0
$\begingroup$

I do not understand how does the result of two vectors acting on a particle require me to take the cross product to find the resultant.

Won't the actual force on the particle be the result of the vectorial addition of the two vectors? How is it that result of two vectors on a charged particle in a magnetic field (the magnetic field vector and the velocity vector of the particle) has the resultant force on the particle in a mutually perpendicular direction?

$\endgroup$
5
  • $\begingroup$ Because electric/magnetic fields are not force fields? $\endgroup$ Commented Oct 30, 2012 at 9:41
  • $\begingroup$ Lets take the electric field... The strength or magnitude of the electric field at a given point is defined as the force that would be exerted on a positive test charge of 1 coulomb placed at that point; the direction of the field is given by the direction of that force. They are 'force fields'... $\endgroup$
    – MoonKnight
    Commented Oct 30, 2012 at 9:57
  • $\begingroup$ @Killercam the charge is not of 1 Coulomb, because such a big charge (or any finite charge, for the matter) would alter charge distributions which could be the source of the electric field, therefore altering the field itself. Instead, you take the force felt by a charge $q\rightarrow 0$ (this limit is not rigorous, because of charge quantization, but this is not relevant for macroscopic purposes)and you divide it by $q$. Besides, none of them is a force field because they are dimensionally not forces. If you mean that their field lines lay above the force lines, this is true only for $\vec E$ $\endgroup$ Commented Oct 30, 2012 at 10:25
  • $\begingroup$ The term 'force field' is 'loose'. A force field is a vector field that describes a non-contact force acting on a particle at various positions in space. I conceed that magnetic force on a charged particle, depends on the particle's velocity as well as its position and is thus not a 'Force Field' - however, due to electrostatics (and the perfectly sound definition I provide above); and the fact that Electric Field Intensity is a vector quantity, the Electric Field is a force field... $\endgroup$
    – MoonKnight
    Commented Oct 30, 2012 at 10:48
  • 1
    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/29133/2451 $\endgroup$
    – Qmechanic
    Commented Oct 30, 2012 at 15:10

2 Answers 2

3
$\begingroup$

I'm not sure if I understood what you mean.

You want to know why the Lorentz force is given by, $$ \vec F = q \vec v \times \vec B $$ instead of $$ \vec F = q\vec v + \vec B\,, $$ am I right? For a start, the second equation is not dimensionally correct. It would if $\vec B$ would be a generic outer force and $q$ would be some fluid friction coefficient, but neither of those quantities are such. $q$ is a charge, $\vec v$ is a velocity and $\vec B$ is a magnetic field. They are not forces, and should be not treated as forces.

This is the reason why the second equation is wrong. As for the reason why the first equation is right, the best answer i can give you is just "because". It is just a fact, an axiom you base electromagnetism on. You could reprhase it in term of other axioms regarding the simmetry of reality between electric forces and magnetic forces, using field theoretical arguments (e.g. introducing the electromagnetic tensor $F_{\mu\nu}$ and its coupling to the 4-current $J_\mu$), but there is no specific reason (as far as we know) by which reality should have these symmetries rather than others.

$\endgroup$
1
$\begingroup$

The magnetic field is not a vector. It is a pseudovector, it transforms differently upon space inversion (mirror reflection if you prefer). The most natural way to construct a vector from a vector and pseudo vector is to take their cross product.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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