Prove that the mechanical power required to move a rod in a uniform magnetic field is converted into electrical power.

(N/A) Let $r$ be the resistance of the movable arm $PQ$ of the rectangular conductor shown in the figure. We assume that the remaining arms have negligible resistance compared to $r$. Thus,the total resistance of the loop is $r$.
The induced electromotive force (emf) is given by $\varepsilon = B v l$.
The current $I$ in the loop is:
$I = \frac{\varepsilon}{r} = \frac{B v l}{r}$ ... $(1)$
Due to the magnetic field,a magnetic force acts on the arm $PQ$. This force $\vec{F} = I \vec{l} \times \vec{B}$ acts in the direction opposite to the velocity of the rod. The magnitude of this force is:
$F = I l B = \left( \frac{B v l}{r} \right) l B = \frac{B^2 l^2 v}{r}$ ... $(2)$
Power delivered by the external force: To move the arm $PQ$ with a constant velocity $v$,an external force equal in magnitude to $F$ must be applied in the direction of motion. The mechanical power required is:
$P_{mech} = F v = \left( \frac{B^2 l^2 v}{r} \right) v = \frac{B^2 l^2 v^2}{r}$ ... $(3)$
Electrical power produced: The electrical power dissipated in the resistance $r$ is:
$P_{elec} = I^2 r = \left( \frac{B v l}{r} \right)^2 r = \frac{B^2 v^2 l^2}{r^2} \times r = \frac{B^2 l^2 v^2}{r}$ ... $(4)$
Comparing equation $(3)$ and $(4)$,we see that $P_{mech} = P_{elec}$. Thus,the mechanical power required to move the rod is exactly converted into electrical power,which is then dissipated as heat in the resistor.