I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle *moving* in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always has a self-inductance $L$ and some resistance $R$ (of the material it's made of - except for an *ideal* one..!) associated with it. Hence, Some work has to be done by *external agencies* in establishing current. This work done is stored as **electromagnetic** potential energy in an inductor. In case of capacitor, it's electrostatic..!

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$
(As @Alfred said, Power $P=VI$ is used here...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$ 


Negative sign shows the opposing nature of the emf (a consequence of Lenz law) Thus, it has somewhat the same expression as that of a capacitor $\frac{1}2CV^2$. The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an *increasing* order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is **cut-off**, the emf now opposes the *decay* of current. (i.e. It goes in a *decreasing* order)

We could conclude that, the work done by these *agencies* is referred to as the energy stored in an inductor. *Now, do you still relate both of them which are completely non-correlated..?*