# A Frictionless Bar Sliding due to a Magnetic Field and Emf

by gsquare567
Tags: field, frictionless, magnetic, sliding
 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 $$\epsilon$$ 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$$^{-B^2 d^2 t / m R}$$ 2. Relevant equations (1a) $$\Phi$$ = B * A (1b) $$\epsilon$$induced = -d$$\Phi$$/dt (2) F = I l x B 3. 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!
 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.
 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!
HW Helper
P: 3,394

## A Frictionless Bar Sliding due to a Magnetic Field and Emf

Most welcome.

 Related Discussions Introductory Physics Homework 10 Introductory Physics Homework 4 Introductory Physics Homework 10 Introductory Physics Homework 1 Introductory Physics Homework 2