Asking about something that I read on electromagnetic force

[hagopbul]

TL;DR

read a paper related to electromagnetic force and it mentioned a force i didnt heard before

Hello All:

read a paper related to electromagnetic force and its applications in acceleration of charges particles , some thing came up in it , they drive a force applied on the particle called Faraday force = [1/2B]* [dB/dt]*m*v

B magnetic field , m mass of the particle , v the velocity of the particle

i couldnt find any reference to this force , and didnt help notice the momentum of the particle in the equation
do any one have references about this force

Best
Hagop

 

[TSny]

Here's a scenario where the force equation $F =\frac 1 {2B} \frac{dB}{dt} mv$ arises.

Suppose there is a uniform B field in a region of space and there is a particle of mass $m$ and charge $q$ that circles the magnetic field lines with speed $v$. The figure is drawn for positive $q$. For nonrelativistic speeds of the particle, the radius of the orbit is determined to be $$r = \frac{mv}{qB}.$$ There is a flux of magnetic field through the circular path equal to $$\Phi = B \pi r^2.$$ If $B$ starts changing at a rate $\frac{d B}{dt}$, then there will be an induced emf in the path of the particle equal to $$\varepsilon =\frac{d \Phi}{dt} = \frac{dB}{dt} \pi r^2.$$ The emf is due to an induced electric field $E$ and the direction of $E$ at the location of $q$ will be in the direction of $\vec v$ if $B$ is increasing in strength. The relation between $\varepsilon$ and $E$ is $$\varepsilon = E \cdot 2 \pi r.$$ Thus, there will be an accelerating electric force on the charge given by $F = qE$.

If you put all of these relations together, you find that the accelerating force is $F = \frac 1 {2B} \frac{dB}{dt} mv$.