# Quick EMF question - Spherical Shell- North/South Pole, Equator

1. Mar 24, 2014

### binbagsss

The question is: a conducting spherical shell of radius a rotates about the z axis with angular velocity ω, in a uniform magnetic field B= B$_{0}$$\hat{z}$ . Find an expression for the EMF developed between:

i) the north pole and the equator (2 marks);
ii) the north pole and the south pole (1 mark).

I'm struggling picturing why there is 0 flux between the north pole and south pole and a non-zero flux between the north pole and the equator.

So first of all, by rotates about the z-axis, I interpret this as any axis passing through the centre of the sphere.

I have attached two diagrams, the first i take the north and south pole to be aligned with the z axis (vertical) and the second the north and south aligned horizontally.

- From the first diagram, I think, I undertand the flux comments above, but in the second diagram , I would get zero flux for both cases...

Questions:

- So by north and south pole do we mean north and south with respect to the z-axis, as the object is spherically symetric so otherwise how do you choose?
- I can also see that I have not used the fact that the object is a shell and not a dense sphere. Am I correct in thinking that the answers to i and ii remain unchanged if I were to replace the spherical shell with a sphere?

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2. Mar 25, 2014

### BruceW

I'm pretty sure it should be the second thumbnail. The line going from south pole to north pole should be in the same direction as the magnetic field. And yeah, the shell rotates around this line. Why do you say there will be zero flux? also, remember that they are asking about the EMF, which is the line integral of the electromagnetic force. This is not the same as the flux. But there is an important theorem relating a closed line integral to a flux.

3. Mar 26, 2014

### binbagsss

Thanks for the reply. Apologies yes the second thumbnail.
Between the North Pole and Equator the flux is non-zero.
But between the North and South pole it is zero, as what goes in comes out, it cancels?

4. Mar 26, 2014

### BvU

Probably not the way you mean, but: yes. The area "vector" points opposite B for the lower half and with B for the upper half.

Another way to put this is "between the North and South pole it is zero, since they are at equal distances from the axis of rotation".

So what's the expression you found under i) ?

5. Mar 26, 2014

### BruceW

to binbagsss: Flux over what surface? The line integral of the Force along a path between the North and South pole is a valid statement. But the flux between the North and South pole is meaningless, because flux is defined over a surface.

6. Mar 26, 2014

### binbagsss

I thought the north and sole pole are just a single point, and since the axis passes through thse two points, the distance from the axis of rotation is 0 in each case?

7. Mar 26, 2014

### binbagsss

Flux over a circular surface of radius a, with its origin at the centre of the sphere, such that the top and bottom pass through the north and south pole.

8. Mar 26, 2014

### BvU

You mean a vertical surface ? No flux !

9. Mar 26, 2014

### BvU

might help.

Last edited by a moderator: Sep 25, 2014
10. Mar 27, 2014

### binbagsss

Oh yeh, the surface area would be parallel to the flux !
So, a circular cross-section though the sphere , horizontally, for case i?
I'm not sure what surface you would take for ii. Does it need to be a different surface than case i?

11. Mar 27, 2014

### BvU

Yes: for pole-equator you go halfway, for pole-pole you go all the way! Or wasn't that what you were asking ?

And no, horizontal rings won't do the trick either.

What did you pick up from the U-tube (with the field going into the whiteboard, a kind of bottom view of your case...)

12. Mar 28, 2014

### binbagsss

I'm really not sure what surface you would then, so not a circular disk?

13. Mar 28, 2014

### binbagsss

Oh it's the flux passing through the spherical/hemi-spherical surface. So equivalent to the 'projection' of the spherical surface onto the B -field, which is the same expression as $\phi$$_{B}$ for a circle of radius a.

The upper and lower hemisphere have opposite area/normal vectors. s.t $\phi$$_{B}$ (between the North pole and Equator) = B$\pi$a$^{2}$ ,

and $\phi$$_{B}$ (between the North pole and South pole) = B$\pi$a$^{2}$ - B$\pi$a$^{2}$=0.

Are these thoughts correct?
Thanks.

14. Mar 28, 2014

### BvU

Normal vectors are perpendicular to the surface, so for a hemisphere every point has a different normal vector. But integrating from North to south makes the vertical components cancel, from North pole to equator they don't.