamics

  1. Sep 6, 2004 #1
    I've been brushing up on electrodynamics before I start grad school when I encountered problem 5.42 in Griffith's Electrodynamics. I can get everything correct except the coefficient to work out. Any one know where I can find a solution to this problem?
    ----------------------------------------
    "Not everyone has the book"

    I never thought of that. Thanks for the advice. Well,

    The problem. A spinning spherical shell with radius "R" and constant charge density "sigma" is rotating with angular velocity "w" in the z direction. What is the magnetic force between the nothern hemisphere and the southern hemisphere (I'm assuming north is on the positive side of the z axis, and south is the negative side of the z axis)?

    Thanx
    BTW has anyone been looking at the Google problems advertised in the Physical Review?
     
    Last edited: Sep 6, 2004
  2. jcsd
  3. Sep 6, 2004 #2

    Tide

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    It might help if you said what the problem is - not everyone would have a copy of that particular text.
     
  4. Sep 21, 2004 #3
    I have a solution manual that I can sell you.

    please email me:

    spock0149@hotmail.com

    thanks.
     
  5. Oct 1, 2004 #4
    Problem:

    Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell. (ex. 5.11)

    My Solution:
    Look at example 5.11 a the book. It is calculated that the field in the spherical shell is uniform:
    [tex]B=\frac{2}{3}\mu_0\sigma R \omega[/tex]
    If you displaced the hemispheres by a tiny bit [tex]\Delta x[/tex]
    you create a gap of volume [tex]\pi R^2 \Delta x[/tex]
    We can assume the field stay more of less uniform. (This is not an approximation when we take the limit later).
    So the energy in the field in the gap is: (took me a while to look up the eqn for T, energy density, in SI unit:)
    [tex]\Delta U=T\times \Delta V=\frac{1}{2}\frac{B^2}{\mu_0}\times \pi R^2 \Delta x[/tex]
    [tex]\Delta U=\frac{1}{2}\times\frac{4}{9}\mu_0 \sigma^2 R^4 \omega^w \pi \Delta x[/tex]
    So the force is:
    [tex]F=-\frac{\Delta U}{\Delta x}=-\frac{2}{9}\pi \mu_0 \sigma^2 R^4 \omega^2[/tex]

    ...Interesting, I didn't get the factor of [tex]\frac{1}{4}[/tex] either...
    Is this what you got?
     
    Last edited: Oct 1, 2004
  6. Nov 2, 2004 #5
    Hi, I'm having trouble with part b of problem 5.46 from the same text. Any help would be nice :-D

    Problem 5.46: Magnetic field on the axis of a ciruclar current loop is far from uniform (it falls off sharply with increasing z). You can produce a more nearly uniform field by using two such loops a distance d apart.

    part b: Find d such that second partial derivative of B with respect to z is zero when z = 0.

    I can get it so that the second partial derivative equals zero, but my answer does not match the book's answer.
     
    Last edited: Nov 2, 2004
  7. Nov 3, 2004 #6
    Hey!

    Here is my solution:


    From a)
    Let u be the permeability of free space
    B=u*I/2*R^2(1/(R^2+(z+d)^2)^3/2+1/(R^2+(z-d)^2)^3/2)
    dB/dz=-3u*I/2*R^2((z+d)/(R^2+(z+d)^2)^5/2+(z-d)/(R^2+(z-d)^2)^5/2)
    For simplity let H=(z+d)/(R^2+(z+d)^2)^5/2+(z-d)/(R^2+(z-d)^2)^5/2
    when dH/dz=0 is d^2B/dB^2=0
    dH/dz=[1/(R^2+(z+d)^2)^5/2+(z+d)*d/dz(1/(R^2+(z+d)^2)^5/2)+
    1/(R^2+(z-d)^2)^5/2+(z-d)*d/dz(1/(R^2+(z-d)^2)^5/2)]=...=
    [1/(R^2+(z+d)^2)^5/2-5(z+d)^2/(R^2+(z+d)^2)^5/2+1/(R^2+(z-d)^2)^5/2-5(z-d)^2/(R^2+(z-d)^2)^5/2]

    Gives
    dH/dz(0)=2/(R^2+d^2)^5/2-10d^2/(R^2+d^2)^7/2=0

    And d=R/2 ->B(0)=u*I/2*R^2(2/(R^2+1/4R^2)^3/2)=8u*I/(5*sqrt(5)*R) And this equals the answer in my book.

    I hope this will help you
     
  8. Nov 3, 2004 #7
  9. Nov 5, 2004 #8
    Speaking of griffiths - if anyone comes up with a stumper nobody here can solve, he's my current physics prof, and I talk to him on a daily basis. You wouldn't believe how much his writing sounds like his talking :)
     
  10. Nov 5, 2004 #9
    Ehh, another Reedie :). . .

    I'm not too sure David'd be happy to hear of these solution manuals floating around in cyberspace
     
  11. Nov 5, 2004 #10
    Indeed. I hope most of the tough parts are left "as an exercise for the reader" :)
     
  12. Nov 6, 2004 #11
    How is Griffiths as a prof anyway? It's kind of cool having him be your professor... he's pretty much cornered the undergrad physics textbook market on QM ad E&M I think.

    I'm glad John David Jackson is retired though. I wouldn't wanna take E&M from him! (That is, if I only wanted a high grade)
     
  13. Nov 6, 2004 #12
    His "Introduction to particle physics" is pretty swell as well.
     
  14. Nov 6, 2004 #13
    He's fun as a prof, explains things in detail and so on, and keeps the informal attitude that's prevalent throughout his books. Sometimes he goes just a bit slow, but I guess that's connected with his wish to be as clear as possible about everything - he does make sure you're grasping what's going on. Funny thing is, though, even though I'm a physics major, I've only really taken what is pretty much a mathematical methods class from him, so I don't know that well what he'd be like in a real physics course. Will find out next year.
     
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