Solving Magnetic Flux Through a Sphere in a Homogeneous Field

[faen]

Homework Statement

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!

Homework Equations

surface integral over B*ez?

The Attempt at a Solution

Not sure how to calculate this..

Thanks a lot for any help!

 

[tiny-tim]

hi faen!

hint: divergence

 

[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.

 

[rude man]

hi faen!

hint: divergence

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

 

[tiny-tim]

You think they meant to include the bottom of the hemisphere?

no

 

[rude man]

no

So - divergence?

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

 

[rude man]

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

 

[faen]

Thanks for the input so far :)

With the integration I got ∫2*0.5 \pi r²sin²\thetacos 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 :)

 

[rude man]

Homework Statement

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!

Homework Equations

surface integral over B*ez?

The Attempt at a Solution

Not sure how to calculate this..

Thanks a lot for any help!

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

 

[faen]

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

I know but I still don't get it.. Can u give another hint please?

 

[rude man]

Thanks for the input so far :)

With the integration I got ∫2*0.5 \pi r²sin²\thetacos 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 :)

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.

 

[faen]

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.

That means that the total flux is zero.. As gauss law for magnetic fields I guess. I still don't know how to calculate half of the sphere though..

 

[faen]

maybe 0.5 * surface area of circle?

 

[rude man]

maybe 0.5 * surface area of circle?

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

 

[faen]

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

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?

 

[tiny-tim]

hi faen!

(just got up :zzz:)

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?

correct!

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

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

uhh? check your calculator

(and where does the 4 come from?)

 

[faen]

hi faen!

(just got up :zzz:)


correct!

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



uhh? check your calculator

(and where does the 4 come from?)

I got it! Thanks a lot! I checked again the formula for circle and it was \pir².. So removing the 4 gave the right answer :)

 

[rude man]

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

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.