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A Frictionless Bar Sliding due to a Magnetic Field and Emf |
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| Dec6-10, 03:53 PM | #1 |
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A Frictionless Bar Sliding due to a Magnetic Field and Emf
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 finally, (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 so, ( ([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] EDIT: 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. Thanks! |
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| Dec6-10, 09:02 PM | #2 |
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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. |
| Dec7-10, 01:49 PM | #3 |
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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! |
| Dec7-10, 04:05 PM | #4 |
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A Frictionless Bar Sliding due to a Magnetic Field and Emf
Most welcome.
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