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A problem dealing with Faraday's Low of Induction

  1. Oct 13, 2005 #1
    Hi,

    My question:

    A 22.0 cm diameter coil consists of 20 turns of circular cooper wire 2.6mm in diameter. a uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.65 x 10^-3 T/s. Determine the current in the loop.

    I'm not sure how to approach this equation. I was thinking I should use the following equations:

    R=((rho)L)/A, to find the resistance

    Emf=-N(magnetic flux)/t, for the voltage

    then using the resistance and voltage that I found from the equations use them to find the current, I.

    I=(V)/R

    Is anyone of this correct:confused:

    Thank You:smile:
     
  2. jcsd
  3. Oct 13, 2005 #2

    mezarashi

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    Homework Helper

    I think there are some typos in your question or I don't understand how something can have 2 diameters at the same time. In addition, no resistivity was given.

    But in general it looks like the right approach. Try working it out and see how the math goes.
     
  4. Oct 13, 2005 #3
    Okay:smile:

    Work:

    Beginning with resistance
    R=((rho)L)/A
    R=((1.68 x 10^-8 ohm x m)(22 x10^-2 m))/(pi(2.6 x10^-3))^2
    R= 6.96 x 10^-4 ohm

    Next Emf for volts
    E=-N(magnetic flux)/t,
    E=-(20)(-8.65 x 10^-3 T/s)((pi(2.6 x10^-3))^2)
    E= 9.19 x 10^-7 volts

    Finally the current
    I=E/R
    I=(9.19 x 10^-7 volts)/(6.96 x 10^-4 ohm)
    I=.001 A

    Is this okay?

    Thank You:smile:
     
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