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

Given, resistance, R =`6.0Omega`
Magnetic field, `B=2.57T`
Length of the rod, `l=1.20m`
The motional emf in the rod is given as, 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`
(i)The magnitude of force F required with be equal to the magnetic force acting on the rod, which oppses the motion.
`therefore" "F=F_(m)=lvB=ilB`
`or" "F=(1.0)(1.2)(2.5)N=3N`
(ii) Rate by which energy is delivered to the resistor is
`P_(1)=i^(2)R=(1)^(2)(6.0)=6W`
(iii) The rate by which work is done by the applied force is,
`P_(2)=Fv=(3)(2.0)=6WrArrP_(1)=P_(2)`