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Faraday's law on circular wire

  1. May 22, 2013 #1
    In my examples on Faraday's law in my book, they use a drawing of a magnet approaching a circular wire. The changing magnetic flux then induces an emf on the circle of wire, which in turn causes a current to flow.

    I'm wondering if a current will flow in that wire without it having an element to provide resistance, or if it's just drawn that way for simplicity. Perhaps the internal resistance of the wire itself is good enough.
  2. jcsd
  3. May 22, 2013 #2
    Current will exist with or w/o R. A changing B results in a Lorentz force on the wire's free charge carriers, namely electrons. As the electrons move around the loop, if no R is present no lattice collisions take place and no photons are emitted. So the wire stays cool. If R is present, collisions result in electrons droppint from conduction band into valence band, a lower energy state. To conserve energy photons are emitted in the infrared region which we feel as heat. In a superconductor there is an induced current but no heat since no R is present.

    The voltage around the loop will be related to the current per the inductive reactance value. In circuit theory lumped parameters are employed. In field theory distributed parameters must be considered because that is the real world.
  4. May 22, 2013 #3
    what would happen if there was a gap in the wire loop?
  5. May 22, 2013 #4
    You would have a small capacitance between the ends of the wire loop, resulting in a small displacement current. The induced voltage would be calculated per law of Faraday, and the current can be computed as the open circuit voltage divided by the loop total impedance. This impedance Zloop, is the wire resistance plus the inductance plus the capacitive reactance. Did I help?

    Last edited: May 22, 2013
  6. May 22, 2013 #5
    How does what you say relate to a magnet approaching a wire loop?
    I think impedance and reactance are associated with AC circuits and cannot see where they help here.
    How would you calculate the impedance of a wire loop with a gap and use it to analyse the effect of an approaching magnet?
  7. May 22, 2013 #6
    R = ρl/A

    C = ε0A/d

    L = μ0N2A/l

    V(open circuit) = -Nd(phi)/dt
    I = Voc/Z
    Z = jXL - jXC + R

    Last edited: May 22, 2013
  8. May 23, 2013 #7
    The formula you have given for inductance relates to a toroidal coil !!!!!
    This question is about a wire loop.
  9. May 23, 2013 #8

    Andrew Mason

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    Inductive and capacitive reactance are associated with any time dependent emf. If there is some small capacitance in the loop some small current will flow for a very brief time due to the fact that the induced emf around the wire loop will not be constant. I think that is all cabraham is saying.

  10. May 23, 2013 #9
    N=1! The wire cross section can be assumed rectangular, if round a correction factor is needed. The equation is valid for a solenoid, 1 to N turns.

  11. May 23, 2013 #10
    The current will exist longer than a "very brief time", but rather continuously. If open circuit voltage Voc is 10 volts, with R of 10 kohm, XC of -j1.0 Mohm, and XL of +j10 kohm, then the current I is given by

    10 / (10k - j1.0M + j10k) = 10.100 microamp with angle 89.42 degree.

  12. May 23, 2013 #11
    These calculations look interesting !!! is A the area enclosed by the loop (I assume this !!!) or the cross sectional area of the wire making up the loop?
    For the toroidal coil A is the cross sectional area of the coil....not the area enclosed by the coil !!
  13. May 23, 2013 #12
    R = ρl/Awire

    C = ε0Awire/d, where "d" is the space between the ends of the wire

    L = μ0N2Aloop/l, where "l' is the length of the solenoid, or height of a single turn loop.

    V(open circuit) = -Nd(phi)/dt
    I = Voc/Z
    Z = jXL - jXC + R

    I hope this helps.

  14. May 23, 2013 #13
    height of a single turn loop
    do you mean diameter of the loop?
  15. May 23, 2013 #14
    Diameter of wire if round, times adjustment factor. Or, for a rectangular wire, height of wire, which is thickness in direction normal to loop plane.

  16. May 23, 2013 #15


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    I assume the magnet is on the polar axis of the loop, as the effect will vary depending on geometry. Classically, we would say that as the magnet approaches the loop, field lines cut the loop and induce an EMF. If there is a "conducting" path around the loop the electrons can move and a current will flow. Post Maxwell, we say that the field of the moving magnet creates an electric field, and this encompasses the conductor of the loop, creating an EMF round the loop. Then, if a conducting path exists around the loop, a current can flow.
    When the electrons start to move they are accelerating, so they will radiate an EM wave.
    The radiated wave will have E and B fields. In addition, the magnetic field associated with the movement of electrons will constitute the magnetic part of the antenna induction near field. The electric field driving the electrons will constitute the electric part of the antenna induction near field. The energy lost to radiation will be permanently taken from the kinetic energy of the magnet. The reaction on the magnet of the induction fields will constitute stored energy which can be given back to the magnet.
  17. May 23, 2013 #16
    Ok... I assumed a wire of cross sectional area 1mm2 with a gap of 1mm.
    This wire forms a loop of rea 1m2
    I used C =ε0A/d and got C = 8.86x10-12F

    I took the area of the loop formed by the wire to be 1m2and the wire to be square 1mm x 1mm (l = 1mm in the equation for L)
    This gives me L = 4∏x10-7x10-3 H = 4πx10-10H

    How do you use these values of C and L to calculate Xc and XL?
  18. May 23, 2013 #17


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    The problem you have is that the EMF induced in the wire in your case is a complex function containing many frequencies, rather than, say, a sine wave which it would be for a loop rotating in a uniform field. In the rotating case, you find the reactance at the rotation frequncy and calculate the current from Ohm's Law. The total reactance of the loop will be Xl - Xc. You find reactances as follows: Xl = 2 pi F L and Xc = 1/2 pi F C.
  19. May 23, 2013 #18
    Of course we should not forget to include loop resistance R as well. The capacitive & inductive reactances are per tech99's post above. R + jXL - jXC = Zloop.

    Last edited: May 23, 2013
  20. May 23, 2013 #19


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    Hi Cabraham
    I was trying to keep it simple, as resistance is small compared with reactance. But if you want to use total impedance you have to add reactance and resistance vectorially,
    so Z = sqrt (R^2 + X^2)
  21. May 23, 2013 #20
    You see the values I got for C and L....do these look reasonable?....how would you calculate Xland Xc so that I can take this further.... I am OK with R
    Edit. I do not see any calculation of XL or Xc !
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