A conducting rod of length t moves on two horizontal frictio

A conducting rod of length t moves on two horizontal, frictionless rails, as in the figure. A constant force of 1.00 N moves the bar at 2.00 m/s through a magnetic field B that is directed into the monitor. What is the current through the 9.00 Q resistor R? 1 A What is the rate at which energy is delivered to the resistor? What is the mechanical power delivered by the force F_app?

Solution

Since F=IlB and = Blv

we have I=/R = Blv/R so B=IR/lv

a) since Fb= I^2*l*R/lv we have I=(Fbv/R)^1/2 =(1*2/9)^1/2 =.4714 A

b)P= I*I*R = .4714*.4714*9 =2W

c) For constant force, P = F v = (1.00 N)(2.00 m/s) =2W