I'm having trouble with the following problem:

A bar of mass m, length d, and resistance R slides without friction on parallel rails. A battery that maintains a constant emf E is connected between the rails, and a constant magnetic field B is directed perpendicular to the plane of the page and out of the page. If the bar starts from rest, show that at time t it moves with a speed:

v = E/Bd(1 - e^{(-B^2)(d^2)t/(mR)})

in diagram the bar is on the left and the battery is set up on the right so the current runs counterclockwise through the circuit.

The first thing that I'm confused about is the force that sets this in motion. From the problem statement and the diagram, the magnetic field is caused by the current running through the wire, so unless some force is applied to the bar it's not going to move.

I begin trying to solve the problem by working with the following equation:

Sum of the Forces in the x direction = -IdB + F_applied = ma
F_applied = mass x some initial acceleration?

Is this the correct approach. I've tried many and I'm got getting the correct answer.

Can anyone nudge me in the right direction?

Thanks

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Doc Al
Mentor
Originally posted by discoverer02
The first thing that I'm confused about is the force that sets this in motion. From the problem statement and the diagram, the magnetic field is caused by the current running through the wire, so unless some force is applied to the bar it's not going to move.
That constant magnetic field coming out of the paper is not caused by the wire. You have a current carrying wire in a magnetic field, which exerts a force on it.

Thanks again for your help Doc Al.

I already tried this approach and didn't seem to lead me in the right direction.

The magnetic force on the rod, in this case, is directed to the left because dF = Ids X B = IdB to the left.

So the forces in the x direction are -IdB = ma

The induced E-field in the rod once it starts moving is v x B = vB.

So the induced emf in the rod once it starts moving is

$$\int Efield \cdot ds$$

which is vBd.

The induced current is vBd/R in the direction opposite the current from the battery because the magnetic flux is increasing in the loop because the area is increasing, right.

How do I treat the current already in the loop? Do I include it in the equation of my forces and subtract from it the current created by the back emf? If I do, I'm not coming up with the right answer.

Doc Al
Mentor
Think of it this way. The EMF across the wire is:

&Delta;V = E - EMFinduced

EMFinduced = DBv

Current through the wire is:

I = &Delta;V /R

Force on the wire is:

F = IDB

Acceleration of the wire is:

a = dv/dt = F/m = IDB/m

Write and solve the differential equation. Make sense?

It makes sense.

Thanks.