# Metal bar on conducting rails

## Homework Statement

This is a Problem 7.7 fom Griffiths Introduction to Electrodynamics (3ed)

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 page, fills the entire region.

If the bar moves to the right at speed v, what is the current in the resistor ?

## Homework Equations

$$\Phi$$=BACos$$\phi$$
$$E$$=$$\frac{d\Phi}{dt}$$

## The Attempt at a Solution

my reasonig is:

magnetic flux is:
$$\Phi$$=BACos$$\phi$$=BA=B(A0+A1).

A0 is initial surface, and A1 is surface which bar makes moving to the right with spead v.
so:

A1=x*l=v*t*l , but v is also function of t, so: A1=v(t)*t*l

I know that equation for A1 is wrong, becouse when I try to get electromotive froce
I get this:

E=d$$\Phi$$/dt=B(0+v'(t)*t+v(t))

in solution manual it's:

E=Bl*dx/dt=Blv

Can someone explain why my reasoning is wrong, it seams logical to me.

## Answers and Replies

Doc Al
Mentor
A1=x*l=v*t*l , but v is also function of t, so: A1=v(t)*t*l
Setting x = v*t assumes that v is constant.

In any case, what matters is the rate at which flux changes, which depends on the speed at the moment in question:
d(A1) = l*v*dt
d(A1)/dt = l*v, even if v is changing.

Last edited:
Ok, than you.
That will help me to answer other question from that problem.