2
$\begingroup$

I was just wondering if this relation for the magnetic force exerted on a moving charge $\vec F = q\vec v\times\vec B$ actually has a derivation or is it just a definition? Im inclided to believe it is just the definition for magnetic force on moving charge that came about from experimental evidence and it does not actually have a mathematical derivation, but im not sure about this. Most of the resources i find on the internet just introduce the formula without any explanation from where it came about.

Also, i was wondering if the formula $\vec F=I\vec L\times\vec B$ for a uniform straight conductor was the first one to be 'discovered' or the one for a moving charge? Thank you

$\endgroup$
5
  • 1
    $\begingroup$ As far as I know $\vec{F} = q\vec{v}\times\vec{B}$ is a formula that synthesizes experimental observations. $\endgroup$ – JTS Apr 11 '19 at 18:47
  • 1
  • $\begingroup$ When you ask whether or not this force has a "derivation", what assumptions do you want to derive it from? $\endgroup$ – ACuriousMind Apr 11 '19 at 20:50
  • $\begingroup$ It defines the B-field. $\endgroup$ – ProfRob Apr 11 '19 at 21:34
  • $\begingroup$ I'm just asking if the equation is simply a definition which just came about or can it be derived through mathematical reasoning from other theories. I don't know if I'm expressing myself right though. $\endgroup$ – Luca Ion Apr 12 '19 at 13:39
0
$\begingroup$

It comes from experience with magnetic field acting on current-carrying wires (experiments and inference of force formulae from them was by Ampere, Biot, Savart, Laplace) and magnetic field acting on cathode rays (electrons) in electric and magnetic fields (experiments by Crookes, J. J. Thomson).

The generalization of the above formulae to the case of force acting on a charged particle was, although partially implied already by Thomson's work, established some years by H. A. Lorentz in his theory of electrons: he postulated that force on charged particle can be expressed as integral

$$ \int_{volume~containing~the~particle} \rho\mathbf d + \mathbf j\times \mathbf h\,dV $$

where $\mathbf d,\mathbf h$ are microscopic electric and magnetic field and $\rho,\mathbf j$ are microscopic electric charge density and current density of the particle.

Today, we often simplify this by neglecting any details of internal structure and write the force as function of total charge, velocity and external fields:

$$ q\mathbf E_{ext} + q\mathbf v \times \mathbf B_{ext} $$ which is more appropriate to description of experiments such as Thomson's cathode rays (we set up the macroscopic fields $\mathbf E_{ext}, \mathbf B_{ext}$, not the microscopic ones, those are hard to control. In this modern notation $\mathbf E_{ext}$ replaces Lorentz's $\mathbf d$ and $\mathbf B_{ext}$ replaces his $\mathbf h$.

$\endgroup$
1
$\begingroup$

Historically, it was simply discovered empirically. However, we now have a more fundamental understanding of why it has to be that way, based on special relativity. (To understand the following, you need to know about four-vectors in special relativity.)

If a particle of charge $q$ has a velocity four-vector $v$, then the force four-vector $F$ acting on the particle has to be a linear function of $q$ and $v$. The most general such relation is $F=q\mathcal{F}v$, where $\mathcal{F}$ is a 4x4 matrix.

Since $\mathcal{F}$ is a 4x4 matrix, it has 16 components, but these are not all independent. There is a relativistic constraint that the acceleration vector has to be orthogonal to the velocity vector (the constraint being necessary because the velocity vector has to stay normalized). This forces $\mathcal{F}$ to be antisymmetric, so it only has 6 independent components. Of these, 3 are identified as the components of the electric field and 3 as the components of the magnetic field. The standard Lorentz force law is the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.