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Rate of Magnetic Field Change

  1. Mar 30, 2006 #1
    A conducting loop of wire has an area of 6.9 10-4 m2 and a resistance of 110[tex]\Omega[/tex] . Perpendicular to the plane of the loop is a magnetic field of strength 0.18 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.18 A?

    here's what i've done so far:

    [tex]\theta[/tex] = 0

    i used Ohm's Law V = IR to find V which is the same as EMF, which came out to be 19.8 V. Then to find initial flux i used [tex]\phi_{i}[/tex] = (.018T) (6.9 x 10[tex]^-4[/tex]) (cos 0) and got 1.242 x 10[tex]^-4[/tex]

    this is where i think i might have gone wrong:

    i'm assuming they are talking about a change over 1 second, so [tex]\Delta[/tex]t = 1s

    and N = 1 since it originally says "A conducting loop of wire"

    so i set up the Emf formula like this: 19.8 V = ([tex]\phi_{f}[/tex] - 1.242 x 10[tex]^-4[/tex]) and solved for [tex]\phi_{f}[/tex] and of course i'm not getting the right answer... any suggestions??
    Last edited: Mar 30, 2006
  2. jcsd
  3. Mar 30, 2006 #2

    Doc Al

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    Staff: Mentor

    Start by writing Faraday's law, which relates induced EMF with the rate of change of the flux through the loop.
  4. Mar 30, 2006 #3

    i did: 19.8 = 1 (([tex]\phi_{f}[/tex] - 1.242E-4)/(1))

    [tex]\mid[/tex]E[tex]\mid[/tex] = 19.8 V
    N = 1
    [tex]\phi_{i}[/tex] = 1.242E-4 T
    [tex]\Delta[/tex]t = 1s
    Last edited: Mar 30, 2006
  5. Mar 30, 2006 #4
    anyone else?
  6. Mar 30, 2006 #5


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    You are fiding the final flux but that`s not what they are asking. They want the rate of change of the magnetic field!
    Use that [itex] {d \phi \over dt} = A cos \theta {d B \over dt} [/itex]. In your case cos theta = 1. Set this equal to the emf induced and dolve for dB/dt and see if you get the correct answer. As far as I can tell, the initial B field is not needed.

  7. Mar 30, 2006 #6
    ahh yes I don't why I thought finding the final flux was finding the rate of change :redface: thank you for clearing that up
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