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Maximum emf of Rotating Loop

  1. May 9, 2007 #1
    1. The problem statement, all variables and given/known data

    A rectangular loop of wire has area A. It is placed perpendicular to a uniform magnetic field B and then spun around one of its sides at frequency f. The maximum induced emf is:
    • BAf
    • BAf
    • 4(pi)BAf
    • 2(pi)BAf
    • 2BAf

    Note: No, it's not my mistake that BAf appears as an option twice.
    The correct answer (verified by WebAssign) is 2(pi)BAf.

    2. Relevant equations


    [tex]d\phi_B=\vec{B}\cdot \vec{A}[/tex]

    3. The attempt at a solution

    Looking for maximum emf, so the dot product B*A simplifies to BA.
    The crux of the problem will be finding out how A varies with time.
    Frequency is given, and A will be very closely related to frequency.

    And then I didn't know where to go from there, so I guessed, and got lucky.

    Help with a systematic approach to this problem, please?
  2. jcsd
  3. May 9, 2007 #2


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    Can you write A in terms of an angle? Think of the standard wave function.... :wink:
  4. May 9, 2007 #3
    Basic wave function: [tex]y(x,t)=y_m\sin{(kx-\omega t)}[/tex]

    • y(x,t) will be the area perpindicular to the magnetic field at a given time
    • y_m=amplitude=actual area of the loop=A
    • [tex]kx-\omega t[/tex] is the angle that the loop makes with the magnetic field... I think...
    I feel like an idiot:cry:
    I've also been doing nothing but physics for a couple of days, please excuse me.
    :the sort of emoticon that looks all normal then spontaneously explodes, sending tiny bits of itself all over the screen, but it explodes in a sphere, of course, because space is isotropic:
    Last edited: May 9, 2007
  5. May 9, 2007 #4


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    I know the feeling, I've just done a QM exam today :grumpy: . But don't despair, your very close. Perhaps my hint of the wave function was a little unfair. Since you've already posted your homework I don't mind showing you this derivation. Okay, so you know Gauss' law;

    [tex]\Phi_B = \int \vec{B}\cdot d\vec{A}[/tex]

    Now, recall the definition of the dot product, namely [itex]\vec{A}\cdot\vec{B} = |A||B|\cos\theta[/itex] thus for a coil of area A and N loops we can write;

    [tex]\Phi_B = \int \vec{B}\cdot d\vec{A} = BAN\cos\theta[/tex]

    Now, from Faraday's law;

    [tex]\xi = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}\left( BAN\cos\theta \right)[/tex]

    Recall that [itex]\omega = \theta/t \Rightarrow \theta = \omega t[/itex]; hence,

    [tex]\xi = -\frac{d}{dt}\left( BAN\cos\omega t \right)[/tex]

    [tex]\xi = BAN\omega\sin\omega t[/tex]

    Since [itex]\omega = 2\pi f[/itex] we arrive at our desired result;

    [tex]\xi = 2BAN\pi f\sin(2\pi f t)[/tex]

    I hope this helps. Don't be to hard on yourself, if you've been doing nothing but physics lately, your probably very tired and very bored of physics. Try forgetting about physics for a few hours and doing something different, you'll probably find you'll come back refreshed :smile:
  6. May 9, 2007 #5
    No, it wasn't unfair at all. If anything, it was a leetle too fair o:)

    Thank you! Thank you! Thank you!
    I followed what you said, but deriving it myself would have been slow and tortuous work.

    by Piet Hein

    Do you know that weary feeling
    when your mind is strangely strangled
    and your head is like a ball of wool
    that's very, very tangled;
    and the tempo of your thinking
    must be lenient and mild,
    as though you were explaining
    to a very little child.
  7. May 9, 2007 #6


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    It was my pleasure. Don't worry about the derivation, its just practise :smile:
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