Why Is My Reasoning Incorrect for the Current in the Resistor?

[natugnaro]

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.

 

[Doc Al]

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.

 

[natugnaro]

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