Solve Cyclotron Motion: m\frac{d\hat{v}}{dt}=q\hat{v}\times\hat{B}

In summary, ian2012 attempted to solve the equation of motion for charged particle in a magnetic field, but was stuck on the little part where he integrated the cross product. He was familiar with the cross product, but incorrectly wrote the equation of motion. He found an easier way to solve his problem after reading and understanding the help provided by the chatbot.
  • #1
ian2012
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Homework Statement



Given the equation of motion for charged particle in a magnetic field, with mass m and charge q moves in the y-z plane under the influence of a uniform magnetic field [tex]\hat{B}=B_{0}\hat{x}[/tex] is given by:

[tex]m\frac{d\hat{v}}{dt}=q\hat{v}\times\hat{B}[/tex]

Find the solution to the equation of motion, the cyclotron motion.

(This isn't the full problem, I am stuck on this little part of it)

Homework Equations





The Attempt at a Solution



so what i have done is treat the vector cross product as a multiplication sign. I have separated the variables and integrated:

[tex]\int\frac{d\hat{v}}{\hat{v}}=\frac{qB_{0}}{m}dt[/tex]

Upon integration, you get an exponential velocity, which makes no sense at all? I am not sure how else you'd find the cyclotron velocity solution?
 
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  • #2
Hi ian2012,

As you've discovered yourself, it is not correct to treat the cross product as simple multiplication. In fact, you should be able to show in this problem that the acceleration is orthogonal to the velocity. How familiar are you with the cross product? The most basic way to approach this problem is to write the equation of motion in coordinates rather than sticking with the vector notation.
 
  • #3
I am familiar with the cross product. If you compute the cross product with the direction of the field B and the velocity in the y-z plane, you will get:
[tex]\hat{v}\times\hat{B}=\hat{v}B_{0}[/tex]
But then I am again integrating the same expression.
 
  • #4
ian2012 said:
I am familiar with the cross product. If you compute the cross product with the direction of the field B and the velocity in the y-z plane, you will get:
[tex]\hat{v}\times\hat{B}=\hat{v}B_{0}[/tex]
But then I am again integrating the same expression.

Sorry, I don't know your conventions, but if you take B to point in the x direction and v to lie in the y-z plane, then the cross product is not proportional to v. In fact, it is orthogonal to v, as I said earlier [tex] \vec{v} \cdot ( \vec{v} \times \vec{B} ) = 0 [/tex]. Physically, the magnetic force should be perpendicular to the velocity.
 
  • #5
Sorry, my hats represent vectors. I beg your pardon, if we say:
[tex]\vec{v}=(0, v_{0}, v_{0})[/tex]

then after taking the cross product, we get:
[tex]m\frac{d\vec{v}}{dt}=qB_{0}(0, v_{0}, -v_{0})[/tex]

I then did:
[tex]m\frac{d}{dt}(0, v_{0}, v_{0})}=qB_{0}(0, v_{0}, -v_{0})[/tex]

I split the problem up into the y, z component DE's:
[tex]dv_{0y}=\frac{qB_{0}}{m}v_{0y}dt[/tex]
[tex]dv_{0z}=-\frac{qB_{0}}{m}v_{0z}dt[/tex]

Therefore the solution is:
[tex]\vec{v}=(0, v_{0y}, v_{0z})=(0, C_{1}e^{\frac{qB_{0}t}{m}}, -C_{2}e^{-\frac{qB_{0}t}{m}})[/tex]

Don't know if this is correct?
 
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  • #6
You're almost there, you just have some confusion about the velocity components. Write [tex] \vec{v} = (0, v_y, v_z) [/tex] and then do the cross product. You should find that the equations of motion are slightly altered from what you wrote.
 
  • #7
But then how do you integrate the expressions:
[tex]\frac{dv_{y}}{dt}=\frac{qB_{0}}{m}v_{z}[/tex]
[tex]\frac{dv_{z}}{dt}=-\frac{qB_{0}}{m}v_{y}[/tex]

you can't integrate v_{z} w.r.t dv_{y} and visa versa?
 
  • #8
Try differentiating the first equation with respect to time and using the second equation to produce an equation containing only one velocity component.
 
  • #9
I managed to get:
[tex]v_{y}=c1sin\Omega t+c2cos\Omega t[/tex]
[tex]v_{z}=c1cos\Omega t-c2sin\Omega t[/tex]

[tex]\Omega = \frac{qB_{0}}{m}[/tex]
 
  • #10
Looks good to me.
 
  • #11
Oh, i just realized, i don't have to calculate the cross product. I found an easier way to solve my problem. Thanks for your help!
 

FAQ: Solve Cyclotron Motion: m\frac{d\hat{v}}{dt}=q\hat{v}\times\hat{B}

What is cyclotron motion?

Cyclotron motion is the circular motion of a charged particle in a uniform magnetic field. It is characterized by the particle's velocity being perpendicular to the magnetic field and the particle's acceleration being perpendicular to both the velocity and the magnetic field.

What is the equation for cyclotron motion?

The equation for cyclotron motion is m

What does the equation for cyclotron motion represent?

The equation m\frac{d\hat{v}}{dt}=q\hat{v}\times\hat{B} represents Newton's second law for a charged particle in a magnetic field. It describes the relationship between the particle's mass, charge, and motion in a magnetic field.

How is cyclotron motion used in scientific research?

Cyclotron motion is used in a variety of scientific research, including particle accelerators, mass spectrometry, and nuclear magnetic resonance imaging. It allows scientists to manipulate and study the behavior of charged particles in a controlled environment.

What are some real-life applications of cyclotron motion?

Some real-life applications of cyclotron motion include medical imaging technologies such as PET scans and proton therapy for cancer treatment. It is also used in industrial applications for the production of radiopharmaceuticals and other radioactive materials.

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