Question

Two conducting rails 39.2 cm apart rest on a 4.11° ramp. They are joined at the bottom by a 0.520 Ω resistor, and at the top a copper bar of mass 0.0450 kg is laid across the rails. The whole apparatus is immersed in a vertical 0.580 T field. What is the terminal (steady) velocity of the bar as it slides frictionlessly down the rails?

Answer 1

power into the bar=power in the resistor.

rate of change of GPE=V^2/R
Velocity*sinTheta*mg= (B dA/dt)^2 /R
velocity*sinTheta*mg=( B* dArea/dt )^2/R

Area=2W+2L
dARea/Dt=2 *velocity

so if my algebra is right (check it)
velocity= Resistance*sinTheta*mg/B^2

check that again.