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Current induced in moving rectangular wire

  1. Jun 4, 2016 #1
    1. The problem statement, all variables and given/known data
    As shown in the figure below, a current I passes through a straight conducting wire located within this page. Rectangular circuit ABCD on the page is moving away from the straight conducting wire with speed v, while side AB remains parallel with current I. The length of side AB is 2a, and the length of side AD is 2b. The resistance of the circuit is R. How much current flows through circuit ABCD when the center of the circuit (O) is distance r from the straight wire?
    Untitled_zpsyc5eku6y.png


    2. Relevant equations
    B = μ0 I / (2πd)

    V = I.R

    Faraday's and Len's Law

    3. The attempt at a solution
    emf will be produced because there is change in magnetic field experienced by the rectangular wire as it is moving away from the straight wire but I don't know how to find the rate of change of magnetic field.

    I have tried to find the magnetic field at AB and CD:
    B at AB = μ0 I / (2π (r - b))
    B at CD = μ0 I / (2π(r+b))

    Difference of magnetic field between AB and CD = μ0 I 2b/ (2π(r2-b2))

    I don't even know why I found those things above, I just tried anything. And also, what is v? is it 2b/t? or is it b/t?

    Thanks
     
  2. jcsd
  3. Jun 4, 2016 #2
    Assuming r is the distance at time t=0, then at a time t the circuit will occupy the region of space from ##x_1(t)=r+vt-b ## to ##x_2(t)=r+vt+b## and the magnetix flux passing through it will be ##\phi(t)=2\alpha\int\limits_{x_1(t)}^{x_2(t)}f(x)dx##. Can you see what the function f(x) is?
     
  4. Jun 5, 2016 #3
    I understand until this part
    I don't get this part. The formula of magnetic flux that I know is Φ = BA cos θ

    What is α? And also, magnetic flux is integration of position function?

    Thanks
     
  5. Jun 5, 2016 #4
    Sorry I used the greek letter for a, cause that's what is used in the diagram of the original post.

    The formula you know for flux is valid only if the magnetic field B is the same across all surface A. But in our case the magnetic field changes in the x-direction (though it doesn't change in the y-direction). If you haven't been taught the magnetic flux as a double integral over the x,y dimensions then I am not sure how to proceed on this problem.

    Another approach maybe is to use the formula ##E=Bvl## for the pieces of wire (AB) and (DC) (at BC and AD we ll have little to no voltage generated (why?))...
     
  6. Jun 5, 2016 #5
    Looks to be the simplest way to solve this problem.
     
  7. Jun 5, 2016 #6
    E at AB means emf (potential difference) between A and B and E at DC means potential difference between D and C. The value will be different because the value of B is different for the two sections so it means for same value of resistance R, the value of current flowing through AB and DC will differ?

    I also don't understand why at BC and AD we'll have little to no voltage generated. There will be emf induced in the rectangular coil because there is change in magnetic flux experienced by it. There is induced emf means there will be current flowing and the current should be the same at all sections (AB, BC, CD and DA) so the voltage at BC and AD will be at a considerable level, no?

    Thanks
     
  8. Jun 5, 2016 #7
    Sorry I didn't state it accurately, there will be voltage at BC and AD but no EMF generated by these segments.

    The current in the circuit will be everywhere the same, AB and DC will act as EMF sources with different values (but same internal resistance).
     
  9. Jun 5, 2016 #8
    There will be no EMF induced at section BC and AD because the orientation of the wire is parallel to the direction of motion. Am I correct?

    But if we see the formula E = Bvl or E = Bvl sin θ (θ is angle between B and v), there is B experienced by BC and AD and θ is 90o so there should be induced EMF at BC and AD?

    If I use formula B = μ0 I / (2πd) to calculate the value of B at BC and AD, what value of d should I use?

    How is voltage produced at BC and AD?

    Thanks
     
  10. Jun 5, 2016 #9
    Voltage in BC and AD is purely the result of Ohm's law V=IR. The EMF generated in BC and AD will be zero( if we treat BC and AD as one dimensional line segments with zero thickness) and that is exactly for the reason you stating (orientation of wire parallel to the direction of motion).

    if you check the derivation of the formula ##E=Bvlsin\theta## you ll see that the orientation of wire is vertical to the direction of motion.
     
  11. Jun 5, 2016 #10
    BC and AD still experience magnetic field from the straight wire, right? What value of distance should I use if I want to calculate the value of B on BC and AD?

    Thanks
     
  12. Jun 5, 2016 #11
    For a given time (frozen time) t, which means for specific distance r from the source wire with current I, the magnetic field changes over BC (or AD) as we move from B to C from ##\mu\frac{I}{2\pi(r-b)}## at B to ##\mu\frac{I}{2\pi(r+b)}## at C. This variation in the spatial coordinate doesn't mean that it induces EMF in the wire. If we put## r=r(t)## in the above formulas then the field is time-dependent (and that is because of the movement of the wire) but still the EMF induced is zero in BC and AD , simply because the orientation of wire is parallel to the direction of motion.
     
  13. Jun 5, 2016 #12
    Oh ok. It means that to find the average B we need to use integration.

    Thanks a lot for the help
     
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