A Frictionless Bar Sliding due to a Magnetic Field and Emf

[gsquare567]

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) $$\epsilon$$induced = -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 $$\epsilon$$induced:
using (1a) and (1b),
$$\Phi$$ = B * A = Bxd (where x is the width of the area enclosed. it will change with velocity v.)
$$\epsilon$$induced = -d$$\Phi$$/dt = -Bd (dx/dt) = -Bvd
finally,
(3) $$\epsilon$$total = $$\epsilon$$ - 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$$\int$$du/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!

 

[Delphi51]

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.

 

[gsquare567]

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!

 

[Delphi51]

Most welcome.

 

[rwisz]

Your approach seems to be on the right track. However, there are a few mistakes that may be causing your incorrect result. Here are some suggestions to help you correct your solution:

1. In your equation (3), the total emf should be the sum of the applied emf (\epsilon) and the induced emf (\epsiloninduced), not the difference. So the correct equation should be: \epsilontotal = \epsilon + \epsiloninduced = \epsilon - Bvd.

2. In your equation (3), the total current should be the sum of the applied current (I) and the induced current (Iinduced), not the difference. So the correct equation should be: Itotal = I + Iinduced = I - (Bvd)/R.

3. In your equation (3), you have used the wrong sign for the induced emf. Since the induced emf is proportional to the rate of change of flux, it should be: \epsiloninduced = -Bd(dx/dt). This will change your final equation for velocity.

4. When you integrate with respect to time (dt), the limits of integration should be from 0 to t, not from 0 to v. This will give you the correct result for the integral.

5. When you back-substitute and simplify, you should get the following equation: t = mRln(\epsilon/Bd\epsilon - Bvd). Then, raising both sides by e and simplifying should give you the correct result for velocity: v = \epsilon/Bd(1 - e^{-B^2d^2t/mR}).

I hope this helps you to correct your solution. Keep in mind that when you are dealing with induced emf and current, the signs and directions are very important. Pay close attention to them and make sure you are using the correct equations and signs in your calculations. Good luck!