(a) A changing flux introduces an EMF given by Faraday’s law:
E =\oint E \cdot d l=-\frac{d \Phi}{d t} \Rightarrow E 2 \pi b=-\frac{d \Phi}{d t} \Rightarrow \quad E =-\frac{1}{2 \pi b} \frac{d \Phi}{d t} \hat{\phi} .
The work done when the charge moves a distance dl = v dt around the ring is
d W=q E d l=-\frac{q}{2 \pi b} \frac{d \Phi}{d t} v d t=-\frac{q v}{2 \pi b} d \Phi .
But v/b is the angular velocity ω, so \frac{d W}{d \Phi}=-\frac{q \omega}{2 \pi} .
(b) The mechanical angular momentum is (Equation 4.192 and foonote 62)
\frac{d\langle r \rangle}{d t}=\frac{1}{m}\langle( p -q A )\rangle (4.192).
L _{\text {mechanical }}=( r \times p )-q( r \times A )= L -q\left(r \frac{\Phi}{2 \pi r}\right)(\hat{r} \times \hat{\phi})= L -q \frac{\Phi}{2 \pi} \hat{k} .
Therefore the z component of the mechanical angular momentum is
\left[ L _{\text {mechanical }}\right]_{z}=L_{z}-\frac{q \Phi}{2 \pi}=-i \hbar \frac{d}{d \phi}-\frac{q \Phi}{2 \pi} .
Acting on \psi_{n}=A e^{i n \phi} (Equation 4.203),
\psi=A e^{i \lambda \phi} (4.203).
L _{[\text {mechanical }]_{z}} \psi_{n}=\left(-i \hbar \frac{d}{d \phi}-\frac{q \Phi}{2 \pi}\right) \psi_{n}=\left(\hbar n-\frac{q \Phi}{2 \pi}\right) \psi_{n} .
we see that \psi_{n} is an eigenstate with eigenvalue
\hbar\left(n-\frac{q \Phi}{2 \pi \hbar}\right) .
(c) Taking the derivative of E_n (Equation 4.206), we get
E_{n}=\frac{\hbar^{2}}{2 m b^{2}}\left(n-\frac{q \Phi}{2 \pi \hbar}\right)^{2}, \quad(n=0, \pm 1, \pm 2, \ldots) (4.206).
\frac{d E_{n}}{d \Phi}=\frac{\hbar^{2}}{2 m b^{2}} 2\left(n-\frac{q \Phi}{2 \pi \hbar}\right)\left(-\frac{q}{2 \pi \hbar}\right)=-\frac{L_{\text {mechanical }}}{I} \frac{q}{2 \pi} .
where I = mb² is the moment of inertia, and since L = I ω we have
\frac{d E_{n}}{d \Phi}=-q \frac{\omega}{2 \pi} .
foonote 62:
That is to say, \langle r \rangle d\langle r \rangle / d t , etc ,are unchanged. Because Λ depends on position \langle p \rangle (with p represented by the operator -i \hbar \nabla ) does change, but as you found in Equation 4.192, p does not represent the mechanical momentum (mv) in this context (in Lagrangian mechanics p =m v +q A is the so-called canonical momentum).