Help with Griffiths Electrodynamics

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Homework Help Overview

The discussion revolves around problems from Griffiths' Electrodynamics, specifically problem 5.42 concerning a spinning spherical shell with a constant charge density and its magnetic force interactions. Participants are exploring the calculations related to magnetic fields and forces in electrodynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem involving a rotating charged spherical shell and the magnetic force between its hemispheres. There are attempts to derive expressions for magnetic fields and energy densities, with some questioning the accuracy of coefficients in their calculations. Another participant raises a separate issue regarding problem 5.46, seeking clarification on achieving a uniform magnetic field with circular current loops.

Discussion Status

The discussion is ongoing, with various participants sharing their attempts and insights. Some have provided partial solutions and calculations, while others are seeking further clarification or assistance. There is no explicit consensus on the correctness of the approaches or results presented.

Contextual Notes

Some participants note that not everyone has access to Griffiths' text, which may limit the discussion. Additionally, there are references to solution manuals and concerns about their availability, indicating a broader context of academic integrity in homework help.

sinyud
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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?
 
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It might help if you said what the problem is - not everyone would have a copy of that particular text.
 
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:
B=\frac{2}{3}\mu_0\sigma R \omega
If you displaced the hemispheres by a tiny bit \Delta x
you create a gap of volume \pi R^2 \Delta x
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:)
\Delta U=T\times \Delta V=\frac{1}{2}\frac{B^2}{\mu_0}\times \pi R^2 \Delta x
\Delta U=\frac{1}{2}\times\frac{4}{9}\mu_0 \sigma^2 R^4 \omega^w \pi \Delta x
So the force is:
F=-\frac{\Delta U}{\Delta x}=-\frac{2}{9}\pi \mu_0 \sigma^2 R^4 \omega^2

...Interesting, I didn't get the factor of \frac{1}{4} either...
Is this what you got?
 
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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.
 
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iMook said:
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.
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
 
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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 :)
 
Ehh, another Reedie :). . .

I'm not too sure David'd be happy to hear of these solution manuals floating around in cyberspace
 
  • #10
Duarh said:
Ehh, another Reedie :). . .

I'm not too sure David'd be happy to hear of these solution manuals floating around in cyberspace

Indeed. I hope most of the tough parts are left "as an exercise for the reader" :)
 
  • #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 want to take E&M from him! (That is, if I only wanted a high grade)
 
  • #12
His "Introduction to particle physics" is pretty swell as well.
 
  • #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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