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Induced Current Question

  1. Apr 22, 2017 #1
    1. The problem statement, all variables and given/known data
    A 5.0-cm-diameter coil has 20 turns and a resistance of 0.50Ω. A magnetic field perpendicular to the coil is B=0.020t+0.010t^2, where B is in tesla and t is in seconds.

    Find an expression for the induced current I(t) as a function of time.

    2. Relevant equations

    3. The attempt at a solution

    [tex] \varepsilon = \frac{d\phi}{dt} [/tex]
    [tex] \varepsilon dt = d\phi [/tex]
    [tex] \int\varepsilon dt = \int d\phi [/tex]
    [tex] \phi = (0.020t + .010t^2) * \pi r^2 [/tex]
    [tex] \varepsilon \int dt = \int d\phi [/tex]
    [tex] \varepsilon t = \pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)] [/tex]
    [tex] \varepsilon = \frac {I}{R} [/tex]
    [tex] I = \frac{\pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)]}{tR} [/tex]
  2. jcsd
  3. Apr 22, 2017 #2


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    OK, except you haven't taken into account that you have 20 turns in the coil.

    Don't do this.

    Use your expression ## \phi = (0.020t + .010t^2) * \pi r^2 ## in ## \varepsilon = \frac{d\phi}{dt} ## along with the correction for the 20 turns.
  4. Apr 22, 2017 #3
    So [tex] I = \frac{20\pi r^2 [ (1/2) (0.020t^2) + (1/3)(.010t^3)]}{R} [/tex]

    So like this?
  5. Apr 22, 2017 #4


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    No. The numerator in your expression for ##I## does not equal ##\varepsilon##.

    If ## \phi = (0.020t + .010t^2) * \pi r^2 ##, then what is ##\frac{d\phi}{dt} ##?
  6. Apr 23, 2017 #5
    Oh ok I get it now. The answer should be [tex] \frac {20\pi r^2(0.020+.020t)}{R}[/tex]
  7. Apr 23, 2017 #6


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    Looks good.
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