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Faraday's Law and electron force

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    For the situation described in Figure P31.32, the magnetic field changes with time according to the expression B = (8.00t^3 - 4.00t^2 + 0.800) T, and r2 = 2R = 5.00 cm.

    Figure P31.32
    (a) Calculate the magnitude and direction of the force exerted on an electron located at point P2 when t = 2.00 s.

    2. Relevant equations
    E = -0.5r(dB/dt)
    F = qE + (qv X B)

    3. The attempt at a solution
    E = -0.5r(dB/dt)
    F = qE + (qv X B)

    Alright, the force equation should simplify to just F = qE because B is zero outside the circle.

    When I plug in and get:
    F = 0.5qr (24t^2-8t) and plug in for the values:
    q = 1.602 x 10^(-19)
    r = 0.05m
    t = 2

    I get: 3.204 x 10^(-19) which is wrong...

    What am I doing wrong here?
  2. jcsd
  3. Oct 30, 2008 #2


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    Homework Helper

    Hi Ithryndil,

    I believe this is where your error is; this equation is not correct. How did you get it, and do you see what it needs to be?
  4. Oct 30, 2008 #3
    The equation is right out of my book.
  5. Oct 30, 2008 #4


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    Homework Helper

    My guess is that in your book it is part of an example problem; in your diagram for example that equation would apply to the point P1, but not to point P2.

    You need to derive an equation that applies to point P2, starting with Faraday's law.
  6. Oct 30, 2008 #5
    Alright, yeah looks the equation applies to inside the circle, not to the outside...alright, well it's late, I will work on that derivation tomorrow after some sleep. Thanks for the insight.
  7. Oct 30, 2008 #6
    Thank you. I got it. The expression of E I needed is:

    E = [R^2(24.0t^2 - 8.00t)]/(2r) where R is the radius of the circle and r is the radius of the electron's orbit.
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