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Solinoid EMF question

  1. May 14, 2009 #1

    takumi_91

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    heres the question:

    A 110 turn coil has a radius of 4.1 cm and a resistance of 25 Ohm. The coil is in a uniform magnetic field that is perpendicular to the plane of the coil. What rate of change of the magnetic field strength will induce a current of 4.3 A in the coil? What rate of change of the magnetic field strength is required if the magnetic field makes an angle of 20° with the normal to the plane of the coil?




    2. Relevant equations

    i know that i need to use faradays equation:E = -n(dtheta/dt), and Ohm's law:V =IR

    i dont know how to get 'dt' for my equation. i know there is another way to do this problem but i cant really come up with anything at the moment
     
    Last edited: May 15, 2009
    takumi_91, May 14, 2009
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  3. May 15, 2009 #2

    waht

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    you are asked to find the rate of change of the magnetic field strength (I assume B not H since it is hard to figure this out from the context)


    [tex] \frac{dB}{dt} [/tex]

    we know that

    [tex] \mathcal{E} = -N \frac{d\Phi_B}{dt} [/tex]

    so what is?

    [tex] \Phi_B[/tex]
     
    waht, May 15, 2009
  4. May 16, 2009 #3

    overt26

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    Given the question refers to magnetic field strength rather than flux density, then
    I guess you could (for the purpose of obtaining a solution) "safely" assume it's an air cored coil and free of the influence of any nearby magnetic material.
     
    overt26, May 16, 2009
  5. May 16, 2009 #4

    Triangulum

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    [tex]\PhiB[/tex] = B * A (dot product)
     
    Triangulum, May 16, 2009
  6. May 16, 2009 #5

    Triangulum

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    Magnetic Flux = B dot A

    Because the Area is constant
    d[tex]\Phi[/tex]/dt = dB/dt * A cos 20

    so emf = N d[tex]\Phi[/tex]/dt

    We know emf because of ohm's law V=iR

    Therefore,

    iR = dB/dt * N * A cos20

    dB/dt = (iR)/(NA cos20)
     
    Last edited: May 16, 2009
    Triangulum, May 16, 2009
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