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Homework Help: Finding the resistance, magnetic flux, and current of a loop section

  1. Aug 29, 2014 #1
    ...based on the angle.

    1. The problem statement, all variables and given/known data

    Figure 30-78 shows a wire that has been bent into a circular
    arc of radius r = 24.0 cm, centered at O. A straight wire OP can be
    rotated about O and makes sliding contact with the arc at P.
    Another straight wire OQ completes the conducting loop. The
    three wires have cross-sectional area 1.20 mm2 and resistivity
    1.70 10^-8 Ωm, and the apparatus lies in a uniform magnetic
    field of magnitude B = 0.150 T directed out of the figure.Wire OP
    begins from rest at angle θ = 0 and has constant angular acceleration
    of 12 rad/s2. As functions of θ (in rad), find (a) the loop’s
    resistance and (b) the magnetic flux through the loop. (c) For what
    u is the induced current maximum and (d) what is that maximum?

    The image of the problem: http://i.imgur.com/TJVSmAK.jpg

    2. Relevant equations

    Flux = ∫BdA
    ε = -d(Flux)/dt
    B = μ_0*i/2pi*r

    3. The attempt at a solution

    For part a), I solved using the resistivity equation, rho = R*A/l, with l = 2*r+ rθ, and solved for R.

    For part b), I started with Flux = ∫BdA, took B outside of the integral (since it is uniform). I went to the integral solution for a circle, so I could just plug in the angle to get the area of the circle, which gave me 0.24^2/2 * θ = 2.88E-2*θ. That's my guess of where to go, anyhow.

    What's throwing me off is I'm given the angular acceleration of the wire itself. I am at a loss of where to go. From worked out problems in the book, the velocity of the loop itself figures into the problem. But that's usually because they're trying to find the emf of the loop itself. Am I overthinking this second part? Should the solution just be Flux = B*2.88E-2*θ?
  2. jcsd
  3. Aug 29, 2014 #2
    Looks good to me. The angular acceleration of the wire is only important when you have to figure out what the rate of change of the magnetic flux is with respect to time.
  4. Aug 30, 2014 #3
    I'm confused about part c) now. Hah.

    I'm certain I have to take the equation I found for the flux and take the derivative of that to find the induced EMF. But all I have is the angular acceleration for the wire, which is 12 rad/s. So, the double integral of that is 6t^2. I put that into the equation I found for flux based on angle, take the derivative of it and now I have an equation for EMF based on time.

    If I take the equation I found for resistance, and replace theta with 6t^2 and divide EMF/R, I can find the induced current. Correct? All I have to do is take the derivative of that, set to zero and solve?

    Afterwards, I can take what that equation gives me for t, put it into theta = 6*t^2 and solve.
  5. Aug 30, 2014 #4
    Could you include a picture of the problem statement given in the book?
  6. Aug 30, 2014 #5
  7. Aug 30, 2014 #6
    All that sounds good.

    I wasn't quite sure what you meant by u in this:
  8. Aug 30, 2014 #7
    Okay, so I think I went down the wrong track. I got 2 radians for when the current is at a maximum, but when I solved for current I got 6.0E9 Amps which seems a tad high.
  9. Aug 30, 2014 #8
    Your result for θ matches mine, but your value for the current does not :wink:

    I usually convert all given values to SI base units before I plug anything in. You probably just need to go through them again.
  10. Aug 30, 2014 #9
    Okay, I found my mistake. I am not used to this new scientific calculator, so it did something I wasn't expecting. The current is now approx 6.0E-15, which sounds more realistic.

    Thank you so much for your help!
  11. Aug 30, 2014 #10
    Hmm, I get a maximum value of approximately 2.20 A.
  12. Aug 30, 2014 #11
  13. Aug 30, 2014 #12
    If you want to be sure, you can write out your expression for the current and the values you plug into it so we can go through it together.
  14. Aug 30, 2014 #13
    How is that? EMF has an approx magnitude of 1E-2, and R has an approx magnitude of 1E-12. Any value of t I enter in will give me something around 1E-15.
  15. Aug 30, 2014 #14
    i = ε/R

    ε = -5.18E-2*t

    R = 4.8E-12 + 1.44E-11*t^2

    t = 0.577439 s

    That is what I have.
  16. Aug 30, 2014 #15
    It helps if we start with the symbolic expression. Does this:
    i(t) = \frac{\alpha \, r A_w B \, t}{\rho(\alpha t^2 + 4)}
    where ##A_w## is the cross-sectional area of the wires and ##\alpha## is the angular acceleration of ##OP##, match yours?
  17. Aug 30, 2014 #16
    No. I have

    i(t) = B*r^2*θ'*A / 2ρ(2r+rθ)

    For θ', I have 12*t, and θ I have 6t^2.

    I'm not sure why you have angular acceleration *and* time in the equation.
  18. Aug 30, 2014 #17
    Sorry, ##\alpha## is the constant value of angular acceleration.

    If you insert your expression for θ' and θ into i(t) and simplify a bit, your expression will match mine.

    Are these:
    r = 0.24 \, \mathrm{m}\\
    A_w = 1.20 \times 10^{-6} \, \mathrm{m^2}\\
    \rho = 1.70 \times 10^{-8} \, \mathrm{\Omega \cdot m}\\
    B = 0.150 \, \mathrm{T}\\
    \alpha = 12 \, \mathrm{\frac{rad}{s}}
    the values you're substituting into your expression for i(t)?
  19. Aug 30, 2014 #18
    It wasn't originally. I converted the area of the wire incorrectly. I simply had 1.2E-3, instead of E-6.
  20. Aug 30, 2014 #19
    Assuming t = 0.577439 is the correct time, I ran the equation again with the correct wire area size and I'm still getting a ridiculous number for i.
  21. Aug 30, 2014 #20
  22. Aug 30, 2014 #21
    I'm beginning to think I don't know how to use this calculator, and how it wants expressions entered into it.

    I found one error on my part (didn't include r^2 in the top expression), and now my answer is close to what yours is, at i = 3.78 A.
  23. Aug 30, 2014 #22
    Also, I've been using the wrong time as well. I now have the answer you got. Woof.

    Thank you again!
  24. Aug 30, 2014 #23
    You're welcome.
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