How do I integrate Gaus's Law for magnetism for surface of half sphere?

[Jacobim]

The problem shows a picture of a surface of a half sphere. It is labeled surface 1 being the disk of the top of the half sphere. Surface 2 is the remaining surface of the half sphere. R is the radius.

The magnetic field is uniform and makes an angle theta with the vertical (or with the normal of the disk).

First I am asked to find the flux through surface 1.

The B field makes an angle theta with the normal of the flat surface. This is an easy integration resulting in

[itex]\Phi[/itex] = B cos θ ∏ R^2


The next part asks to find the magnetic flux through surface 2. This is more complicated because the angle between the normal of each dA is different. I have not been able to think of a way to relate the angle to the area in order to integrate.

Thank you for any clues.

 

[Dick]

The problem shows a picture of a surface of a half sphere. It is labeled surface 1 being the disk of the top of the half sphere. Surface 2 is the remaining surface of the half sphere. R is the radius.

The magnetic field is uniform and makes an angle theta with the vertical (or with the normal of the disk).

First I am asked to find the flux through surface 1.

The B field makes an angle theta with the normal of the flat surface. This is an easy integration resulting in

[itex]\Phi[/itex] = B cos θ ∏ R^2


The next part asks to find the magnetic flux through surface 2. This is more complicated because the angle between the normal of each dA is different. I have not been able to think of a way to relate the angle to the area in order to integrate.

Thank you for any clues.

You don't have to do the second integral at all if you know the divergence theorem. Sure you don't have something like that?

 

[vela]

The normal to an area element on the spherical part is just ##\hat{r}##. Can't you just write down what the dot product will equal explicitly?

 

[Jacobim]

I looked at the divergence theorem on wikipedia. In this problem, surface 2 is not a closed surface. The boundary is the circle where the sphere is cut in half.

 

[HallsofIvy]

Use polar coordinates [itex]\theta[/itex], [itex]\phi[/itex], and [itex]\rho= R[/itex]. Taking [itex]0\le \theta\le 2\pi[/itex], [itex]0\le \phi\le \pi/2[/itex] gives the upper half sphere. And the differential of surface area will be [itex]R^2 sin^2\phi d\theta d\phi[/itex].

 

[Jacobim]

There is no position vector in the definition of magnetic flux.

It is just the integral of B dot dA

 

[Muphrid]

I looked at the divergence theorem on wikipedia. In this problem, surface 2 is not a closed surface. The boundary is the circle where the sphere is cut in half.

Dick is making a broader point. You can use the divergence theorem on the hemispherical volume and divide the surface that bounds it into the two integrals that you've been asked to calculate.

[tex]\int_{S_1} B \cdot dA_1 + \int_{S_2} B \cdot dA_2 = \int \nabla \cdot B \; dV[/tex]

You know the first integral on the left, and Maxwell's equations should tell you the value of the integral on the right. You can then solve the value of the second integral without using the techniques of integration. The problem reduces to one of simple algebra.Nevertheless, the integral should not be difficult to compute directly. As vela said, on the hemispherical surface, the normal vector [itex]dA[/itex] is in the direction of [itex]\hat r[/itex]. HallsofIvy has ponited out to you what the magnitude of the normal vector should be. You should then be able to compute [itex]B \cdot dA[/itex].

 

[Jacobim]

Thank you for all the replies. I will have to study this information to understand it.