# Magnetic flux sphere

1. Mar 17, 2013

### faen

1. The problem statement, all variables and given/known data

In homogeneous magnetic field the magnetic flux density vector is B = (0, 5T)ez (ez is unit vector in z direction). There is a sphere of radius R = 2cm with its centre at the origin. Find the flux crossing that half of the
sphere surface for which z larger or equal to 0!

2. Relevant equations

surface integral over B*ez?

3. The attempt at a solution

Not sure how to calculate this..

Thanks a lot for any help!

2. Mar 17, 2013

### tiny-tim

hi faen!

hint: divergence

3. Mar 17, 2013

### rude man

Consider a differentially thin strip of area dA on the sphere's surface at z = z0. Let r be a line connecting a point on the strip and meeting the z axis at a right angle. Let angle θ be the polar angle, i.e the angle between the z axis and a line connecting a point on the strip with the origin. So the strip can alternatively be defined by θ = constant.

So r = Rsinθ where R is the radius of the sphere. What is the area dA of this strip?

So every strip dA has its own constant θ and therefore B * n is constant thruout the strip, where n is the normal to the strip but B * n will be a function of θ only. (I am using * to denote the dot-product).

Then integrate B * n from θ = 0 to pi/2.

Last edited: Mar 17, 2013
4. Mar 17, 2013

### rude man

You think they meant to include the bottom of the hemisphere? I assumed not.

5. Mar 17, 2013

### tiny-tim

no

6. Mar 17, 2013

### rude man

So - divergence?

Ooh, wait - like the bottom surface integral = -top surface integral? Clever!

Last edited: Mar 17, 2013
7. Mar 17, 2013

### rude man

Just had another thought - how about Stokes? In which case never mind my 1st post.
EDIT: Scratch that.

Last edited: Mar 17, 2013
8. Mar 17, 2013

### faen

Thanks for the input so far :)

With the integration I got ∫2*0.5 $\pi$ r2sin2$\theta$cos 90-$\theta$ d$\theta$

I'm not sure if I could integrate it, but I think I'd get some points for this solution still.

I still don't know how I can use divergence to solve the problem.. I think I need further hints.. If you could please post the solution for me it would be nice, since I have exam tomorrow early and need to go to sleep :)

9. Mar 17, 2013

### rude man

Tiny tim has given you - us! - a clever hint! Making the computation extremely easy compared to what I posted.

10. Mar 17, 2013

### faen

I know but I still dont get it.. Can u give another hint please?

11. Mar 17, 2013

### rude man

What is div B at any point in space? Remember div B = 0?

So what does the divergence theorem say? What is the total flux thru the hemisphere including the bottom planar area? And if all the flux entering at the bottom has to leave at the top (the curved surface), how can you easily compute the latter?

We cannot just 'post the solution'. That violates our rules. We can give hints only. You won't learn anything by having someone give you the sol'n on a silver platter. I'm giving you plenty hints already.

12. Mar 17, 2013

### faen

That means that the total flux is zero.. As gauss law for magnetic fields I guess. I still dont know how to calculate half of the sphere though..

13. Mar 17, 2013

### faen

maybe 0.5 * surface area of circle?

14. Mar 17, 2013

### rude man

Why 0.5? All that goes in at the bottom must pop out at the top!

15. Mar 17, 2013

### faen

0.5 is the magnetic field density.. So i was thinking if that is multiplied with the surface area of a circle (bottom of the object in question) then that would be the same amount of flux leaving at the top?

I still didn't get the right answer, it's supposed to be 0, 6283 mVs

and I calculated it to be 0.5*4*pi*0.02^2 = 0.025

I'm not supposed to turn a surface integral into a volume integral (gauss theorem) or am I?

16. Mar 18, 2013

### tiny-tim

hi faen!

(just got up :zzz:)
correct!

the divergence is zero, so you can choose any surface with the same boundary

(and where does the 4 come from?)

17. Mar 18, 2013

### faen

I got it! Thanks a lot! I checked again the formula for circle and it was $\pi$r2.. So removing the 4 gave the right answer :)

18. Mar 18, 2013

### rude man

OK, I get confused with the European way of writing the decimal point ... yes, the answer is B*A where A is the area of the flat bottom of the hemisphere.