How Does the Feynman Paradox Apply to Solenoid Flux Calculations?

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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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