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Current in a Loop

  1. Mar 5, 2007 #1
    1. The problem statement, all variables and given/known data
    This is the third part of a problem: The loop has a resistance of 0.20, an area of 200cm^2, and consists of 400 turns. What is the magnitude of the current in the loop at t=20s?

    2. Relevant equations
    Magnetic Flux=BAcos(theta)
    Faraday's Law=-N(Magnetic Flux)/time

    3. The attempt at a solution

    I know I am missing an equation I need to figure this out but I'm not sure what it is. I have the N, A, t, and i have the resistance. I need to use the resistance to find the B, but I'm not sure how to do that. Please help. Thanks
  2. jcsd
  3. Mar 5, 2007 #2
    Are you given B(t)?
  4. Mar 5, 2007 #3
    No so that's why I'm confused...in the first part of the problem it gives me a graph and at the time it says the B looks to be about -0.66 but I'm not sure if I'm supposed to use that or figure out the actual B using the resistance??
  5. Mar 5, 2007 #4
    So you are given a graph of B versus t?
  6. Mar 5, 2007 #5
    yes in the first part of the problem...i don't know how to insert pictures or i'd post that...for the time 20s it looks like it is -.66
  7. Mar 5, 2007 #6
    Remember that the induced emf depends on the time rate change of the B field. What does that mean graphically?
  8. Mar 5, 2007 #7
    Well I know the induced current is the slope of the line of the B vs t graph, but I'm not sure how to figure out what to do with the resistance for this part of the problem..??
  9. Mar 5, 2007 #8
    No that is not quite right. The slope of B vs. t is a changing flux density (area density). Multiply it by the area to have the change in flux per time. Faraday found something out about this changing flux...
  10. Mar 5, 2007 #9
    Ok now I'm really confused. I thought that the induced current was proportional to the negative slope of the magnetic flux graph..? Sorry if it's dumb, we just learned this tonight and have this homework due Wednesday so I'm trying to sort through it. Thanks for your help.
  11. Mar 5, 2007 #10
    The induced emf is proportional to the change in flux, here the area times the slope of the B vs. t, and I stress that works here.(Changing area is possible too)Now the loop is ohmic so that implies what?
  12. Mar 5, 2007 #11
    Is the induced emf the same thing as the induced current? We learned that the induced current is proportional to the neg slope of the graph. We also learned that the motional emf is -vLB. I don't know what ohmic is??
  13. Mar 6, 2007 #12
    Induced emf is not the same as induced current. The induced emf provides a potential which can move charge. How much moves? Well it depends; for an ohmic device the current is proportional to the voltage (emf). I will let you find out the proportionality constant.
    The current is proportional to the slope for this problem, I wouldnt say that is a general statement.
    Motional emf doesnt really apply here.
    Google ohmic.
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