# Doesn't $F=ILB$ violate the conservation of energy?

Assume we have a current carrying conductor of finite-length $$l$$, carrying a current of $$I$$ (w/ a negligible cross-sectional area) under the influence of a Magnetic field, whose constant Flux density is $$B$$ (units ignored) All under a vacuum. Now, we know the famous equation, $$F = ILB$$ Where a conductor suffers a Force in the orthogonal direction to the Magnetic field.

Suppose now, we increase the length of this conductor, $$l' = kl \, \text{where}\, k\in R^+$$.

Thus, by our equation above, we will get $$k$$ times more Force than before.

Now, I'm well aware that $$\text{Force}\neq\text{Energy}$$, But $$W=Fs$$ where $$s\rightarrow \text{displacement}$$ and as I said, the entire system is in a vacuum. Therefore, after extending my conductor by $$k$$, We get:

$$W' = (kF).s = kW$$

So just by extending the conductor, I've obtained new, $$k$$ times more energy in the system. This conductor had negligible resistance (But $$\neq 0$$) so the loss of the current $$I$$ throughout the conductor would be negligible. $$B$$ is fixed. Nothing has changed except $$l' = kl$$, which shouldn't introduce new energy in the system at all (After all, we're just introducing free electrons to carry the current - like adding more pans in a conveyor belt waiting to be filled with sauce).

This would violate the conservation of energy, so I'm obviously wrong. But where?

Edit: To clarify, the conductor is a simple, straight one where current is flowing. The overall circuit is of little matter (if you made it around in a loop, then at a given instant you will observe a moment/torque, which would be $$2F$$ - yielding the same result but doubled. However, for the context of this question one can assume a simple, straight CCC)

• What is generating your external $B$ field? Is it possible that this is providing a source of energy? The scenario with $kL$ is no more a violation of conservation of energy than the scenario without $k$. Where did the energy transferred to the wire come from in the first case? It's safe to say the energy in the second case comes from the same source, it simply draws more energy from that source
– Jim
Commented Nov 25, 2022 at 14:46
• The geometry of this is unclear. Are you intending a straight conductor? If so are you allowing charge to build up at the end or do you have a return path of some sort that you are neglecting? Or are you intending the wire to form a loop? In which case a different formula would be more appropriate. Please edit the question to clarify
– Dale
Commented Nov 25, 2022 at 15:34
• @Jim The source of the external $B$ could be set constant - i.e, there is a constant amount of energy going into the system in the form of the $B$ electric field. You could generate it with a variety of apparatus. What matters it that the energy input-ted is the same. The definition of $B$ is defined per unit length of the conductor, so you can assume that is indeed constant. Commented Nov 25, 2022 at 17:24
• @neelg $B$ is not a measure of energy directly. To generate a magnetic field using, say, a wire is easy. However, if that magnetic field would interact with another object such that there's a transfer of energy, that acts like increased resistance in the wire. You need to pump in more voltage (energy) to run it. Having a constant $B$ is not the same thing as a constant input of energy
– Jim
Commented Nov 25, 2022 at 18:53