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Square loop, find the charge in the resistor

  1. Jul 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A square loop of wire is made up of 50 turns of wire, 45 cm on each side. The loop is immersed in a 1.4T magnetic field perpendicular to the plane of the loop. The loop of wire has little resistance but it is connected to two resistors in parallel as shown. (a) When the loop of wire is rotated by 180°, how much charge flows through the circuit?
    (b) How much charge goes through the 5.0Ω resistor?

    2. Relevant equations
    F = I(LxB)
    B= uNI/L
    V = IR
    flux = BAcostheta

    3. The attempt at a solution
    I'm stuck at this question and don't have a clue how to start. I think I have to somehow find the current flowing through the resistors, and somehow use that to find the charge. Can someone show me the steps to complete this problem?
    My partial attempt:

    flux = BA
    = 1.4*0.45^2
    =0.2835

    emf = flux*N/t
    =0.2835*50
    14.175
     
    Last edited: Jul 24, 2013
  2. jcsd
  3. Jul 23, 2013 #2

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    You might want to crop your image (it's taking up a lot of space.)

    Is this for a calculus based physics class? I'm guessing it's for a calculus based class.

    For this problem, there are actually several different places to start. This problem combines several different aspects of electricity and magnetism (such induced emf, equivalent resistance, charge vs. current relationship). And realistically, you could start on any one and come back to it later, combining things.

    The good news is that if you understand this problem, you'll get a pretty good understanding of a large part of the whole subject.

    So I'm not going to give much away. I'd like you to think about things as you go.

    In a manner of speaking, yes. You won't necessarily need to find the current directly, but a current variable might show up in one or two of the forthcoming equations.

    Yes. You will need to find the charge.

    Don't forget your units. But anyway, yes, that's the flux when the loop is at an angle θ = 0o.

    But let's not throw in numbers just yet (that's best done at the end).

    What's the flux as the angle θ changes? Can you express the flux as a function of A, B and θ? You'll end up needing that.

    In general that equation isn't true. That's the emf for the special case where the flux is changing linearly, at a constant rate [Edit: and only if the flux is 0 at t = 0]. But that's not necessarily the case for this problem. (Also, you forgot a negative sign.)

    Can you express the emf in more general terms? [Use calculus based notation such as "d()" and "dt".]
     
    Last edited: Jul 23, 2013
  4. Jul 23, 2013 #3
    Hi, this is actually a trig based physics class.

    If the flux changes 180 degrees would it be BAcos0-BAcos180 = 2BA = 0.567Tm^2. But whats the point of finding the flux if you can't use that equation.

    The most general way I can describe emf is emf = IR but I don't see how that's useful in starting this problem.
    I know how to find equivalent resistance, its just the first part of the problem I'm having trouble trying to work through.
     
  5. Jul 23, 2013 #4

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    That's good to know. I'll keep that in mind.

    And forget what I said then about your emf equation. But do change your '[itex] t [/itex]' to [itex] \Delta t [/itex], and change your [itex] BA [/itex] to [itex]\Delta (BA) [/itex] in that equation. The change in flux over the change in time. (And there is a minus sign).

    Good! You've found the change in flux. You will need that. Keep it in terms of BA for now though. (i.e. the change in flux is 2BA). We'll come back to this.

    Now what is the emf, in terms of B, A and [itex] \Delta t [/itex]?

    What is the relationship between current, I, and charge, Q?
     
  6. Jul 23, 2013 #5
    emf = -N(2BA)/t

    I = q/t

    I = emf/R (R = equivalent resistance) R equivalent = 10/3
    =-N(2BA)/(tR)=q/t

    q=-N(2BA)/(R)
    q = -50(2*1.4*0.45^2)/(10/3)
    q = -8.5 C

    Am I going in the right direction?
     
    Last edited: Jul 23, 2013
  7. Jul 24, 2013 #6

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    Yes. You're doing great. :smile:

    Yes you are going in the right direction! Very nice. :approve:

    I should correct myself on something though. Before, I agreed with you that the change in flux was 2AB. But if you think about it, the flux starts at AB, goes to 0, then continues to -AB. So the total change in flux is -2AB.

    What effect does that have?

    Carrying it through, it means that q = +8.5 C , not negative. There's two negatives that make a positive. Sorry if I confused you on that.

    -----------------

    So the next step is to figure out how much of that total charge goes through the 5 Ω resistor. I'll give you a hint: more than half of the total does. but can you figure out how much?
     
  8. Jul 24, 2013 #7
    Thanks for a correction, I realize now the signs should be switched.
    I think I just use the same equation for part B)
    so q=-N(-2BA)/(R)
    = 50(2*1.4*0.45^2)/(5)
    = 5.67

    Thanks so much for your help, I learned a lot with your explanations.
     
    Last edited: Jul 24, 2013
  9. Jul 24, 2013 #8

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    Very good! :smile:
     
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