A Frictionless Bar Sliding due to a Magnetic Field and Emf

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Homework Help Overview

The problem involves a bar sliding on parallel rails in a magnetic field, with a focus on deriving the speed of the bar over time given certain parameters such as mass, length, resistance, and electromotive force (emf). The context includes concepts from electromagnetism and dynamics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the speed of the bar by integrating a differential equation related to the forces acting on it. They express concern about their integration steps and the resulting equation not matching the expected form.
  • Some participants question the integration method used and suggest checking standard integral forms to clarify the process.
  • Others note the importance of ensuring that exponents in the derived equations are dimensionless, highlighting a potential oversight in the original poster's calculations.

Discussion Status

The discussion is ongoing, with the original poster actively seeking clarification on their approach and expressing gratitude for the feedback received. There is a collaborative atmosphere as participants offer insights into the integration process and the mathematical relationships involved.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. The original poster is grappling with the integration of a differential equation and the implications of their assumptions regarding the forces at play.

gsquare567
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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!
 
Last edited:
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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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