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Induced EMF in metal loop near RC circuit

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data

    (Challenge Problem 80, Ch33) A rectangular metal loop with 0.050 ohms resistance is placed next to one wire of the RC circuit shown in the below figure. The capacitor is charged to 20 V with the polarity shown, then the switch is closed at t = 0s.
    [url=http://postimg.org/image/bjo7lwzh5/][PLAIN]http://s4.postimg.org/bjo7lwzh5/P33_80_Figure.jpg[/url][/PLAIN]

    What EMF (in V) is induced in the metal loop at t = 5.0 μs? (Assume that only the circuit wire next to the loop is close enough to produce a significant magnetic field.)



    2. Relevant equations

    I=(V0/R)e^(-t/RC)
    dø=B . dA ?
    ε = dø/dt



    3. The attempt at a solution

    The RC circuit discharging equation gives me the current as a function of time. I=(V0/R)e^(-t/RC)

    First I tried just solving this for t= 5.0μs and got I = 6.065. Then I integrated to solve for the magnetic flux through the loop at that current and got 8.5x10^-8 T. This is where I am stuck now since I cant use ε = dø/dt to calculate the EMF with only ø and I cant differentiate this with respect to time with no time variable.

    So now I'm thinking that I need to keep the time variable in my equations instead of just solving for a specific time at the beginning so i can end up with dø/dt. I calculated dI/dt and tried to use this in my integral instead of I, only now I have no idea how to solve this integral with only first year calculus. Now I'm thinking that I may have missed/overlooked something that helps to simplify this problem?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 30, 2014 #2

    gneill

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    Staff: Mentor

    Note that magnetic field strength is given in T (Tesla), while the flux is in Tm2, or Wb (Webers).

    You're thinking along the right path.

    When you integrate to find the flux at some instant of time, the time variable can be treated as a constant. Easier: Just leave the current expressed as I(t) in your symbolic integration over the loop area. You can plug in the actual function of time for I(t) after the dust from the integration settles.
     
    Last edited by a moderator: May 6, 2017
  4. Mar 30, 2014 #3

    rude man

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    Homework Helper
    Gold Member

    While the intent of the problem may be to determine the emf generated by the circuit wire's B field as though the loop resistance were infinite, this is not the case here.

    The finite loop current generates its own B field which represents a counter emf around the loop. So there are two sources of emf around the loop: that generated by the circuit wire's B field and the counter-emf generated by the loop current.

    To compute the total emf in the loop requires knowing the self-inductance of the loop.

    To make matters more complex still, the circuit wire current is not simply what it would be in the absence of the loop. So there is also mutual inductance to be considered. Fortunately, not enough information is given to extend the problem to that level.

    BTW consider the circuit wire length to be infinite.
     
  5. Mar 30, 2014 #4
    Thanks for the help. Got the answer when I integrated this time.
     
  6. Mar 30, 2014 #5

    rude man

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    Did you include the loop resistance in your computations?
     
  7. Mar 31, 2014 #6
    No not in those calculations but the next question was to calculate the induced current in the loop so I used it then. Was I supposed to use it in the first part too?
     
  8. Mar 31, 2014 #7

    rude man

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    Well, as I said, I suspect the answer is no, because in an introductory course the only B field they want you to consider is the externally applied field (thhe one generated by the circuit wire). So you don't use R of the loop until you compute the current which will be a very simple procedure.

    But, as I said, the real picture is that the loop current generates its own B field which tends to counter the wire B field and generates its own emf ("back emf"), and the net emf in the loop is the algebraic sum of the two separate emf's. However, in order to compute the back emf numerically you'd need the self-inductance of the loop which is not at all easy to compute. So that's ususally encountered in an advanced e-m course. Either that or the self-inductance is given in the problem (there are websites that can compute it too). And, as I further said, the picture is even more complicated since the circuit wire current gets modified by the loop current via something called 'mutual inductance'.

    Bottom line: reality is often more complicated than can be ascertained by a simple formula! So just consider this a sample of what lies ahead if you decide to pursue an advanced e-m course. Meanwhile, what you did was what the problem intended, I'm sure.
     
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