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Electrodynamics question - Induced EMF

  1. Nov 13, 2013 #1
    > A very long straight wire carries a current I. A plane rectangular coil of high resistance, with sides of length $a$ and $b$, is coplanar with the wire. One of the sides of length $a$ is parallel to the wire and a distance $D$ from it; the opposite side is further from the wire. The coil is moving at a speed $v$ in its own plane and away from the wire.

    >(a) Find the e.m.f. induced in the coil.

    >(b) Let R be the resistance of the coil. Calculate the force needed to move the coil with speed v as described, and show that the mechanical power used to move it is equal to the rate at which heat is generated in the coil.


    I have included my workings/thoughts.



    I know that i first have to calculate the magnetic field of the wire:
    $$B(y)=\mu_0 I/2πy $$
    Then the emf,
    $$\mathcal{E}=-\frac{dφ(B)}{dt}= -\frac{d}{dt} \int_D^{D+b}B\cdot ds = -\frac{d}{dt}(a\cdot \int_D^{D+b} B\cdot dy)$$


    I have been given the answer of
    $$\mathcal{E}=\frac{\mu_0Ivab}{2\pi D\left(D+b\right)}$$

    What I am having trouble with is the intermediate step getting from the integral to the above answer.

    I know that $P_{mech}=F.v$ and that $P_{heat}= v^2/R = \mathcal{E}/R$, but I do not know how to calculate the force on the coil.
     
  2. jcsd
  3. Nov 13, 2013 #2

    TSny

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    Hello mrmoriarty and welcome to PF!

    Can you show us your attempt at evaluating this integral?

    There is a basic formula for calculating the magnetic force on a current in a magnetic field.
     
  4. Nov 13, 2013 #3

    rude man

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    Do you know the Blv law? It's usually better than computing the flux change within the coil. In fact, sometimes where moving media are concerned this approach fails completely.
     
  5. Mar 6, 2014 #4
    Apologies for resurrecting the thread, but I have the exact same question set so it seemed pointless to make a new topic.

    I have done the first part (calculating the emf), but do not know how to find the force.

    I have tried using F = I dl x B with I = emf/R.

    [itex]F = \frac{\mu_0Iva}{2\pi D}\frac{a}{R}\frac{\mu_0I}{2\pi D} + \frac{\mu_0Iva}{2\pi(D+b)}\frac{a}{R}\frac{\mu_0I}{2\pi(D+b)}[/itex]

    [itex]F = (\frac{\mu_0Ia}{2\pi})^2\frac{v}{R}[\frac{1}{D^2}+\frac{1}{(D+b)^2}][/itex]

    This does not lead to the correct answer. Any help would be appreciated :)
     
  6. Mar 6, 2014 #5

    TSny

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    Hello, Zatman. Let's break it down.

    Let ##I'## be the current induced in the loop. Find an expression for the force on the side of the loop closest to the long wire in terms of ##I## and ##I'## where ##I## is the current in the long wire. Don't bother yet to substitute for ##I'##. Just express the force in terms of ##I, I', a## and ##D##.
     
  7. Mar 6, 2014 #6
    Hi TSny, thanks for your reply. I have:

    [itex]F = \displaystyle\int^a_0 I'\frac{\mu_0I}{2\pi D}\ dl[/itex]

    [itex]F = \frac{\mu_0aII'}{2\pi D}[/itex]
     
  8. Mar 6, 2014 #7

    TSny

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    Great. Now do the same for the far side of the loop. Also, think about the directions of the forces. Then combine the two forces.
     
  9. Mar 6, 2014 #8
    Okay, so the only things that will be different are
    (1) the sign, since the current is in the opposite direction.
    (2) D --> D+b

    So the force on the far side of the loop is

    [itex]F_2 = -\frac{\mu_0aII'}{2\pi (D+b)}[/itex]

    The net force is then the sum of the two forces:

    [itex]F_{net} = \frac{\mu_0aII'}{2\pi}(\frac{1}{D}-\frac{1}{D+b})[/itex]

    [itex]F_{net} = \frac{\mu_0aII'b}{2\pi D(D+b)}[/itex]
     
  10. Mar 6, 2014 #9

    TSny

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    Fantastic. Now, how would you express the mechanical power required to move the loop at speed v?
     
  11. Mar 6, 2014 #10
    [itex]P = \frac{dW}{dt} = \frac{d(Fx)}{dt} = F\frac{dx}{dt} = Fv[/itex]

    [itex]P = F_{net}v = \frac{\mu_0IvabI'}{2\pi D(D+b)}[/itex]

    Now I think I can substitute the expression for I':

    [itex]I' = \frac{emf}{R} = \frac{\mu_0Ivab}{2\pi RD(D+b)}[/itex]

    Which gives the correct answer. I see my mistake now; I somehow managed to have two different currents in each section of the wire.

    Thanks for your help! :)
     
  12. Mar 6, 2014 #11

    TSny

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    Good work!

    One minor point where you wrote

    [itex]P = \frac{dW}{dt} = \frac{d(Fx)}{dt} = F\frac{dx}{dt} = Fv[/itex].

    The middle expression is not actually correct, but the other expressions are correct.
     
  13. Mar 6, 2014 #12
    Yes, it is only valid for a constant force? Which in this question is not actually the case, I notice.
     
  14. Mar 6, 2014 #13

    TSny

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    It would be valid for a constant force, but I don't think it's a good idea to write W = d(Fx)/dt even for that case. During a small time dt, the work done is Fdx. So the rate of doing work is (Fdx)/dt = F(dx/dt) = Fv.
     
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