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The Back-EMF generated by an inductor in Volts is: \$V = -L\dfrac{di}{dt}\$
Where \$L\$ is the self-inductance and \$\dfrac{di}{dt}\$ the rate of current change.
Now, let's say I want to store those Volts on a capacitor \$(F, V_0)\$ where \$F\$ is the capacitor capacity in Farads and \$V_0\$ is the initial voltage on the capacitor.
How can I find out the final voltage on the capacitor? How much energy stored on the Back-EMF field?
The instantaneous energy stored in an inductor is
$$E = \frac{1}{2}L I^2$$
The energy stored in a capacitor is
$$E = \frac{1}{2}C V^2$$
You can see that there's a tradeoff between the capacitance value and the voltage required to store a particular amount of energy.
Start with current I0 flowing through inductor L.
Assume there is a perfect diode between L and C, and that C is large.
Switch off the current source, and $$\frac{dI}{dT}$$ will be negative, so the EMF V will be positive.
At V=V0, the diode turns on, so $$\frac{dI}{dT} = \frac{-V_0}{L}$$.
(approximately constant because we assumed C is large)
At I=0 there is no more energy stored in the inductor so the diode turns off, occurring at \$t = I_0 \cdot L/V_0\$.
Charge transferred Q is the integral of the current transferred over time,
$$Q = \frac{I_0 \cdot t}{2}$$
$$Q = \frac{I_0^2 \cdot L}{ (2 \cdot V_0)}$$
which will add dV = Q/C to the capacitor voltage.
(if dV is large, our assumption about C was incorrect)
The energy transferred in this cycle is simply $$E = Q \cdot V_0$$ or as Dave Tweed says, $$E = \frac{I_0^2 \cdot L}{2}$$
