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Induced current in a two-loop circuit

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data


    q.png



    I am using these variables:
    L for the square length - so the rectangle on the left is L/2 by L
    B' = rate at which field increases
    λ = Resistance/Length

    I1 = Current on the right (so the top/right/bottom of square) - goes counterclockwise

    I2 = Current on the left (so the far left and the top/bottom on the little rectangle piece) - counterclockwise

    I3 = current through the middle wire (what the question is asking for) - unknown direction, I assumed downwards


    2. Relevant equations

    V = IR

    Induced emf = derivative of flux

    flux = ∫ B dA


    3. The attempt at a solution


    I have tried setting up loop-rule equations and then just using matrices to calculate the values of I.


    Loop rule for the square:
    V = B' * L2 = I1 * 3Lλ + I3 * Lλ

    Loop rule for the whole outer rectangle
    V = B' * 1.5 * L2 = I1 * 3Lλ + I2 * 2Lλ

    Loop rule for the left rectangle:
    V = B' * .5 * L2 = I2 * 2Lλ + I3 * Lλ

    Using the sum of currents in = sum of currents out

    I1 = I2 + I3


    So I got 4 equations. I remember from non-magnetic field type circuit analysis questions with 3 unknown currents I had to use 2 voltage equations + sum of current equations, because the 3 voltage equations would not give enough information to solve the system. My guess is that any 2 voltage equations and the current equation would be all I need.


    I am getting the wrong answer with this set-up. Is there an error with my equations, or is this the completely wrong approach?
     
    Last edited: Apr 4, 2012
  2. jcsd
  3. Apr 4, 2012 #2

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    Hello OtherWindow,

    There are actually a variety of methods to solve the problem in terms of Kirchhoff's circuit laws. Using different methods leads to different equations. However, they will all ultimately come up with same answer in the end (if done correctly and with self-consistency).

    In my attempt to help below, I will try to stick as closely as possible to the methods in which you have already started. Just keep in mind, the hints that I describe are not the only way to solve the problem.

    I will stick with the above definitions.

    (And this is what I meant by my first paragraph. You could have defined the current loops differently, but the final answer would still come out the same. That said, the above definitions are fine the way they are, so I will use those.)
    Okay, that's a valid way to approach the problem so far.
    The above equation is inconsistent with the rules that you outlined above. You defined a loop on the left, and a loop on the right. Never did you define I1, or I2, or any other In in terms of going around the whole outer rectangle. The above equation doesn't belong according to your own definitions of I1, I2 and I3.

    (By the way, you could have defined your loops such that one of them is defined around the whole outer rectangle. Doing so means that you would be able to get rid of either one of your I1 or I2, since you wouldn't need both of them anymore.)

    (As a general guideline, once all the components in a circuit have at least one current loop associated with it, stop there! Defining more loops than are necessary doesn't help.)
    Which direction is I3 pointed? Are you summing in the same direction as I3 or the opposite? That has an impact on the sign of the I3 * Lλ term. Don't always assume everything is positive. If you're going against the direction of your own definition of a given current, the sign becomes negative.
    Okay that looks fine.
    Only 3 of your 4 equations are linearly independent (well, that and one has a minor mistake and the other doesn't even apply given the rules that were set up). You you really only have 3 independent equations.

    From here I suggest using substitution, linear algebra, or whatever your favorite method is for solving simultaneous equations.
     
  4. Apr 4, 2012 #3
    Thank you for your help. In an earlier attempt I had included negative signs only to remove them. I see why you would need them.


    So, I'll use the equation for the square:
    V = B' * L2 = I1 * 3Lλ + I3 * Lλ

    and a fixed version of the left rectangle

    V = B' * .5 * L2 = I2 * 2Lλ - I3 * Lλ


    I have these three equations set up now:
    ## (3L\lambda) I_1 + (0)I_2 + (L\lambda) I_3 = B' \, L^2 ##
    ## (0) I_1 + (2L\lambda)I_2 + (-L\lambda) I_3 = \frac{1}{2} B' \, L^2 ##
    ## (1)I_1 + (-1)I_2 + (-1)I_3 = 0 ##

    Trying this led me to an incorrect answer still, so I'm not quite sure what could have gone wrong.
     
  5. Apr 4, 2012 #4

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    I haven't spotted anything wrong with your equations. As far as I can tell, they look correct to me.

    But I know that solving simultaneous equations can be a tedious, error prone process. If you show your work maybe I can help figure out what went wrong.

    (I've solved the problem using two different methods of defining my loops and got the same answer. And in one of the methods, my equations are equivalent to the ones you wrote above, unless I really missed something. So it could be that the mistake that led to your incorrect answer was made in combining/solving the simultaneous equations.)
     
  6. Apr 5, 2012 #5
    Here is what I have gotten:
    scan0001-2.jpg


    My answer is boxed, it was cut off a bit by the scanner, it is in microAmperes.
    I know 448.1267 is not the answer.

    I suppose if this is still wrong I will have to show how I'm plugging in the numbers. (Wolfram|alpha makes typing in the matrices kind of a pain, unfortunately)
     
  7. Apr 5, 2012 #6

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    For what it's worth, your answer is the same answer that I got. Maybe it's a mistake in the book?
     
  8. Apr 5, 2012 #7
    That's reassuring to hear. This is actually from an online homework system done through the University of Texas (this is just for a high school class though).

    I do recall there being errors in the past, though. I'll talk to my teacher (he has access to the solutions) on Monday,.

    Thanks for the help, I definitely think I understand this better now.
     
  9. Apr 5, 2012 #8

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    I hope everything works out.

    Here is what I have to say about the confidence of your solution. If you plug your answers for I1, I2 and I3 back into your original equations, the equalities hold. That means your solved the simultaneous equations correctly.

    (By the way though, there are ways to make the process of solving the equations much easier than the way you did them. For example, in the first two equations you could have divided both sides by . And with hardly any effort, you could have substituted your third equation into the first two [for example, anywhere in the first two equations where there is an I1, substitute in I2+I3], thus reducing the number of simultaneous equations from 3 to 2. Two equations and 2 unknowns are easier to work with than 3. Whatever the case, your approach was valid and obtained the correct answers, assuming the original equations were correct.)

    So were the original equations correct? I think so. I've double checked a few different ways, and I don't find anything wrong with them (except that you can simplify them a bit if you wanted to -- but they're still correct).
     
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