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Griffiths - example 5.5 Electrodynamics - Calculate B Field

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Greetings,

    In Griffiths, Introduction to Electrodynamics, example 5.5 (page 216), calculating the B field a distance “S” away from a current carrying wire.

    l' (dl’) is the horizontal current carry wire – will be segmented to dl’

    [tex] tan(\theta) = \frac{l’}{s}[/tex]

    In the next step, it is stated

    [tex] dl’ = \frac{s}{cos^2(\theta)} d(\theta)[/tex]

    I am stuck on this - I do not see how to get from the first equation to the second. Is there an approximation required?



    2. Relevant equations



    3. The attempt at a solution

    [tex] tan(\theta) = \frac{l’}{s}[/tex]

    and
    [tex] r^2 = l^2 + s^2[/tex]

    [tex] l' = s tan(\theta) [/tex]

    I could square both sides

    [tex]l^2 = s^2 tan^2(\theta)[/tex]

    [tex]l^2 = s^2 \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

    [tex]l^2 = (r^2-l^2) \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

    [tex]\frac{l^2}{(r^2-l^2)} = \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

    [tex]sin^2(\theta) - 1 = \frac{sin^2(\theta)}{cos^2(\theta)}[/tex]

    I do not have this in terms of l' anymore.


    bottom line how does Griffiths get to

    [tex] dl’ = \frac{s}{cos^2(\theta)} d(\theta)[/tex]

    from

    [tex] tan(\theta) = \frac{l’}{s}[/tex]

    Thanks for the help

    -Sparky_
     
  2. jcsd
  3. Feb 18, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    It's easy if you recall the formula for the derivative of the tangent function.
     
  4. Feb 18, 2013 #3
    ah crap - chain rule

    I was thinking straight algebra type identity or small angle approximation or some such

    did not see the forest for the trees

    THANKS!!
    Sparky_
     
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