I Asking about something that I read on electromagnetic force

AI Thread Summary
The discussion centers on the Faraday force equation, F = (1/2B)(dB/dt)mv, which relates to the acceleration of charged particles in a magnetic field. The equation arises when considering a particle of mass m and charge q moving in a uniform magnetic field B, where changes in B induce an electromotive force (emf) that accelerates the particle. The induced emf is linked to the changing magnetic flux and results in an electric field that exerts a force on the particle. The participants seek references or further clarification regarding the derivation and implications of this force equation. Overall, the conversation emphasizes the relationship between electromagnetic forces and particle dynamics in varying magnetic fields.
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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
 
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Here's a scenario where the force equation ##F =\frac 1 {2B} \frac{dB}{dt} mv## arises.
1674015562204.png

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##.
 
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