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  1. #1
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    Flux integral

    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.
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Croc View Post
    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.

     -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,

     \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,  F = 2xy \hat i + yz \hat j + 1 \hat k has the divergence such that,

     div F = 2y + z + 1

    Our flux is therefore,

     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!).

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

    Where  y = r sin \theta


     Flux = \int_0^{2 \pi } d \theta \int_0^r r dr \int_{-h}^{h} (2rsin \theta + z + 1) dz
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  3. #3
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    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, \vec{k}. The "vector differential of surface area" is just \vec{n}dS= \vec{k}dxdy. Since \vec{F}(x,y,z)= 2xy\vec{i}+ yz\vec{j}+ \vec{k}, \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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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    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, \vec{k}. The "vector differential of surface area" is just \vec{n}dS= \vec{k}dxdy. Since \vec{F}(x,y,z)= 2xy\vec{i}+ yz\vec{j}+ \vec{k}, \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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