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Induced EMF and a square copper loop

  1. Nov 13, 2004 #1
    Hello,
    I don't know if you will find this question silly or not because I know it shouldn't be that hard but for some reason I keep getting it wrong. Here it is:

    A square copper loop, with sides of length 10.4 cm, is located in a region of changing magnetic field. The direction of the magnetic field makes an angle of 34.0 degrees with the plane of the loop. The time-changing (increasing) field has the following time dependence: B(t)=0.600 T+(3.30×10-3 T/s)t.
    Find the magnitude of the induced emf in the copper loop for times t>0.

    Alright, so I was thinking this should be an easy faraday's law problem and I tried faraday's law: E = - d (flux)/dt = (B)(A)(cos(theta))*dB/dt
    so I did: E = .6 * .104^2 * cos(34) * 3.3*10^-3 and I got 1.775 * 10^-5, which is wrong. The correct answer is 2.00 * 10^-5. Any ideas to what I did wrong?

    Thanks
     
  2. jcsd
  3. Nov 13, 2004 #2

    dduardo

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    is T just a constant?
     
  4. Nov 13, 2004 #3
    If you mean time, no I don't think it's constant but I don't think the answer depends on it. I know they give an equation for B(t) but they don't give any specific time, so the answer can't depend on it.

    The magnetic field is not constant and they give the rate of change in the equation (in units of T/s), which I used but I don't know why I get the answer wrong.
     
  5. Nov 13, 2004 #4

    dduardo

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    I just realized T is Telsa. ok, give me a sec.
     
  6. Nov 13, 2004 #5

    dduardo

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    ok, I got the answer. It is 1.95 *10^-5 V

    I got this by using the formula:

    Vemf = - surface_integral [ ( partial B-> / partial t ) dot product (dS)]

    Here are the steps:

    1. Take the square loop plane on Y=0
    2. Draw the B vector at a 35 degree angle to the plane
    3. Since your trying to find out the portion of the magnetic field that goes through the loop the only component of the B field that goes through is By. Therefore:

    By -> = (.6+3.3*10^-3 *t)*sin(34)

    4. Since the formula wants: partial B /partial t , then:

    dBy/dt = 3.3*10^-3 *sin(34) ay

    (ay is the unit vector in the y direction)

    5. Since the the plane is on X=0 and Y=0 the area dS points in the y direction. Therefore dS = dx*dz *ay

    6. The dot product of dBy/dt and dS is: 3.3*10^-3 *dx*dz

    7. Integrate the surface from x=0 to x=10.4cm and z=0 to z=10.4cm

    (I really need to learn latex :yuck: )
     
    Last edited: Nov 13, 2004
  7. Nov 13, 2004 #6

    dduardo

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  8. Nov 13, 2004 #7
    First, I want to thank you dduardo for this very fast response. I don't really understand your explanation for the answer very well, but I noticed that you used sin(34) not cos(34). I drew the picture and figured out why you used sin (and felt like an idiot for not noticing that). Anyways, I kept trying to solve it with sin instead of cos and came up with this: .104^2 * sin(34) * 3*10^-3 = 1.9959*10^-5. I am pretty sure that is the right answer, but I still don't know why this works. Also, why did they give me the .6 T in the problem?

    The equation says that (B->.A->)*(change in flux over time) = EMF

    What I came up with is just the area * cos(theta) *the change in flux over time which gave me the right answer. What happened to the original B?

    PS. I am in calculus two (don't know partial derivatives yet)
    Thanks
     
  9. Nov 13, 2004 #8

    dduardo

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  10. Nov 13, 2004 #9

    dduardo

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    All the partial derivative means is take the derviative of the function in terms of one variable and let everything else be a constant.
     
  11. Nov 13, 2004 #10

    dduardo

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    The equation:

    (B->.A->)*(change in flux over time) = EMF

    is wrong

    The actual equation is:

    Vemf = - d flux / dt
     
  12. Nov 13, 2004 #11
    Alright, I think I get it. Thanks very much :biggrin:
     
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