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Magnetic field problem

  1. Mar 22, 2005 #1
    A square conducting loop of side 2a lies in the z=0 plane and carries a current in the counterclockwise direction. Show that at the center of the loop
    H = sqrrt2*I/pi*a in the z direction.

    I am stuck in this problem. Heres what Ive got. I placed the center of the loop at the origin. Im using Biot-Savarts Law. I found an R vector to be sqrt x^2 + y2 in the rho direction and cos x/sqrt x^2 + y2 in the phi direction. Heres where Im stuck. When I go to do the integral, my dl depends on the side im integrating. So if im my dl is dx, i dont know how to cross that why R because R has the unit vectors rho and phi.
  2. jcsd
  3. Mar 22, 2005 #2
    I didn't quiet understand your approach. Find the magnetic field due to one wire of length 2a at a distance a from it's center. Contribution from each wire is same. So B = 4 * B(A).

    | a

    From the center of the wire at a distance l choose current element dl.
    Distance between dl and A is r.

    Using biot law,

    [tex] B(A) = \frac{\mu_0}{4\pi}\int_{-a}^{a} \frac{I dl X n}{r^2}[/tex]

    [tex] B(A) = \frac{\mu_0 I}{4\pi}\int_{-a}^{a} \frac{dl sin \theta}{a^2+l^2}[/tex]

    [tex] sin \theta = a / \sqrt (a^2 + l^2)[/tex]

    Do the integration.

    [tex] B(A) = \frac{\mu_0 I}{4\pi a}* \sqrt 2 [/tex]

    [tex] H = 4* H(A) = \frac{I}{\pi a} * \sqrt 2 [/tex]
  4. Mar 22, 2005 #3
    where does the sin theta come from in that second step
  5. Mar 22, 2005 #4

    Andrew Mason

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    You only have to do one side because they all add together. Call them sides 1, 2, 3, 4.

    [tex]B_1 = \mu_0I \int \frac{d\vec L \times \vec r}{4\pi r^2} = \frac{\mu_0I}{4\pi} \int \frac{rdlsin\theta}{r^2}[/tex]

    In polar coordinates:

    [itex]dL = cos\theta d\theta[/itex]
    [itex]r = a/cos\theta[/itex]

    [tex]B_1 = \frac{\mu_0I}{4\pi} \int_{-\pi/4}^{\pi/4} \frac{(a/cos\theta)cos\theta d\theta sin\theta}{(a/cos\theta)^2}[/tex]

    [tex]B_1 = \frac{\mu_0I}{4\pi} \int_{-\pi/4}^{\pi/4} \frac{asin\theta cos^2\theta}{a^2}d\theta [/tex]

    Since [itex]B = 4B_1[/itex]:

    [tex]B = \frac{\mu_0I}{\pi a} \int_{-\pi/4}^{\pi/4} sin\theta cos^2\theta d\theta [/tex]

    You have to use some trigonometric identities to integrate that. Better check my math too.

  6. Mar 22, 2005 #5
    In my first integral, n is the unit vector along the direction of vector r. [tex]sin \theta [/tex] comes in when you take the cross product of dl X n. [tex]sin \theta [/tex] is the angle between r and dl.

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