# Help with Griffiths Electrodynamics

1. Sep 6, 2004

### sinyud

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?
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"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. Sep 6, 2004

### Tide

It might help if you said what the problem is - not everyone would have a copy of that particular text.

3. Sep 21, 2004

### robousy

I have a solution manual that I can sell you.

spock0149@hotmail.com

thanks.

4. Oct 1, 2004

### mathfeel

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?

Last edited: Oct 1, 2004
5. Nov 2, 2004

### iMook

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
6. Nov 3, 2004

### eys_physics

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.

7. Nov 3, 2004

### stuffy

Last edited by a moderator: Apr 21, 2017
8. Nov 5, 2004

### Jenga

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 :)

9. Nov 5, 2004

### Duarh

Ehh, another Reedie :). . .

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

10. Nov 5, 2004

### Jenga

Indeed. I hope most of the tough parts are left "as an exercise for the reader" :)

11. Nov 6, 2004

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)

12. Nov 6, 2004

### Kalimaa23

His "Introduction to particle physics" is pretty swell as well.

13. Nov 6, 2004

### Duarh

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.