Is the electric force that drives current equivalent to the Lorentz force acting on those currents?

Current flow ($$I$$) and the electric force responsible in moving the charges are proportional to the Lorentz force acting on those charge due to the magnetic field that the current flow produces, all based on the definition of the Lorentz force

$$F_L = IL \times B$$

However, I'm curious if they are equal or greater/less than one another. My assumption is it all depends on the design of the circuit and the resistance defined. The circuit can be very very low in resistance(yielding in high conductivity) and the force required(and work) to move the charge could be low.

I'm trying to see if I'm missing something with the following statement:

$$F_E \propto F_L$$

...Are the magnitudes necessarily equal? Or only proportional?

Example system:

Rectangular loop with current flow, with 4 sides, how side 1 is experiencing a force from the external magnetic fields of 2,3,4.

• Note: $F_L$ is the magnitude of the force experienced by a current-carrying wire in an external magnetic field. Are we assuming that there's another source of magnetic field present? – probably_someone Feb 12 at 15:43
• Generally the force on the current-carrying particles is discussed in terms of Ohm's Law, is it not? – honeste_vivere Feb 12 at 15:54
• @probably_someone for simplicity, consider a rectangular loop, with 4 sides. How side 1 experiences a force generated by sides 2-4, and etc... (Motion is not important here so, it's not a rail system). – UnkownConstants Feb 14 at 8:55
• @honeste_vivere Initially, yes. However, I wanted to make sure the proportionality are not strictly equal. I think they aren't. – UnkownConstants Feb 14 at 8:55

According to Ohm's Law, a wire with current $$I$$ and resistance $$R$$ has a potential drop of $$V=IR$$ across it. This means that the average electric field magnitude is $$E=\frac{V}{L}=\frac{IR}{L}$$ for a wire of length $$L$$. This, in turn, means that the average force that a charge in the wire feels is $$F_E=qE=\frac{qIR}{L}$$, which is pointed along the wire.
In contrast, the force felt by the wire due to the opposite current-carrying wire is $$F_L=ILB$$, pointed outward from the center of the square. In general, they are neither equal nor proportional to each other.
It's also worth noting that the Lorentz force is actually defined as $$\vec{F}=q\vec{E}+q\vec{v}\times\vec{B}$$. The form you're citing is a special case of this one.