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A Frictionless Bar Sliding due to a Magnetic Field and Emf

  1. Dec 6, 2010 #1
    1. The problem statement, all variables and given/known data
    A bar of mass m, length d, and resistance R slides without friction in a horizontal plane, moving on parallel rails. A battery that maintains a constant emf [tex]\epsilon[/tex] is connected between the rails, and a constant magnetic field B is directed perpendicularly to the plane. Assuming the bar starts from rest, show that at time t it moves with a speed
    v = [tex]\epsilon[/tex]/Bd(1 - e[tex]^{-B^2 d^2 t / m R}[/tex]

    2. Relevant equations
    (1a) [tex]\Phi[/tex] = B * A
    (1b) [tex]\epsilon[/tex]induced = -d[tex]\Phi[/tex]/dt
    (2) F = I l x B

    3. The attempt at a solution
    To get the real voltage, I need to sum [tex]\epsilon[/tex] and [tex]\epsilon[/tex]induced:
    using (1a) and (1b),
    [tex]\Phi[/tex] = B * A = Bxd (where x is the width of the area enclosed. it will change with velocity v.)
    [tex]\epsilon[/tex]induced = -d[tex]\Phi[/tex]/dt = -Bd (dx/dt) = -Bvd
    (3) [tex]\epsilon[/tex]total = [tex]\epsilon[/tex] - Bvd

    Now, I use equation 2 to isolate for v:
    F = I l x B = ma
    from (3), we can get Itotal, which is I + Iinduced = ([tex]\epsilon[/tex] - Bvd) / R
    ( ([tex]\epsilon[/tex] - Bvd) / R ) (d) (B) = m a
    but a is just dv/dt...
    Bd( ([tex]\epsilon[/tex] - Bvd) / R ) = m(dv/dt)

    after separating dv and dt to opposite sides and integrating, i get:
    t = -mR/(b^2d^2) ln( ([tex]\epsilon[/tex] - Bvd) / [tex]\epsilon[/tex] )

    but raising both sides by e does not give me the required result:
    v = [tex]\epsilon[/tex]/Bd(1 - e[tex]^{-B^2 d^2 t / m R}[/tex]

    instead, i get
    v = [tex]\epsilon[/tex]/Bd(1 - e[tex]^{t + mR/(b^2d^2)}[/tex]

    i have been playing around with my math, and here is where I think i am going wrong.
    i separate the equation like so:
    dt = m(R/ (Bd[tex]\epsilon[/tex] - B^2vd^2)) dv

    then i integrate, left side from 0 to t, and right side from 0 to v. letting u = Bd[tex]\epsilon[/tex] - B^2vd^2...
    t = mR[tex]\int[/tex]du/u

    after solving (it becomes ln(u)), i back-substitute and have
    t = mRln( (Bd[tex]\epsilon[/tex] - B^2vd^2) / Bd[tex]\epsilon[/tex] )

    then i separate the ln and raise to the power e:
    e^(t/mR) = 1 - Bvd/[tex]\epsilon[/tex]

    but that still gets me:
    v = [tex]\epsilon[/tex]/Bd(1-e^(t/mR)), which is not the expected result. however, i am much closer. i am missing a B^2d^2 in the exponent =S

    I feel like I must be doing something seriously wrong here, perhaps with the way I am defining the force, because I checked over my integration and it seems fine to me.

    Last edited: Dec 6, 2010
  2. jcsd
  3. Dec 6, 2010 #2


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    Homework Helper

    I have left this as long as I can in hope someone who knows the answer would help. I think I have a wee bit of help for you. It looks to me like something has gone wrong in your integration step. Maybe just use the table which says integral[dx/(a+bx)] = -1/(b(a+bx))

    Differentiating the given answer does work back to your differential equation correctly. Also, the exponent must be dimensionless - the given answer has this feature, but your answer's exponent is in units of seconds.
  4. Dec 7, 2010 #3
    THANK YOU!!! I should look for those rules before using complicated substitutions like I tried. Also, your point about differentiating the answer to find what your integral should look like is a great idea! I will definitely apply those to future integration problems.

    Thanks again!
  5. Dec 7, 2010 #4


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    Homework Helper

    Most welcome.
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