How Does a Conducting Bar on Rails Behave in a Magnetic Field?

[krh68]

A conducting bar of length L = 21.2 cm and mass M = 60.0 g lies across a pair of conducting rails. The contact friction between the bar and the rails is negligible, but there is a resistor at one end with a value R = 30.0 Ohms. Initially the rod is given an initial speed of v0 = 64.0 meters per second. There is a uniform magnetic field perpendicular to the plane containing the rod and rails of magnitude B = 1.3 T.
What is the speed of the rod at time t = 26.068 s?

I know:
v=v0 + at
F=ma
F=iLB
i=(emf)/R
emf = dflux/dt
flux = BA

I know I need to solve for the area to get the flux and the length (L) is constant while the width is changing but I don't understand how to get the integral or set up the integral for the width. Please help ASAP.

 

[kuruman]

v = v₀+at is out of the picture. This is not a constant acceleration situation. All your other equations are relevant.
You need to set up a differential equation and solve it. Start with

F = m (dv/dt)

Replace F with iLB and then replace i with (1/R)(dΦ/dt). The expression for dΦ/dt is proportional to v. So you end up with the differential equation that is essentially

dv/dt = (const)v

You should be able to find what "const" is and to integrate the above equation.

 

[krh68]

Okay, so dv/dt = v(const) or dv/dt = v(LB/mR)
I still don't know how to solve for v.
When I integrate dv/dt, do I get r(LB/mR)? and if so, what is r?

 

[kuruman]

Okay, so dv/dt = v(const) or dv/dt = v(LB/mR)

This is incorrect. Please show how you got it, then I can point out where you went wrong.

I still don't know how to solve for v.

Worry about that later. First get the correct expression for dv/dt.

When I integrate dv/dt, do I get r(LB/mR)?

No, you do not.

and if so, what is r?

I don't know, but r it appears in your expression above. You made it up so you should know what it represents.