One end of a horizontal track of gauge $l$ and negligible resistance, is connected to a capacitor of capacitance $C$ charged to voltage $V_0$. The inductance of the assembly is negligible. The system is placed in a homogeneous, vertical magnetic field of induction $B$, as shown in the figure.

A frictionless conducting rod of mass $m$ and resistance $R$ is placed perpendicularly onto the track. The polarity of the capacitor is such that the rod is repelled from the capacitor when the switch is turned over. Under what conditions is the efficiency of this 'electromagnetic gun' maximal?

The relations in previous part show that the maximum velocity of the rod is proportional to the initial voltage $V_0$ across the capacitor. Thus, the final kinetic energy of the rod is proportional to $V_0^2$ (for given values of $C$ and $m$), i.e. proportional to the initial energy of the system. The coefficient of proportionality can be regarded as the efficiency \eta of the apparatus (considered as an electromagnetic gun), and can be written in the form 

$\eta =\frac{\frac{1}{2}m{v}_{max}^2}{\frac{1}{2}C{V}_0^2}=\frac{1}{{\left(\frac{\sqrt{m}}{Bl\sqrt{C}}+\frac{Ml\sqrt{C}}{\sqrt{m}}\right)}^2}$

The product of the two terms in the brackets is $1$, and from the inequality between arithmetic and geometric means, it follows that their sum is at least $2$. This means that the efficiency of the electromagnetic gun cannot be more than $25$ percent.

Note: If the condition for maximum efficiency $m=C{B}^2{l}^2$ is substituted into the expression for the final charge on the capacitor we find that $Q_{min}=V_{0}C/2$, i.e. only half of the initial charge on the capacitor is left. Thus only one quarter of the initial energy of the capacitor is left; one quarter of it is transformed into the kinetic energy of the rod, and the other half is dissipated in the rod as Joule heat.