# Why does velocity increase force according to F_m = QvB?

• Arcthor
In summary, the increase in velocity of a particle in a magnetic field leads to an increase in the force acted upon it, according to the equation F_m = QvB. This is due to the direct proportionality between the centrifugal force and the square of the velocity, as observed in orbital dynamics. While the concept of frame dragging explains this phenomenon in the case of gravity, the increase in force in an electromagnetic field can be attributed to the velocity-dependent forces and the nature of the magnetic field strength vector. Moreover, the theory of Special Relativity provides a deeper understanding of this relationship between velocity and force in electromagnetism.

#### Arcthor

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?

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".

• vanhees71
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.

## 1. What is the equation F_m = QvB?

The equation F_m = QvB is used to calculate the force experienced by a charged particle moving through a magnetic field. F_m represents the magnetic force, Q is the charge of the particle, v is its velocity, and B is the strength of the magnetic field.

## 2. How does velocity affect force in F_m = QvB?

In the equation F_m = QvB, velocity is directly proportional to the force. This means that as the velocity of the charged particle increases, the force it experiences also increases. This is because a higher velocity means the particle is moving through the magnetic field at a faster rate, resulting in a stronger magnetic force.

## 3. Why is there a magnetic force on a moving charged particle?

According to the Lorentz force law, a charged particle moving through a magnetic field will experience a force. This is because the magnetic field exerts a force on the moving charged particle, causing it to change direction. This force is perpendicular to both the velocity and the magnetic field.

## 4. How does the direction of the velocity affect the force in F_m = QvB?

The direction of the velocity does not affect the magnitude of the force in the equation F_m = QvB. However, the direction of the force will be perpendicular to both the velocity and the magnetic field, causing the charged particle to move in a circular path.

## 5. Can the equation F_m = QvB be used for any type of charged particle?

Yes, the equation F_m = QvB can be used for any type of charged particle, as long as the charge and velocity are known. This includes particles such as electrons, protons, and ions. However, the direction of the force may vary depending on the charge and velocity of the particle.