Determine the force in z-direction on the gyrocenter of a charged particle

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SUMMARY

The force in the z-direction on the gyrocenter of a charged particle in a diverging magnetic field can be determined using the equation \(\vec{F} = -\mu_m \vec{\nabla}B\). This relationship indicates that the effective force is influenced by the magnetic field gradient, \(\frac{\partial B}{\partial z} < 0\). The sign of the force is contingent upon the charge of the particle, as derived from the Lorentz force applied to the guiding center. Understanding this relationship is crucial for analyzing particle dynamics in non-uniform magnetic fields.

PREREQUISITES
  • Understanding of Lorentz force and its application to charged particles
  • Familiarity with magnetic field gradients and their implications
  • Knowledge of magnetic moment (\(\mu_m\)) concepts
  • Basic principles of plasma physics and charged particle motion
NEXT STEPS
  • Research the derivation of the Lorentz force in non-uniform magnetic fields
  • Study the implications of magnetic pressure on charged particles
  • Explore the concept of magnetic moment and its role in particle dynamics
  • Investigate the effects of varying magnetic fields on gyrofrequency
USEFUL FOR

Physicists, plasma researchers, and students studying electromagnetism and charged particle dynamics in magnetic fields.

Logarythmic
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Homework Statement


Determine the force in z-direction on the gyrocenter of a charged particle in a diverging magnetic field.

\frac{\partial B}{\partial z} &lt; 0


Homework Equations





The Attempt at a Solution



Please give me a starter. Could I use the Lorentz force in this case?
 
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Is there a discussion in one's text regarding magnetic force or magnetic pressure on a charged particle?

A short answer - the effective force on a charged particle in a spatially varying magnetic field is given by:

\vec{F}\,=\,-\mu_m\,\vec{\nabla}B

Hopefully one would be able to derive this relationship.
 
I did it with the Lorentz Force on the guiding center and got the equation given by Astronuc but with a plus sign. I guess that the sign is dependent on the charge.
 

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