A Frictionless Bar Sliding due to a Magnetic Field and Emf

AI Thread Summary
A bar sliding on parallel rails in a magnetic field is analyzed to determine its speed over time, with the equation v = ε/Bd(1 - e^(-B^2 d^2 t / m R)) derived from the principles of electromotive force and magnetic induction. The discussion highlights the need to account for both the applied voltage and the induced voltage due to motion, leading to the total voltage equation ε_total = ε - Bvd. The user struggles with integration steps and the correct formulation of the force, suggesting a misunderstanding in the integration process. A suggestion is made to utilize a simpler integral approach to resolve the dimensional issues in the exponent of the derived equation. The conversation emphasizes the importance of careful mathematical manipulation and verification in physics problems.
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Homework Statement


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 \epsilon 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 = \epsilon/Bd(1 - e^{-B^2 d^2 t / m R}


Homework Equations


(1a) \Phi = B * A
(1b) \epsiloninduced = -d\Phi/dt
(2) F = I l x B


The Attempt at a Solution


To get the real voltage, I need to sum \epsilon and \epsiloninduced:
using (1a) and (1b),
\Phi = B * A = Bxd (where x is the width of the area enclosed. it will change with velocity v.)
\epsiloninduced = -d\Phi/dt = -Bd (dx/dt) = -Bvd
finally,
(3) \epsilontotal = \epsilon - 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 = (\epsilon - Bvd) / R
so,
( (\epsilon - Bvd) / R ) (d) (B) = m a
but a is just dv/dt...
Bd( (\epsilon - Bvd) / R ) = m(dv/dt)

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

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

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

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\epsilon - B^2vd^2)) dv

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

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

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

but that still gets me:
v = \epsilon/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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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.
 
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!
 
Most welcome.
 
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