Momentum and Lorentz-force

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In summary, Lorentz-force and momentum are related in a way that p = qBr. However, p is only measured in GeV/c, and not m/s.
  • #1
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[SOLVED] Momentum and Lorentz-force

Homework Statement


Hi all.

I have two questions.

1) Please take a look at page 22 in this document (look at the bottom):

http://walet.phy.umist.ac.uk/P615/Notes.pdf [Broken]

It says that the momentum in GeV/c for a particle which moves in a magnetic field is given by:

p = 0.3*B*r, where B is in Teslas and r is the radius in metres.

I can derive p = qBr, but how does it become p = 0.3Br when p is measured in GeV/c?

2) If I have a proton moving in a magnetic field, where the magnetic field is perpendicular to the velocity-vector of the proton and the proton turns right, does the magnetic field point into or out of the plane?

The Attempt at a Solution


1) Ok, I know that the net-force is mv^2/r = q(E + v x B) = qvB, so p = q*B*r. But this is in m/s. How do I convert to GeV/c to obtain the expression above?

2) I tried using the right-hand-rule: The index finger in the direction of the velocity, the thumb in the direction of F (right, since it is turning this way) and then the middle-finger points upwards, which is the direction of B? Can you confirm this?

Thanks in advance.

Sincerely Niles.
 
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  • #2
You've got the index and the middle fingers mixed up.
1ev=1.6e(-19) joules.
 
  • #3
Thanks on the right-hand-rule.

Can you show me how it is derived? I've spent several hours today trying to do it, but I keep messing it up.
 
  • #4
Ok, I can start myself.

First I have p = qvB, where [p] = kg*m/s = J*s/m.

I want this in GeV/c. We know that 1 eV = 1.602 E -19 J, so I multiply J*s/m with (1,602 E -19)^(-1) eV/J. This is where I am stuck
 
  • #5
Niles said:
Ok, I can start myself.

First I have p = qvB, where [p] = kg*m/s = J*s/m.

I want this in GeV/c. We know that 1 eV = 1.602 E -19 J, so I multiply J*s/m with (1,602 E -19)^(-1) eV/J. This is where I am stuck

First, your derivation of qvB is not correct since everything is relativistic here so p i snot mv but [itex] \gamma m v [/itex] and so on. But I think that your final result is correct.

Now, the equation given in the text does not contain the charge so you know it can't be the complete equation. It seems as if they are assuming a charge equal to the elementary charge. This will give a factor of [tex]1.6 \times 10^{-19} [/tex] which will cancel out with the factor coming from the conversion of eV to Joules. There is a factor of 10^9 coming from the GeV. And finally, don't forget the factor of c, the speed of light (which why there will be a factor of 1/3 approx 0.3).
 
  • #6
kdv said:
First, your derivation of qvB is not correct since everything is relativistic here so p i snot mv but [itex] \gamma m v [/itex] and so on. But I think that your final result is correct.

Now, the equation given in the text does not contain the charge so you know it can't be the complete equation. It seems as if they are assuming a charge equal to the elementary charge. This will give a factor of [tex]1.6 \times 10^{-19} [/tex] which will cancel out with the factor coming from the conversion of eV to Joules. There is a factor of 10^9 coming from the GeV. And finally, don't forget the factor of c, the speed of light (which why there will be a factor of 1/3 approx 0.3).

My mistake..It won't be 1/3, it will simply be 3x10^8/10^9 = 0.3 .
 
  • #7
Niles said:

Homework Statement


Hi all.

I have two questions.

1) Please take a look at page 22 in this document (look at the bottom):

http://walet.phy.umist.ac.uk/P615/Notes.pdf [Broken]

It says that the momentum in GeV/c for a particle which moves in a magnetic field is given by:

p = 0.3*B*r, where B is in Teslas and r is the radius in metres.

I can derive p = qBr, but how does it become p = 0.3Br when p is measured in GeV/c?

2) If I have a proton moving in a magnetic field, where the magnetic field is perpendicular to the velocity-vector of the proton and the proton turns right, does the magnetic field point into or out of the plane?

The Attempt at a Solution


1) Ok, I know that the net-force is mv^2/r = q(E + v x B) = qvB, so p = q*B*r. But this is in m/s. How do I convert to GeV/c to obtain the expression above?

2) I tried using the right-hand-rule: The index finger in the direction of the velocity, the thumb in the direction of F (right, since it is turning this way) and then the middle-finger points upwards, which is the direction of B? Can you confirm this?

Thanks in advance.

Sincerely Niles.

we can't answer your second question without knowing in what direction the proton i smoving!
 
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  • #9
Ok, I know that the units for q*B*r = kg*m/s which equals 1.87*10^18 GeV/c. Where to go from here? I don't know what to do with the q in Coulombs.

If we say that the charge q = 1.602*10^(-19) eV, then I get Br = 0.3 GeV/c - but that's wrong?
 
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  • #10
The charge cannot be in ev. Thats a unit of energy.
 
  • #11
Then what steps am I missing to get 0.3Br?
 
  • #12
Niles said:
Then what steps am I missing to get 0.3Br?

The formula is qBR which, as you pointed out, is in kg m/s which is the same as Joule/(m/s).


Let's start by getting rid of the charge. Let's stay in SI units for now. Let's say we want a formula of the form K BR where K is some numerical constant such that when people plug in the value of B and R in Tesla and in meters, they get the correct momentum in SI units. Then obviously the formula is

[tex] 1.6 \times 10^{-19} B R [/tex]
will give the momentum in SI units if B and R are plugged in in tesla and meters.

Now, this will give the momentum in J/(m/s)

Now let's say that you want people to plug in the values of B and R in meters and in Tesla but you want the answer to come out in GeV/(m/s). Since [itex]1\,GeV = 1.6 \times10^{-10} J[/itex], you must divide the equation by that factor.

So the formula

[tex] \frac{1.6 \times 10^{-19}}{1.6 \times 10^{-10}} BR = 10^{-9} BR [/tex] will give the momentum in GeV/(m/s) if B and R are given in tesla and in meters.

Now you want the final result to be given in GeV/c. Since c is approximately 3x10^8 m/s, you must multiply the equation above by c to get the momentum in GeV/c, So finally, 0.3BR will give the momentum in GeV/c if B and R are given in Tesla and in meters.
 
  • #13
That figures, but only for a unit or test charge...
 
  • #14
chaoseverlasting said:
That figures, but only for a unit or test charge...

Like I said in my post #5, the equation makes only sense if we assume that the particle has an elementary charge [itex] e [/itex]. For particle physics experiments, this makes sense since one would not be working with nuclei (other than a single proton) or atoms.
 
  • #15
Thanks - I see my method wasn't correct.
 

1. What is momentum?

Momentum is a measure of an object's motion, calculated by multiplying its mass by its velocity. It is a vector quantity, meaning it has both magnitude and direction.

2. How is momentum related to velocity?

Momentum and velocity are directly proportional. This means that as an object's velocity increases, its momentum also increases.

3. What is the Lorentz force?

The Lorentz force is the force exerted on a charged particle moving through an electric and magnetic field. It is given by the equation F = q(E + v x B), where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.

4. How does momentum relate to the Lorentz force?

According to Newton's second law, the net force acting on an object is equal to its mass multiplied by its acceleration. In the case of a charged particle moving through an electric and magnetic field, the Lorentz force is the net force acting on the particle, causing it to experience an acceleration and therefore a change in momentum.

5. How does relativity affect momentum and the Lorentz force?

Einstein's theory of relativity states that the laws of physics are the same for all observers in uniform motion. This means that momentum and the Lorentz force are also affected by relativity, as they are both based on an object's motion and the observer's frame of reference.

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