While analysing a force changing the motion of a particle, I always like to imagine the simplest case I can think of: an elastic collision of a ball with a block, and then ask myself in what terms does the system I am focused on vary from the straightforward model:

Straightforward model

When a ball collides elastically with an initially at rest block (isolated system), momentum is conserved and, besides, the change in momentum equals to the force exerted on the block.

$$\vec F = \frac{d \vec p}{dt}$$

enter image description here

Because the magnitude of the velocity of the ball doesn't change after the collision, we note that the final momentum of the ball equals to two times the initial momentum of the ball (in terms of magnitude of course).

Case I am currently focused on

I am analysing the following circuit, which is half-contained in a uniform magnetic field $\vec B$.

enter image description here

I am trying to justify $f_{mag}$ on this diagram:

enter image description here

I see that I cannot justify $f_{mag}$ by Newton's Second Law; the force must be perpendicular to the distance the charge takes (magnetic forces do not work).

Actually it would be nice to have an analogy for $\vec F = \frac{d \vec p}{dt}$ for the magnetic force.

Why cannot we state an analogy? Why $f_{mag}$ behaves like that (i.e. why must be perpendicular to the distance the charge takes)?

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    $\begingroup$ The collision you describe between the ball and the block isn't elastic. Your system has more energy after the collision than before it. $\endgroup$ – BioPhysicist Apr 13 '19 at 21:51
  • $\begingroup$ @AaronStevens I see. For the collision to be elastic $KE_f = KE_i$ and here after the collision the block gains some $KE$ so it doesn't hold. $\endgroup$ – JD_PM Apr 13 '19 at 22:48
  • $\begingroup$ Right, but I don't think it's essential to the heart of the question. I would edit accordingly though $\endgroup$ – BioPhysicist Apr 13 '19 at 22:50
  • $\begingroup$ @AaronStevens If you prefer we can delete these comments and I can edit my question stating this is an inelastic collision. $\endgroup$ – JD_PM Apr 13 '19 at 22:54

(1) "Actually it would be nice to have an analogy for 𝐹⃗=𝑑𝑝⃗𝑑𝑑." You don't need an analogy; 𝐹⃗=𝑑𝑝⃗𝑑𝑑 applies to the magnetic force as it does to any force. It gives the rate of change of momentum of a particle acted on by $\vec F$ alone.

In fact the particles in your set-up are 'free' electrons in ab, but these are acted on not only by $\vec f_{mag}$ but also by resistive forces (due to collisions with ions in the wire) and by electrostatic forces that effectively transmit $\vec f_{pull}$ to the particle. The resultant force is zero, so the electron doesn't gain momentum, but instead 'drifts' through the wire at constant velocity $\vec u,$ while the wire itself is moving at velocity $\vec v,$ giving the electron a resultant velocity $\vec w.$

(2) The 'magnetic Lorentz' force acts at right angles to the electron's velocity $\vec w.$ This is the nature of this particular force. A resistive force would act in the opposite direction to a particle's velocity. The force due to an electric field would act in a direction unrelated to that of the particles's velocity. It is not the role of 𝐹⃗=𝑑𝑝⃗𝑑𝑑 or any analogue thereof to tell us how the force is related – or not related – to the particle's velocity. The actual role of 𝐹⃗=𝑑𝑝⃗𝑑𝑑 is to tell us how the particle responds to a resultant force, as explained in (1) above.

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