How Does the Feynman Paradox Apply to Solenoid Flux Calculations?

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SUMMARY

The discussion focuses on applying the Feynman Paradox to solenoid flux calculations, specifically addressing the relationship between magnetic flux (Φ) and vector potential (A). The key equations include Φ = ∫d²x*B and the Lorentz force equation F = q(E + v × B). The participant aims to derive A(initial) from the observed charge change Δmv = eΦ/2πr, concluding that A(initial) = mΦ/2πr is consistent with the integral relationships involving A and B.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically solenoids and magnetic flux.
  • Familiarity with vector calculus, particularly curl and divergence operations.
  • Knowledge of the Lorentz force and its components in electromagnetic fields.
  • Proficiency in solving differential equations related to electric and magnetic fields.
NEXT STEPS
  • Study the derivation of the Lorentz force equation in detail.
  • Learn about the mathematical properties of vector potentials in electromagnetism.
  • Explore advanced topics in electromagnetic theory, such as Maxwell's equations.
  • Investigate the implications of the Feynman Paradox in various physical systems.
USEFUL FOR

Students of physics, particularly those studying electromagnetism, as well as educators and researchers interested in advanced applications of vector potentials and magnetic flux in theoretical mechanics.

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I am currently going through a friend's Mechanics II notes and homework before I take the course at a different university next semester. I have a few problems that I am having trouble understanding and am posting here for help.

1. Homework Statement

1. A solenoid is on, Flux is Φ.
2. The solenoid is turned off. A charge of Δmv = eΦ/2πr is observed.
3. Find A(initial) assuming A(final) = 0 from this data.
4. Show that A(initial) is consistent with ∫A*dx = ∫d2x∇xA = ∫d2x*B = Φ

Homework Equations


Φ = ∫d2x*B
∇xE = =d/dt∇xA = -dB/dt
dmv/dt-eE = e dA/dt

The Attempt at a Solution


3. dmv/dt-eE = e*dA/dt => mv-e∫E*dx = eA => mv/e-A = ∫E*dx = -dΦ/dt => A(initial) - mv/e = A final = 0 => A(inital) = mv/e
A(initial) = mΦ/2πr
4. I'm not sure where to start here.
 
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First thing the Lorentz force is ##F= q(E+v\times B)##;
and in your three you are missing the curl of A, and btw it should be ##\nabla \times A = B##, without the minus sign as you wrote over there.
 
Another thing, your integral should be with respect to ##dt##, i.e. ##\int curl A dt##.
 

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