# Mass sliding through a metal bar consisting of two conducting rails

## Homework Statement

A metal bar of mass m slides frictionlessly on two parallel conducting rails a distance l apart. A resistor R is connected across the rails and a uniform magnetic field B , pointing into the page , fills the entire region.

a) If the bar moves to the right at speed v, what is the current in the resistor? In what direction does it flow?

b) What is the magnetic force on the bar? In what direction?

c) If the bar starts out with the speed v0 at t=0 , and is left to slide , what is its speed at a later time t

d) the initial KE of the bar was , of course 1/2mv0^2. Check that the energy delivered to the resistor is exactly 1/2mv0^2.

## Homework Equations

I=emf/R

emf=closed integral(fmag*dl)

## The Attempt at a Solution

a)I=emf/R, emf=vBh

in this case h is l. Therefore emf=vBl

I=vBl/R

b) emf=closed integral(fmag*dl)=vBl. Not sure what dl is. I supposed it the length of the metal bar?
If so, then
fmag*l=vBl==> fmag=vB

Didn't really begin on c and d but I will ask about those problems later when I attempt to solved them.

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Think of the magnet flux enclosed by the circuit. The dimension l is a constant. The bar can be thought of as sliding in the x direction, so

emf = -d(flux)/dt = -d(Blx)/dt = -Bldx/dt = -Blv

where the flux is found from the integral of B dot dA.

To find the direction of the current in the bar use

F = qv x B and the right hand rule. Remember, q is a positive charge carrier so the electrons will move in the opposite direction.

The force on the moving bar is found from

F = il x B

where i is the current in the bar. The right hand rule will give the direction of the force or you can use the concept of Lenz's Law to determine the force direction on the bar.