How to Calculate the Cyclotron Magnetic Dipole Moment in a Penning Trap?

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Izzyg
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Hello, this is a question regarding Penning trap design.

I need to calculate the magnetic dipole moment of the cyclotron motion, as a function of the cyclotron quantum number. The result needs to be given in terms of Bohr's magnetron.

The magnetic dipole moment is defined as current x area enclosed by current.
 
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  • #2
Izzyg said:
I need to calculate the magnetic dipole moment of the cyclotron motion, as a function of the cyclotron quantum number. The result needs to be given in terms of Bohr's magnetron.
Just curious. Is this a homework question?
 
  • #3
Indeed it is. So far I have found
{\displaystyle {\boldsymbol {\mu }}={\frac {-e}{2m_{\text{e}}}}\,\mathbf {L} \,,}
from https://en.wikipedia.org/wiki/Electron_magnetic_moment, but I can't see current x area enclosed there.

Cyclotron quantum number must be
{\displaystyle E_{n}=\hbar \omega _{\rm {c}}\left(n+{\frac {1}{2}}\right)+{\frac {p_{z}^{2}}{2m}},\quad n\geq 0~.}
https://en.wikipedia.org/wiki/Landau_levels
 
  • #4
dlgoff said:
Is this a homework question?
Izzyg said:
Indeed it is.
Thread moved to the advanced physics homework forum.
 
  • #5
Izzyg said:
So far I have found
{\displaystyle {\boldsymbol {\mu }}={\frac {-e}{2m_{\text{e}}}}\,\mathbf {L} \,,}
from https://en.wikipedia.org/wiki/Electron_magnetic_moment, but I can't see current x area enclosed there

Consider a particle of mass ##m## and charge ##q## moving in uniform circular motion of radius ##r## and speed ##v##.

Can you express the angular momentum ##L## in terms of ##m##, ##r## and ##v##?

Can you express the current ##I## due to the motion of the charge in terms of ##q##, ##r## and ##v##?
 
  • #6
TSny, thank you for your message, really helpful. I think I've now found what I need: μ = (-e/2m)L = IA
 
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FAQ: How to Calculate the Cyclotron Magnetic Dipole Moment in a Penning Trap?

What is the cyclotron magnetic dipole moment in a Penning trap?

The cyclotron magnetic dipole moment in a Penning trap is a measure of the magnetic moment associated with the cyclotron motion of a charged particle in the magnetic field of the trap. It is an important parameter in the study of particle dynamics and precision measurements in Penning traps.

How do you derive the formula for the cyclotron magnetic dipole moment?

The cyclotron magnetic dipole moment (\(\mu_c\)) can be derived from the relationship between the cyclotron frequency (\(\omega_c\)), the charge of the particle (q), and the mass of the particle (m). The magnetic dipole moment is given by \(\mu_c = \frac{q}{2m}L_c\), where \(L_c\) is the angular momentum of the particle in the cyclotron motion. For a particle in a magnetic field B, the cyclotron frequency is \(\omega_c = \frac{qB}{m}\).

What role does the magnetic field play in calculating the cyclotron magnetic dipole moment?

The magnetic field (B) is crucial in determining the cyclotron frequency (\(\omega_c\)) and, consequently, the cyclotron magnetic dipole moment. The strength of the magnetic field directly influences the frequency of the cyclotron motion, which is used to calculate the magnetic dipole moment. The relationship is given by \(\omega_c = \frac{qB}{m}\), and the dipole moment is proportional to this frequency.

How does the charge and mass of the particle affect the cyclotron magnetic dipole moment?

The charge (q) and mass (m) of the particle are fundamental parameters in the calculation of the cyclotron magnetic dipole moment. The dipole moment is directly proportional to the charge and inversely proportional to the mass of the particle. This relationship is encapsulated in the formula \(\mu_c = \frac{q}{2m}L_c\), indicating that higher charge and lower mass result in a larger magnetic dipole moment for the same angular momentum.

Can you provide an example calculation of the cyclotron magnetic dipole moment?

Sure! Let's consider a particle with charge q = 1.6 x 10^-19 C (like a proton) and mass m = 1.67 x 10^-27 kg in a magnetic field B = 1 Tesla. The cyclotron frequency is \(\omega_c = \frac{qB}{m} = \frac{1.6 \times 10^{-19} \times 1}{1.67 \times 10^{-27}} \approx 9.58 \times 10^7 \, \text{rad/s}\).

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