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Homework Help: Lenz's Law

  1. Nov 7, 2009 #1
    This is not a homework problem due. It's practice. I have the answer of .198mV. I don't know how to get it.

    1. The problem statement, all variables and given/known data


    2. Relevant equations

    [tex]\phi_b = \int \vec B \cdot d\vec A[/tex]

    [tex]E = -\frac{d\phi_b}{dt}[/tex]

    3. The attempt at a solution

    The magnetic field due to the solenoid is:

    [tex]\mu_o i N[/tex]
    [tex](1.26x10^{-6})(1.28 A)(85400 turns/m) = 1.37x10^{-1} T[/tex]

    The flux through the circular loop is:

    [tex]\phi_b = \int \vec B \cdat d\vec A[/tex]
    [tex]\phi_b = BAcos 0 = BA[/tex]

    [tex]B = 1.37x10^{-1}[/tex] [tex]A=6.8x10{-3}[/tex]
    [tex]\phi_b = BAcos 0 = BA = (1.37x10^{1})(6.8x10^{-3})= 9.34x10^{-4} Wb[/tex]

    To find the EMF induced:
    [tex]E = -\frac{d\phi_b}{dt}[/tex]

    I don't know where to go from here. How do I relate that 212rad/s to the problem?
  2. jcsd
  3. Nov 7, 2009 #2


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    Try putting I=I0sin(ωt) to get the magnetic field.

    Then use Φ=NBA

    and then use E=-dΦ/dt
  4. Nov 7, 2009 #3
    So instead of using

    [tex]B=\mu_o i_o N[/tex] we use:

    [tex]B=\mu_o i_osin(\omega t)N[/tex] which leads to

    [tex]\phi = BA = \mu_o i_osin(\omega t)NA[/tex]

    [tex]E = -\frac{d\phi_b}{dt}[/tex]

    [tex]E = -\frac{d}{dt}\mu_o i_osin(\omega t)NA[/tex]

    [tex]E = -\mu_o i_oNA\frac{d}{dt}(sin(\omega t))[/tex]

    Is this how you meant? I think I'm lost.
  5. Nov 7, 2009 #4


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    right yes and what is d/dt(sinωt) ?
  6. Nov 7, 2009 #5
    As far as I can tell, (cos t)(ω)? Maybe??
  7. Nov 7, 2009 #6


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    So then

    [tex]E=\mu_0 I_0 NBA \omega cos(\omega t)[/tex]

    so what is the amplitude?
  8. Nov 7, 2009 #7
    The amplitude of the wave is 1.28 at maximum. But I dont know what that gets me.
  9. Nov 7, 2009 #8


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    no that is for the current

    [tex]E=\mu_0 I_0 NBA \omega cos(\omega t)[/tex]

    What is the amplitude of E?
  10. Nov 7, 2009 #9
    I don't understand.
  11. Nov 7, 2009 #10
    The amplitude is equivalent to the maximum point on the wave that a wave equation describes.

    Remember that cosine (and indeed sine) functions vary between -1 and 1, so the maximum cos value you can get is 1.

    So what is the maximum that E can be in that equation?
  12. Nov 7, 2009 #11
    (I think we have a typo. There shouldn't be a B in the equation, should there?)

    Would it be: [tex]
    E=\mu_0 I_0 NA \omega
    because cos (ωt)=1?
  13. Nov 7, 2009 #12


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    Yes the B should be there.

    so the amplitude would be

    [tex]E= \mu_0 I_0 NBA\omega[/tex]

    EDIT: sorry you are right, it is [itex]E=\mu_0 I_0 NA \omega[/itex]
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