1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    Homework Helper

    Most welcome.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook