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Homework Help: Magnetic Field at the Center of a Wire Loop

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A piece of wire is bent to form a circle with radius r. It has a steady current I flowing through it in a counterclockwise direction as seen from the top (looking in the negative z direction).

    What is B_z(0), the z component of B at the center (i.e., x = y = z = 0) of the loop?

    Express your answer in terms of I, r, and constants like mu_0 and pi.

    2. Relevant equations



    3. The attempt at a solution

    I know this equation:

    [tex]\frac{(\mu_0)I}{2(\pi)r}[/tex]

    but there is a hint that says I need to find the Integrand.

    Thank You.
     
  2. jcsd
  3. Mar 7, 2008 #2
    Integrate the magnetic field around the circular path of radius r.
    [tex]\oint \vec B \cdot d\vec r = ?[/tex]
     
  4. Mar 7, 2008 #3

    Doc Al

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    Staff: Mentor

    Biot-Savart law

    That's the magnetic field from an infinite straight current-carrying wire.

    Look up the Biot-Savart law. That will give you the field from a current element.

    Right. Once you have the field from a current element, you'll need to integrate around the entire loop. (Since you are only asked to find the field at the center of the loop--as opposed to some arbitrary location--the integral will turn out to be quite doable.)
     
  5. Mar 7, 2008 #4
    Isnt the equation I posted the Biot-Savart law?
     
    Last edited: Mar 7, 2008
  6. Mar 7, 2008 #5

    Doc Al

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    No. As I said, the equation you posted is the field from a long current-carrying wire. Look up the Biot-Savart law.
     
  7. Mar 7, 2008 #6
    Sorry, about that, I was looking at the wrong equation in my book.

    B = [tex]\frac{\mu_0}{4\pi}[/tex] [tex]\frac{q(v X r}{r^2}[/tex]

    since its circular motion B = [tex]\frac{qmv}{r}[/tex] <=Would I need to ingetrate this equation?
     
  8. Mar 7, 2008 #7

    Doc Al

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    The one you want is in terms of current:
    [tex]d\vec{B} = \frac{\mu_0 I d\vec{\ell}\times \hat{r}}{4 \pi r^2}[/tex]

    Figure out what that is for a point in the center of the loop, then integrate around the loop.

    Not relevant; No circular motion here.
     
  9. Mar 8, 2008 #8
    Would it be

    [tex]\vec{B} = \frac{\mu_0 I d}{4 \pi r^2}[/tex]

    and then integrate that?
     
  10. Mar 9, 2008 #9

    Doc Al

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    Almost. After taking care of the vector product, it would be:

    [tex]d\vec{B} = \frac{\mu_0 I}{4 \pi r^2}\;d\ell[/tex]

    Integrate that around the loop. (It's easy!)
     
  11. Mar 9, 2008 #10
    is the [tex]d \ell[/tex] distance*length or the derivative of length.

    Then I would [tex]\oint \vec{B} dr[/tex] like Reshma said?
     
    Last edited: Mar 9, 2008
  12. Mar 9, 2008 #11

    Doc Al

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    Neither. [tex]d \ell[/tex] is an element of length around the circumference of the circle. (That should tip you off as to what the integral is. :wink:)

    No. Integrate the expression I gave in the last post, which is the field at the center due to a small element of the current, over the complete loop.
     
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