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Magnetic Flux and Faraday's Law

  1. Oct 29, 2008 #1
    A figure has a 3.0T magnetic field that is normal to the plane of a conducting, circular loop with a resistance of 1.5 ohms and a radius of .024m. The magnetic field is directed out of the paper. The area of the non-circular portion of the wire is considered negligible compared to that of the circular loop.

    a. What is the magnitude of the average induced emf in the loop if the magnitude of the magnetic field is doubled in .4s

    b. What is the average current around the loop if the magnitude of the magnetic field is doubled in .4s

    c. If the magnetic field is held constant at 3.0T and the loop is pulled out of the region that contains the field in .2s, what is the magnitude of the average induced emf in the loop?

    d. If the magnetic field is held constant at 3.0T and the loops is pulled out of the region that contains the field in .2s, at what rate is energy dissipated in R?



    For a. I thought to use the equation E = Q/T and since Q=BCos(angle)A I found the answer to be .027V

    Then on B I got stuck. I am using the equation E = L (I/T), therefore I = Et / L but I am getting the wrong answer. I think I might be using the wrong equation.

    For C I used the equation E = N (Q/T) and Q = BA = 3.0T x (3.14 x .024^2) = .0054 and if you divide that by .2s I got the answer for C to be 2.7 x 10^-2 V

    Then on D I am stuck again. I am using the equation P = (B^2 x V^2 x L^2) / 1.5 ohms = 2.8 x 10^-7, however this is not one of my answer choices so I am also doing this problem wrong.

    If I could get any help on b and d I would really appreciate it!!
     
  2. jcsd
  3. Oct 30, 2008 #2

    Hootenanny

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    For (b) your over complicating thinks a little. You've already calculated the induced emf (voltage) in part (a) and you are given the resistance in the question.

    Can you think of a simple equation linking voltage, resistance and current?

    We'll have a look at (d) when you've got (b) correct.
     
  4. Nov 1, 2008 #3
    I used the equation V = IR and I did not get any of the answers that are provided for me. I did .027 / 1.5 = .018A and that is not any of the answers I have. This is making me think I have part a wrong for the voltage.
     
  5. Nov 2, 2008 #4

    Hootenanny

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    The value of the emf which you calculated is indeed incorrect. Perhaps if you showed us you're calculations we could spot where you went wrong.
     
  6. Nov 2, 2008 #5
    This is what I did for part a

    E = Q / T
    Q = BCos(angle)A = 6.0T x 3.14 (.024)^2 = .01085
    E = .01085 / .4s = .027 V.

    This is one of my answer choices however from part b I know that this one is wrong but I am not quite sure as to where it is wrong.
     
  7. Nov 2, 2008 #6

    Hootenanny

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    Becareful here, the actual equation should be written:

    [tex]\varepsilon = -\frac{dQ}{dt}[/tex]

    That is, the induced emf is equal to the negative rate of change of the magnetic flux. So you need to work out the change in the magnetic flux.
     
  8. Nov 2, 2008 #7
    I guess I am not quite sure as to what the change in flux is then. I thought the initial was zero because I am only given one magnetic field value and one time value. The only other thing I can think of is the initial magnetic field is 3.0 and the final is 6.0 so then I get
    Bi - Bf = (.00543 - .01086) = -.00906 / .4s = .0136 V

    So then for B I would take .0136V / 1.5 = .00905 or 9 x 10^-3
    But which direction is this current in, is it counterclockwise using the right hand rule. since the magnetic field is out of the paper if you curl your fingers they move counterclockwise.
     
  9. Nov 2, 2008 #8

    Hootenanny

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    You're right, the B field changes by three units. This answers is closer, but you need to recheck you calculations, you're out by a factor of ten. I also noticed in your previous post that your using an approximation for pi, it would be much better to use the actual value for pi stored in your calculator rather than an approximation.
     
  10. Nov 2, 2008 #9
    I rechecked my calculations and I am still getting the same thing. I will work it out step by step. Also, on my calculator I am using the symbol for pi, i just wrote 3.14 on my answer because it is easier to type.

    Here is how I calculated A

    Qi = 3.0 x pi (.024)^2 = 3.0 x .0018096 = .0054287
    Qf = 6.0 x pi (.024)^2 = 6.0 x .0018096 = .0108576

    Qf - Qi = .0108576 - .0054287 = .0054289

    E = Q / t
    E = .0054289 / .4s = .01357V

    Then for B
    I = E/R
    I = .01357 / 1.5 = 9.0 x 10^-3
     
  11. Nov 2, 2008 #10

    Hootenanny

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    Sorry, that's my bad. I thought the time was 4 seconds, not 0.4 seconds. Your answers seem okay to me
     
  12. Nov 2, 2008 #11
    Thank you for your help!


    Also, I am unsure of how to do D still. I can not find an equation that will give me the rate of energy that is dissipated.
     
  13. Nov 2, 2008 #12

    Hootenanny

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    HINT: Just treat the coil as a resistor. How would you calculate the power dissipated by a resistor?
     
  14. Nov 2, 2008 #13
    Ok I found an equation that is P = IV and from c. I already know V is 2.7 x 10^-2 so I would have to calculate I, which is E/R so this is what I got

    I = 2.7 x 10^-2 / 1.5 = .018

    P = IV
    P = (.018)(2.7 x 10^-2) = 4.9 x 10^-4

    Hopefully this is correct!
     
  15. Nov 2, 2008 #14

    Hootenanny

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    I've not checked your arithmetic, but your method looks good :approve:
     
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