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Helmholtz Coils

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data
    How do I take the first and second derivatives of the helmholtz equation? The equation is as follows:

    Bx= ((uR^2NI)/2) * ( [1/((x+a)^2+R^2)^(3/2) ] + [1/((x+a)^2+R^2)^(3/2) ] )

    heres what I have to prove
    First Derivative
    http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=First.jpg

    Second Derivative
    http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=Second.jpg

    2. Relevant equations
    Bx= ((uR^2NI)/2) * ( [1/((x+a)^2+R^2)^(3/2) ] + [1/((x+a)^2+R^2)^(3/2) ] )

    http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=First.jpg

    http://s199.photobucket.com/albums/aa160/stryker213/?action=view&current=Second.jpg

    3. The attempt at a solution
    I just started multivariable calculus so I'm not quite sure what to do. I tried to take the derivative with respect to x and treated the rest of the variables as constants. However, I believe that you would need to do the chain rule to figure this problem out. Any suggestions?
     
  2. jcsd
  3. May 14, 2007 #2

    malawi_glenn

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    This is a one variable calculus problem. If you have taken those courses in one variable calculus, these two relations should be very familiar:

    https://www.physicsforums.com/attachment.php?attachmentid=10053&stc=1&d=1179119518

    I use always the first one, so this problem could be written as:

    constant * [((x+a)^2+R^2)^(-3/2) + ((x-a)^2+R^2)^(-3/2)]

    (you forgot one minus-sing in the denominator..)

    And now you just do the thing as usual:

    ((x+a)^2+R^2)^(-3/2) becomes : (-3/2)(2(x+a))((x+a)^2+R^2)^(-5/2) and then the same thing for the other term. Remember to do the inner derivate (Chain-rule).

    And yes, the second is just treat all other things as constants..

    good luck =)
     

    Attached Files:

  4. May 14, 2007 #3
    Hello,

    Just take first order and second order partial derivative of the equation.
    I obtained the results by some calculation.
    The first order derivative vanishes at x=0, which is consistent with the fact that some simple experiments reach approximate uniform magnetic field at the center of the two coils.
    The second derivative does not vanish at x=0.


    Regards
     
  5. May 14, 2007 #4
  6. May 14, 2007 #5

    malawi_glenn

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    why?

    differentiate this one for example: whith respect to x, all other is constants.

    (d/dx){[(x+a)^2+ T]^(3/2)} = (3/2)*2(x+a)*[(x+a)^2+ T]^(1/2)

    You should probably practice a little bit more on how to differentiate a bit more complex functions with one variable, the product rule, the chain rule etc.
     
    Last edited: May 14, 2007
  7. May 14, 2007 #6

    malawi_glenn

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    did you get it now? =)
     
  8. May 14, 2007 #7
    yes, I understand most of it now. Although, I am having trouble with finding the second derivative. I'm not sure if I should use the product rule or quotient rule, but I'm sure I can figure it out

    Thank you to everyone who responded to this thread.
     
  9. May 14, 2007 #8

    malawi_glenn

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    you do the same thing, but now you have things like:

    (d/dx){(x+a)*[(x+a)^2+ T]^(-5/2)}

    And here we use the product rule OR the quotient rule, as I said earlier that I always uses the first one, it is the easiest to remember, so i always uses that one and think of the equatation(function) as a product, as x/t^3 = x*t^-3
     
  10. May 16, 2007 #9
    i have the same problem, and I just wanted to tell you guys this help was amazing. Thanks!
     
  11. May 16, 2007 #10

    malawi_glenn

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    =)
    i hope stryker123 made it now.

    it is of high imoportance to master calculus and linear algebra in the world of phyics.
     
  12. May 16, 2007 #11
    This was actually a very easy problem, but just looks very intimidating, and takes some time.
     
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