Why does velocity increase force according to F_m = QvB?

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An increase in the velocity of a charged particle in a magnetic field leads to a greater magnetic force, as described by the equation F_m = QvB. This relationship highlights that the magnetic force is dependent on the particle's velocity and the magnetic field strength. In contrast to gravitational forces, which behave differently, the electromagnetic force increases with velocity due to the nature of magnetic interactions. The phenomenon can be better understood through the lens of Special Relativity, which explains the behavior of electromagnetic forces. Ultimately, the observed effects are consistent with experimental findings in physics.
Arcthor
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This is probably a very noob question, but I can't seem to get my head around it.

Why does an increase in velocity of a particle in a magnetic field, increase the force acted upon it, according to F_m = QvB?

For example in orbital dynamics, an increased velocity of a body simply breaks the orbit, the force does not magically increase, correct?

Could anyone guide me in the right direction? What does the F=QvB actually mean?
 
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This phenomenon is similar in case of any orbital motion of a particle(object) against gravitational force or electromagnetic force. Both centripetal force and the centrifugal force must be balanced sharply at circular orbital motion. The centrifugal force is born due to change of angular momentum of the moving body and it is directly proportionate to square of velocity. If velocity is increased the centrifugal force too is increased simultaneously and the object is driven off the orbit due to unbalance of forces.
 
Arcthor said:
For example in orbital dynamics, an increased velocity of a body simply breaks the orbit, the force does not magically increase, correct?

Centrifugal force is equated as-
FC = mv2/r

So yeah, as the velocity increases, it breaks orbit and the force increases as well.
 
Arcthor said:
Why does an increase in velocity of a particle in a magnetic field, increase the force acted upon it, according to F_m = QvB?
"Why" questions are tough. I would say that it is a result of applying Special Relativity to Coulomb's law.
http://physics.weber.edu/schroeder/mrr/mrr.html

But if you don't know relativity then it is probably a useless answer, and a more useful answer would simply be that this is what has been observed to happen experimentally.

Arcthor said:
For example in orbital dynamics, an increased velocity of a body simply breaks the orbit, the force does not magically increase, correct?
Gravity behaves differently from electromagnetism. However, there are in fact analogous effects for gravity, the most famous being called "frame dragging".
 
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Arcthor said:
Why does an increase in velocity of a particle in a magnetic field, increase the force acted upon it, according to F_m = QvB?

I don't pretend to be able to give an answer which you'll find satisfying, but here are a few points which may help…

(1) Velocity-dependent forces aren't that uncommon. Air resistance on a moving body is a familiar example, though this acts in the opposite direction to the body's velocity. When a plane is flying in a curved path the forces from the air are velocity-dependent, and, what's more, they have a component at right angles to the plane's velocity.

(2) The magnetic field strength vector \mathbf{B} is defined by the magnetic force it exerts on a moving charged particle, namely \mathbf{F} = q\mathbf{B} \times \mathbf{v}.

(3) The electromagnetic force acting on a charge is for convenience divided up into two parts, (a) q\mathbf{E}, which acts whether the charge is stationary or moving, and in which \mathbf{E} is the so-called electric field and (b) q\mathbf{B} \times \mathbf{v} which acts only if the particle is moving (in our reference frame). The values of \mathbf{E} and \mathbf{B} depend on your reference frame. As Dalespam implied, Special Relativity theory (the easy one!) sheds a lot of light on electromagnetism.
 
Philip Wood said:
F=qB×v

Very sorry. Should have been \mathbf{F} = q\mathbf{v} \times \mathbf{B}. Don't know why I wrote what I did.
 
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