Conversation started Aug 12, 2019 at 10:37.
John Rennie
John Rennie
Aug 12, 2019 10:37
@Nobodyrecognizeable yes
Nobody recognizeable
Nobody recognizeable
59 mins ago, by Nobody recognizeable
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YUSUF HASAN
YUSUF HASAN
@JohnRennie Thank you sir... This info was a gr8 help for me... Wish you a good day! :)
Nobody recognizeable
Nobody recognizeable
57 mins ago, by Nobody recognizeable
Now magnetic charge densities are : surface charge density $k=M.\hat{n}$ and volume charge density is $\rho =\nabla. M$
John Rennie
John Rennie
If the ball has a uniform magnetisation desn't that mean the volume charge density is zero i.e. all the charge is on the surface?
Nobody recognizeable
Nobody recognizeable
@JohnRennie yep.
John Rennie
John Rennie
Aug 12, 2019 10:39
Ah, yes, it would because if $M$ is constant then div M is zero.
Nobody recognizeable
Nobody recognizeable
@JohnRennie you've taken my words.
I don't know which direction the magnetization is?
So surface charge for surface charge density what should I do?
Or should I think its radial?
John Rennie
John Rennie
I would divide the ball into cylinders centred on the axis of the ball. Then the charge will be the charge density multiplied by the area of the end of the cylinder, and the dipole will be the charge multiplied by the length of the cylinder.
I can draw a diagram if it will help.
Nobody recognizeable
Nobody recognizeable
@JohnRennie OK lemme read.
@JohnRennie but can you fit the cylinder so that it covers the sphere?
@JohnRennie ah crap. Fine sorry. Go on with the discussion. No need for the figure.
@JohnRennie are you here?
John Rennie
John Rennie
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Nobody recognizeable
Nobody recognizeable
@JohnRennie there we go...
John Rennie
John Rennie
Aug 12, 2019 10:49
Something like this. This is a cross section so you're seeing the top and bottom slices through a cylindrical shell of inner radius $x$ and thickness $dx$
Nobody recognizeable
Nobody recognizeable
@JohnRennie yep.
John Rennie
John Rennie
The charge at the end will be $dQ = dA \sigma$ and the dipole will be $dp = y dQ$
If you work out an expression for $dp$ you then just integrate from $x = 0$ to $x = r$ to get the total dipole.
Nobody recognizeable
Nobody recognizeable
@JohnRennie so $dp=ydA\sigma$
John Rennie
John Rennie
No, $dp = y dQ = y \sigma dA$
And $\sigma = M \cdot n$
Nobody recognizeable
Nobody recognizeable
@JohnRennie so $ y\sigma dx dy$
@JohnRennie so is M radial?
John Rennie
John Rennie
Aug 12, 2019 10:53
No, as I've draw the diagram $\mathbf M$ is horizontal.
Nobody recognizeable
Nobody recognizeable
@JohnRennie so $\sigma =Mcos\theta$
John Rennie
John Rennie
Yes. And $dA$ is going to be something like $2\pi x dx /cos\theta$
The $\cos\theta$ will cancel so $dQ = 2\pi M x dx$
Nobody recognizeable
Nobody recognizeable
@JohnRennie what about the y?
John Rennie
John Rennie
You get $y$ from the equation for a circle. Actually let's call the length of the cylinder $2y$, because then we have $x^2 + y^2 = r^2$.
So $y = \sqrt{r^2 - x^2}$ and therefore the expression for $dp$ is:
Nobody recognizeable
Nobody recognizeable
@JohnRennie OK.
John Rennie
John Rennie
Aug 12, 2019 11:00
$$ dp = 4 \pi M \sqrt{r^2 - x^2} x dx $$
Nobody recognizeable
Nobody recognizeable
@JohnRennie so $4\pi Mr$
John Rennie
John Rennie
Hmm, I'm not entirely convinced by the form of that equation ut it's worth integrating it to see what happens.
Nobody recognizeable
Nobody recognizeable
Sorry.
@JohnRennie yep you are a champion. One day I hope to be a problem solver like you.
2
John Rennie
John Rennie
$$ 4 \pi M \tfrac{1}{3} (r^2 - x^2)^{3/2} $$
Nobody recognizeable
Nobody recognizeable
It's $4\pi Mr^3/3$ which exactly matches the answer.
 
Conversation ended Aug 12, 2019 at 11:04.