Derive the expression for the mechanical power required to move a conducting rod of length $l$ with a constant velocity $v$ in a uniform magnetic field $B$.

(N/A) When a conducting rod of length $l$ moves with a constant velocity $v$ perpendicular to a uniform magnetic field $B$,an induced electromotive force $(EMF)$ is generated across the rod,given by $\epsilon = Blv$.
If the rod is part of a closed circuit with resistance $R$,the induced current $I$ is given by $I = \frac{\epsilon}{R} = \frac{Blv}{R}$.
The magnetic force acting on the rod is $F_m = IlB$. Substituting the value of $I$,we get $F_m = \left(\frac{Blv}{R}\right)lB = \frac{B^2l^2v}{R}$.
To maintain a constant velocity,an external mechanical force $F_{ext}$ must be applied equal and opposite to the magnetic force,so $F_{ext} = F_m = \frac{B^2l^2v}{R}$.
The mechanical power $P$ required is given by $P = F_{ext} \cdot v$.
Substituting the expression for $F_{ext}$,we get $P = \left(\frac{B^2l^2v}{R}\right)v = \frac{B^2l^2v^2}{R}$.