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Emf induced in pendulum

  1. Apr 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle 2 theta. The earth's magnetic field component in the direction perpendicular to swing is B. The maximum potential difference induced across the pendulum is



    2. Relevant equations
    MAGNETIC FLUX=BAcos theta

    3. The attempt at a solution
    Area traversed by pendulum in 2 theta=theta * L^2
    Magnetic flux=BA=B*Theta*L^2
    EMF=d(B*Theta*L^2)/dt
    =BL^2*d(theta)/dt
    =BL^omega
    =BL^2*(g/L)^1/2

    But the answer is BL*sin (theta/2)*(gl)^1/2
     
  2. jcsd
  3. Apr 10, 2015 #2
    Mgl(1-cosθ) = 1/2 ml2 ω2
    Which gives ω= (2g(1-cosθ)/l)½
    Which gives E = Bωl2 /2
    E = Bl2sin(θ/2) (g/l)½
     
  4. Apr 10, 2015 #3
    I think this is the maximum emf produced because when the pendulum rotates through an angle θ the area swept by it first decreases then increases again for the other half of the motion. Therefore the maximum emf would be produced when it sweeps through an angle θ.
     
  5. Apr 10, 2015 #4
    You use ##\dot{\theta}=w## while ##\dot{\theta}## is something different here as pointed out earlier - we are not assuming uniform circular motion here remember.
     
  6. Apr 10, 2015 #5
    Hey
    Is my method right?
     
  7. Apr 10, 2015 #6
    You are right about there being a maximum emf, the original commenter assumes a constant ##\dot{\theta}## which is actually a function of time for the harmonic oscillator, the maximal value of this derivative depends on the initial angle of release as the first commenter pointed out.
     
  8. Apr 10, 2015 #7
    But if I replace the area to be transversed as theta, then is my approach correct?

    Area traversed by pendulum in theta=(theta * L^2)/2
    Magnetic flux=BA=B*(Theta*L^2)/2
    EMF=d(B*Theta*L^2)/dt
    =(BL^2)/2*d(theta)/dt
    =(BL^2)/2omega
    =BL^2/2*(g/L)^1/2

    But the answer is BL*sin (theta/2)*(gl)^1/2
     
  9. Apr 10, 2015 #8
    Coffee_ said:
    You use θ˙=w while θ˙ is something different here as pointed out earlier - we are not assuming uniform circular motion here rememb

    I suppose omega is constant in pendulum motion. It is equal to (g/L)^1/2
     
  10. Apr 10, 2015 #9
    ##w## is indeed constant, but for a harmonic oscillator ##\dot{\theta}(t)## is not the same as ##w##. Remember that ##\theta(t)=Acos(wt+\phi)## and so the derivative is still a function of time and not constant. You are correct up to the point of ##EMF=\frac{BL^{2}}{2}\frac{d\theta}{dt}## from this point on you make a mistake. What you have to do is realize that EMF is a function of time because ##\frac{d\theta}{dt}## is also a function of time. So what you have to do now is find the maximal value of ##\frac{d\theta}{dt}## - easiest way to do so is to use conservation of energy from the initial position.

    EDIT : To do so consider the relationship ##r\dot{\theta}=v## , and when v is maximal, ##\dot{\theta}## is maximal as well.
     
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