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Divergence Theorem Proof

  1. Feb 18, 2005 #1
    Im having a bit of a problem understanding the crucial part of the divergence theorem from Electromagnetic Fields and Waves by Lorrain and Corson. Ill try descibe the set up of the problem 1st and see if anyone can help me in any way before i continue with the electromagnetism course im doing as i want to be comfortable with the vector algebra.

    The book considers the outward flux through a closed surface, in this case an infinitessimal volume dx dy dz and a vector B whose components Bx, By, Bz are functions of x,y,z. The value of Bx at the centre of the right-hand face may be taken to ve the average value over that face. Through the right-hand face of the volume element, the outgoing flux is

    dΦR = (Bx + (dBx/dx)*(dx/2))dydz

    This is the bit im puzzled at, i understand dydz is the area element and why its the x-component of the vector we use in the line integral but im not getting where the "+ (dBx/dx)*(dx/2)" is coming into the scene. Any suggestions?​
  2. jcsd
  3. Feb 18, 2005 #2


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    Shouldn't it come from:
    [tex] \frac{B_{x}(x+dx)-B_{x}(x)}{2} [/tex] which is indeed the average of the magnetic field over the interval (x,x+dx)...?


    P.S.And the Taylor expansion to first order.BTW,it should be partial derivative wrt "x"...
  4. Feb 18, 2005 #3
    Thats what ive seen elsewhere, how do i do the Taylor expansion? Yeah i know its a partial derivative but im unsure how to do equations etc in my posts so can only keep it basic
  5. Feb 18, 2005 #4


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    Well,you can approximate the Taylor series to the first term:
    [tex] B_{x}(x+dx,y,z)-B_{x}(x,y,z)\approx \frac{\partial B_{x}(x,y,z)}{\partial x} dx [/tex]

    and i believe that's how it's done.

  6. Feb 18, 2005 #5
    Thanks think ive got a better idea now
  7. Feb 18, 2005 #6
    just the 1 more question. what happens to the factor of 2. If the average magnetic field over the interval is B(x+dx)-B(x)/2 then do the Taylor expansion, would that not make the x-component of the flux 1/2(dBx/dx)dV or is this only the outgoing flux through the right hand surface of the volume element, and the outgoing flux through the left side gives 1/2(dBx/dx)dV also
  8. Feb 18, 2005 #7


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    I don't have my copy of Lorrain and Corson handy... but is "B(x)" being evaluated at the center of the box or at a corner of the box of size (dx) x (dy) x (dz)? This might explain the factor of 2.

    In addition, you are setting up an area-integral [which can be read expressed as a volume-integral].

    Note that only the x-component flows through the area element dydz.
    (In other words, the normal to that area element is along the x-axis.)
  9. Feb 18, 2005 #8


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    The interval is centered around x. The interval goes from x-dx/2 to x+dx/2.
    The average value of the x-component of the magnetic field is just [tex]B_x(x)[/tex].

    If you take the taylor series of [tex]B_x(x)[/tex] about x...
    [tex]B_x(x+h) = B_x(x) + B_x'(x)h +....[/tex]

    So you can plug in h=dx/2 or h=-dx/2 to get [tex]B_x[/tex] at the right hand surface or left hand surface.
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