# Velocity of a relativistic particle in a uniform magnetic field

• B Rylen
In summary: The speed could be expressed in terms of any two of these four parameters. (If you want to use a different coordinate system, such as cylindrical or spherical coordinates, then you will need to choose different parameters.)In summary, the problem involves solving for the velocity of a relativistic particle in a magnetic field using the equations F = q(v x B) and F = dp/dt. Depending on the chosen coordinate system, the speed can be expressed in terms of the radius of the orbit, the period or frequency of the orbit, or the angular frequency.
B Rylen
Homework Statement
I am trying to solve for the velocity of a relativistic particle in a magnetic field using magnetic force (F = q(v x B)) and F = dp/dt (where p = ɣmv) and equating the two equations to each other. However, I stuck on how to isolate v in this setup.
Relevant Equations
F = dp/dt = d(ɣmv)/dt
F = q(v x B)
ɣ = 1/sqrt(1-(v/c)^2)
d(ɣmv)/dt = qvB
(dɣ/dt)mv + ɣm(dv/dt) = qvB
Substituting gamma in and using the chain rule, it ends up simplifying to the following:
ɣ^3*m(dv/dt) = qvB

Now, I am confused on how to solve for v.

Delta2
B Rylen said:
(dɣ/dt)mv + ɣm(dv/dt) = qvB
Your problem is one of using scalars instead of vectors. The force is orthogonal to velocity so the speed is constant. Hence (dɣ/dt) = 0.

B Rylen, Delta2, vanhees71 and 1 other person
It should also be mentioned that dv/dt (where v is the speed and not velocity) is also equal to zero.

Delta2, vanhees71 and malawi_glenn
Another possibility is to solve the manifestly covariant equations using the proper time ##\tau## as the parameter of the trajectory/world line. This usually simplifies such calculations!

B Rylen said:
Homework Statement:: I am trying to solve for the velocity of a relativistic particle in a magnetic field using magnetic force (F = q(v x B)) and F = dp/dt (where p = ɣmv) and equating the two equations to each other.
Are you assuming that the particle is orbiting in a plane perpendicular to the field? If so, you will need to choose which parameters you want to use for expressing the speed v. Besides q, B, and m there is the radius R of the orbit, the period T of the orbit, and the frequency f (or angular frequency ω) of the orbit.

Last edited:

## 1. What is the formula for calculating the velocity of a relativistic particle in a uniform magnetic field?

The formula for calculating the velocity of a relativistic particle in a uniform magnetic field is v = c * tanh(qBt/mc), where v is the velocity, c is the speed of light, q is the charge of the particle, B is the magnetic field strength, t is the time, and m is the mass of the particle.

## 2. How is the velocity of a relativistic particle affected by the strength of the magnetic field?

The velocity of a relativistic particle is directly proportional to the strength of the magnetic field. As the magnetic field increases, the velocity of the particle also increases.

## 3. What is the role of the charge of the particle in the velocity formula?

The charge of the particle, represented by q in the formula, determines the direction of the velocity. If the charge is positive, the particle will move in one direction, and if the charge is negative, the particle will move in the opposite direction.

## 4. Can the velocity of a relativistic particle exceed the speed of light?

No, according to Einstein's theory of relativity, the speed of light is the maximum speed that any particle can attain. Therefore, the velocity of a relativistic particle in a uniform magnetic field cannot exceed the speed of light.

## 5. How does the mass of the particle affect its velocity in a uniform magnetic field?

The mass of the particle, represented by m in the formula, has an inverse relationship with the velocity. As the mass increases, the velocity decreases. This is because a heavier particle requires more energy to accelerate to the same velocity as a lighter particle.

Replies
3
Views
2K
Replies
1
Views
1K
Replies
20
Views
1K
Replies
15
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
16
Views
3K
Replies
1
Views
2K
Replies
12
Views
833
Replies
2
Views
2K