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Homework Help: Electromotive force

  1. Dec 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Two vertical conducting rods separated by a distance d are connected to a capacitor of capacitance C. Another horizontal conducting rod of mass m is released at position y=y0 at time t=0 always in contact with the two vertical ones. The set is immersed in a uniform magnetic field B perpendicular to both rods.

    a) Calculate the acceleration; the velocity and the distance delta y at time t.
    b) What is the induced current?
    c) Analyse the balance of energy of the system.


    2. Relevant equations
    - standard electromagnetism formulas

    3. The attempt at a solution
    a)
    - calculate the emf:

    [tex]
    \varepsilon=\oint f_s \cdot dl
    [/tex]
    [tex]
    F_s=q(v\times B)
    [/tex]
    [tex]
    f_s=v\times B
    [/tex]
    [tex]
    \varepsilon=vBd
    [/tex]

    [tex]
    \varepsilon=V
    [/tex]
    Is the signal correct?

    - calculate the acceleration:

    [tex]
    a=g - F_M/m=g-[q(v\times B)]/m
    [/tex]

    Here I'm stuck: How can I get rid of q?

    [tex]
    v=\int_0^t a dt
    [/tex]

    [tex]
    y=y_0+vt+(1/2)at^2
    [/tex]

    b)

    [tex]
    I=dq/dt=CdV/dt=C\frac{d}{dt}(vBd)=CBd\frac{dv}{dt}=CBda
    [/tex]

    c)

    - formulate energy balance:

    [tex]
    mgy=(1/2)mv^2+(1/2)CV^2
    [/tex]
     
    Last edited: Dec 25, 2009
  2. jcsd
  3. Dec 25, 2009 #2

    diazona

    User Avatar
    Homework Helper

    Signal?

    If you're asking about [itex]\varepsilon = vBd[/itex], sure, that looks right. I would have used [itex]\varepsilon = -\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}[/itex] to get it, but the answer is the same either way.

    Try [tex]\vec{F}_M = I\vec{L}\times\vec{B}[/tex] instead... at least, that's all I can think of. If that's what they're after, it seems a little strange that the problem asks you to calculate acceleration before induced current, unless they want you to leave [itex]a[/itex] in terms of [itex]I[/itex].
     
  4. Dec 25, 2009 #3
    Oh, I see. More generally
    [tex]
    \overrightarrow{F}=I\oint_C \overrightarrow{dl}\times \overrightarrow{B}
    [/tex]
    Thank you.
     
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