# Using the Lorentz force equation twice to model to different phenomena?

In a rail system, can I apply Lorentz force separately to:

• Derive the work required to move the charges(motion of current);

• Derive the kinematics relative to the rod's motion;

All stemming from:

$$F = qE + qv\times B$$

Deriving the electric force from:

$$F_E = qE$$

Deriving the Magnetic force:

$$F_M = qv \times B = IL \times B$$

The original equation is used to model each phenomena, but when do I need to combine them? When becomes important to consider the whole force $$F_E + F_M$$?

Only to know the total force acting on the charges?

In a certain sense you are already combining them to solve the system. The fact that makes it so simple is that for instance the charges that move in the (fixed part of the) wire are not too much influenced by the magnetic field, which only pushes them towards one side of the section of the wire. In the same way a difference of potential in the circuit would not move in any relevant way the rod if the magnetic field was absent.

Of course these are only approximations and in reality there will be many effects of backreaction (fields generated by the moving chaeges for instance), but the problem can be treated as in the textbooks as long as the external magnetic field and the difference of potential in the battery are taken as fixed values (which can be an almost arbitrarily good approximation).

There is a lot of confusion on this topic caused by misunderstanding of what the Lorentz force concept means. The Lorentz force is a force on microscopic charged particle, such as one of charge carriers in conductor. Its magnetic part refers to velocity $$\mathbf v_i$$ of such particle. The total Lorentz force on particle of charge $$q$$ is given by

$$\mathbf F_{Lorentz} = q\mathbf E + q\mathbf v_i\times \mathbf B ~~~(1)$$ It is a tiny force. Lorentz force does not give macroscopic EM force on the rod directly.

The whole Lorentz force (electric and magnetic part) gives $$total$$ external EM force acting on a charged particle, not on the rod or any macroscopic body.

• work required to move the charges(motion of current)

This is a more complicated question than what the Lorentz force is supposed to be for. The work done by the power source (battery in your picture) depends on ohmic resistance of the wires and the rod, and velocity of the rod $$\mathbf u$$ and its length $$L$$. If we ignore ohmic resistance, the work is due to having to work against the force of motional emf in the rod.

This emf is present whenever conductor moves in magnetic field. It is true that this emf can be derived from the microscopic forces that the Lorentz force formula is about. In this derivation, however, the velocity of the rod must be taken into account, not only velocity of the charged particles. Also, in addition to magnetic forces, constraint forces due to rod boundaries have to be taken into account (else the charged particles would jump out of the rod). When the rod moves perpendicular to magnetic field, the result of this analysis is that the electromotive force in the rod, called motional emf, is given by:

$$emf = -uBL$$ (the sign signifies that it opposes the current). Notice that the velocity there is that of the rod, not that of the charge carriers! If the rod stands still, there is no emf, even if current flows.

• Derive the kinematics relative to the rod's motion;

I am not sure what you mean by this. If the motion of the rod is what you're after, then you need to know all macroscopic forces that act on the rod. Ignoring friction, this boils down to macroscopic magnetic force on the rod. Macroscopic electric force is negligible in the depicted situation, because there is no strong external electric field.

This magnetic force is sometimes called ponderomotive force or Ampere force or Laplace force, as opposed to Lorentz force, because now we are dealing with macroscopic object.

This macroscopic magnetic force is given by the formula

$$\mathbf F = I \mathbf L \times \mathbf B~~~(2)$$

It is a different force from (1), both in its subject of action and in magnitude. Notice that (2) depends on current $$I$$, that is on velocity of the charge carriers, not on the velocity of the rod! There is net magnetic force on the rod even if the rod stands still.

Now, it is true that the force (2) can be derived considering the microscopic forces (1) and constraint forces on the constituent particles of the rod, but the derivation is not trivial.