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Velocity of a bar on parallel conducting rails

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data

    A pair of parallel conducting rails a distance l apart in a uniform magnetic field B⃗ . A resistance R is connected across the rails. The bar has mass m and is initially at rest. A constant force F⃗ e to the right is applied to the bar. Find its velocity as a function of time in terms of R. l, B, m, Fe, and t

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    First part of the question asked to formulate Newton's second law for the bar, and my answer was m(dv/dt) = Fe - (B^2*l^2*v)/R.

    This is correct, but trying to get v(t), i move Fe to one side, everything else to the other, then factor v out (not sure if that is allowed...) and then divide the Fe by what is left. I get:

    (Fe*R*t)/(m*R+B^2*l^2*t)

    Any help would be greatly appreciated, thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 10, 2014 #2

    BvU

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    Hello Bear and welcome to PF,

    Good thing you know your first part is OK. I agree.

    Next, bringing Fe to one side gets you an equation with v and dv/dt on the righthand side, a real differential equation ! v and dv/dt are different beasts and how you factor them out is a mystery to me.

    Any experience with differential equations ?
     
  4. Apr 10, 2014 #3
    Thanks for the reply, i was way too tired yesterday and not thinking very well... And i have some, but i really struggle with them (been awhile since i took calculus, and i only took the first half). If Fe is on one side, is the next step to evaluate m(dv/dt) from vo to v, and since it is initially at rest it leaves just mv? then for (B^2*l^2*v)/r from 0 to v(t) gives (-2B2lv^2)/r

    This still does not seem correct to me though...
     
  5. Apr 10, 2014 #4

    BvU

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    No, to me neither. You can check if what is left over satisfies the differential equation ...which was
    m(dv/dt) = Fe - (B^2*l^2*v)/R. Of the form $$ {dv\over dt} + \alpha \, v =\beta, \quad \quad \alpha = {B^2\,l^2\over mR},\quad \beta = {F_e\over m} $$ give it a go, good to freshen up calc!
     
  6. Apr 10, 2014 #5
    Would the answer to that be v(t) = β/(1 + [itex]\alpha[/itex]t)?
     
  7. Apr 10, 2014 #6

    BvU

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    Well, what do you get when you fill it in in the differential equation ?
    $$ {dv\over dt} + \alpha \, v = {-\alpha \beta \over (1+\alpha t)^2} \ + {\alpha \beta \over 1 + \alpha t} $$ which doesn't look like ## \beta ## to me.

    Did you ever get to solve something like $$ {dv\over dt} + \alpha \, v = 0 $$in the past ?
     
  8. Apr 11, 2014 #7
    Hey, well the assignment is over now, the correct answer was (Fe*R/B^2*l^2)(1-e^(-b^2*l^2*t/Rm)

    My final attempt was Fe/m - e^(B^2*l^2*t/Rm)

    I still got an 88 on the assignment, so its not the worst, thanks for the help though
     
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