Figure shows the to views of a rod that can slide without friction. The resistor is `6.0Omega` and a `2.5T` magnetic field is directed perpendicularly downward into the paper. Let `l=1.20m`.

a. Calculate the force `F` required to move the rod to the right at a constant speed of `2.0 m/s`,
b. At what rate is energy deliered to the resistor?
c. Show that this rate is equal to the rate of work done by the applied force.

Explanation

The motional emf in the rod `e=Bvl` or `e=(2.5)(2.0)(1.2)v=6.0V`
The current in the circuit `i=e/R=6.0/6.0=1.0A`
a. The magnitude of force `F` required will be equal to the magnetic force acting on the rod, which opposes the motion.
`:. F=F_m=ilB` or `F=(1.0)(1.2)(2.5)N=3N`
b. Rate by which energy is delivered to the resistor is
`P_1=i^2R=(1)^2(6.0)=6W`
c. The rate by which work is done by the applied force is
`P_2=F.v(3)(2.0)=6W`
`P_1=P_2`