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A Frictionless Bar Sliding due to a Magnetic Field and Emf

by gsquare567
Tags: field, frictionless, magnetic, sliding
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Dec6-10, 03:53 PM
P: 15
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.

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Dec6-10, 09:02 PM
HW Helper
P: 3,394
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
P: 15
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
HW Helper
P: 3,394
A Frictionless Bar Sliding due to a Magnetic Field and Emf

Most welcome.

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