Flux integral

Apr 2010
11
0
Hi

Having a bit of trouble with this question. Any ideas? Im thinking its going to be the integral of F.N? With N being the normal...using jacobian. Sorry, just throwing out ideas! Cheers.

Calculate the flux of the magnetic field​
B = (2xy, yz, 1) through the surface of

the planar circle of radius
R centred at the origin and located in the x, y-plane.
 
Apr 2010
384
153
Canada
Hi​


Having a bit of trouble with this question. Any ideas? Im thinking its going to be the integral of F.N? With N being the normal...using jacobian. Sorry, just throwing out ideas! Cheers.

Calculate the flux of the magnetic field B = (2xy, yz, 1) through the surface of
the planar circle of radius R centred at the origin and located in the x, y-plane.
Before we continue, we should note a few things. The first of which is our "circle" will actually become a cylinder in 3 space. When we extend this upward and downward along the z plain we will get a cylinder bounded by heights -h and h. We label these as such because the flux out of an infinately large shape is going to be infinity, so we bound our z such that our result is actually useful.

\(\displaystyle -h \le z \le h \)

We can do this 1 of 2 ways. The first of which is to compute the flux integrals for the 3 distinct sides of the cylinder (top, bottem and side) or use the divergence theorem.

I will use the divergence theorem,

\(\displaystyle \iiint_D div F dV= \iint F \cdot \hat N dS \)

There should be a closed symbol on the right side but I dont know the latex code for this :p

Our Function, \(\displaystyle F = 2xy \hat i + yz \hat j + 1 \hat k \) has the divergence such that,

\(\displaystyle div F = 2y + z + 1 \)

Our flux is therefore,

\(\displaystyle Flux = \iiint (2y + z + 1) dxdydz \)

Noting our dz bounds are the bounds we defined for z at the start of this problem, and our x and y bounds can be most easily expressed in cylindrical co-ordinates (because it is a cylinder!).

\(\displaystyle Flux = \int_0^{2 \pi } d \theta \int_0^r r dr \int_{-h}^{h} (2y + z + 1) dz \)

Where \(\displaystyle y = r sin \theta \)


\(\displaystyle Flux = \int_0^{2 \pi } d \theta \int_0^r r dr \int_{-h}^{h} (2rsin \theta + z + 1) dz \)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I'm afraid AllanCuz has completely misunderstood the problem. The problem does not ask about the flux through the sides of a cylinder, only about the flux through a disk.

Croc, the unit normal to the xy-plane is, of course, \(\displaystyle \vec{k}\). The "vector differential of surface area" is just \(\displaystyle \vec{n}dS= \vec{k}dxdy\). Since \(\displaystyle \vec{F}(x,y,z)= 2xy\vec{i}+ yz\vec{j}+ \vec{k}\), \(\displaystyle \vec{F}\cdot \vec{n}dS= dxdy\).

That is, since the disk is in the xy-plane, only the z-component cause a flux through it. And since that component is 1, the flux is just the area of the circle.
 
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Apr 2010
384
153
Canada
I'm afraid AllanCuz has completely misunderstood the problem. The problem does not ask about the flux through the sides of a cylinder, only about the flux through a disk.

Croc, the unit normal to the xy-plane is, of course, \(\displaystyle \vec{k}\). The "vector differential of surface area" is just \(\displaystyle \vec{n}dS= \vec{k}dxdy\). Since \(\displaystyle \vec{F}(x,y,z)= 2xy\vec{i}+ yz\vec{j}+ \vec{k}\), \(\displaystyle \vec{F}\cdot \vec{n}dS= dxdy\).

That is, since the disk is in the xy-plane, only the z-component cause a flux through it. And since that component is 1, the flux is just the area of the circle.
I assumed the question meant a cylinder because I've been given problems such as the one above and we were meant to interpret/bound it as a cylinder.

My bad on that one
 
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