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Homework Help: Maximize the magnetic field

  1. Nov 22, 2005 #1
    you are given a wire of length L and that carries a uniform current i through it.
    the wire can bent into either a circle or a square
    which shape gives the maximum magnetic field at its center?

    for the circle
    [tex] B = \frac{\mu_{0}}{4 \pi} \int \frac{i ds \cross r}{r^3} [/tex]
    im not qutie sure about the s part since ds = L, riught? so ds is constant value... and so is R so we should get
    [tex] B = \frac{\mu_{0} i}{4 \pi R^2} [/tex]
    where [tex] R = \frac{L}{2 \pi} [/tex]
    so [tex] B = \frac{\mu_{0} i \pi}{L^2} [/tex]

    now for the squar

    the problem is ds is a cosntant value, but r is not because it varies from s/2 to [itex] \frac{s}{\sqrt{2}} [/itex]
    so what do i do? Can i use the square as an Amperian loop and solve it like so
    [tex] B (s^2) = \mu_{0} i [/tex]
    since 4S = L, s = L/4 so
    [tex] B = \frac{16 \mu_{0} i}{L^2} [/tex]
    thus the magnetic field at the center due to th square loop is greater becasue 4 > pi?

    Also i am a bit confused as to when i use the Ampere's law and Biot Savart Law, can the Ampere's law be used for the above described situations or not?

    Please help! Your advice is greatly appreciated!
  2. jcsd
  3. Nov 22, 2005 #2


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    Homework Helper

    Quick reference to http://maxwell.byu.edu/~spencerr/websumm122/node70.html

    shows that the solution to the Biot-savart integral for a loop is

    [tex] B = \frac{\mu_o I}{2R}[/tex]

    Note that for the infinite wire, the solution is the same as that of Ampere's Law.

    If you need to apply the integral to a square. You can do it once for one side then multiply by 4. I don't follow your method. Can you write out your integral limits more clearly.
    Last edited by a moderator: Apr 21, 2017
  4. Nov 22, 2005 #3
    well for the square can i just use the equation for the finite wire of length L where d is the distance between teh point int eh center and wire, and multiply by 4?

    the formula im ean to use is [tex] B = \frac{\mu_{0} i}{4 \pi d} \frac{L}{\sqrt{\frac{L^2}{4} + d^2}}} [/tex]
    would this yile dthe required answer
    of course L in the formula does not mean the length of the wire
  5. Nov 22, 2005 #4


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    Homework Helper

    That formula does not look very familiar. Can you show the derivation?
  6. Nov 22, 2005 #5
    its in my textbook and it would be fine if i used it i believe
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