A motor drives a large metal flywheel with moment of inertia $J$ to a high angular speed ${\omega }_0$. A magnetic field $B$ is constant over the area of the flywheel and in a direction perpendicular to the plane of the wheel. At a particular time the motor is disconnected from the wheel and sliding electrical contacts are made to the rim of the wheel and to the metal spindle driving the wheel. The circuit between the contacts is completed with a load of resistance $R$. What is the initial current through the load and how long does it take for the current to reduce to one-half of the initial value?

At the instant the drive is disconnected and the electrical circuit complete the initial current is 

$I_0=\frac{1}{2R}{\omega }_{0}B{a}^2$

A current $I$ through the load $R$ at time $t$ dissipates energy at the rate $I^{2}R$. This must be provided at the expense of the rotational energy, $E$, of the flywheel. Hence 

$\frac{dE}{dt}+I^{2}R=0$ …………(1)

If the rotational speed at time $t$ is $\omega$, the rotation energy is $E=\frac{1}{2}J{\omega }^2$ and the current is $I=\omega B{a}^2/2R.$

Thus, $E=\frac{2J{R}^2{I}^2}{B^2{a}^4}$ ………………..(2)

After differentiating equation (2) and substituting in equation (1) we get, 

$\left(\frac{4JR}{B^2{a}^4}\right)\frac{dI}{dt}+I=0$

The solution to this differential equation is

$I=I_0{e}^{-t/\tau }$…………..(3)

where $\tau =\frac{4JR}{B^2{a}^4}$

The current thus falls to one-half its initial value in a time $\left(\frac{4JR}{B^2{a}^4}\right)$ times $\ln 2$