= Solution
Let the <electric charge> be $q=-e$, where $e>0$ is the <elementary charge>. <Minimal coupling> to a tangential <magnetic vector potential> gives
$$
H=\frac{(p-qA_x)^2}{2m}.
$$
A constant tangential <magnetic vector potential> is an available gauge on the ring, so $a=qA_x$. By <Stokes theorem>, its circulation is the oriented <magnetic flux>
$$
\Phi_C=\oint_{+x}\mathbf A\cdot d\boldsymbol\ell=LA_x,
\qquad \boxed{aL=q\Phi_C=-e\Phi_C.}
$$
The orientation matters here. In the original diagram, increasing $x$ goes anticlockwise, whose associated positive normal points out of the page, whereas the indicated positive <magnetic flux> $\Phi$ points into the page. Hence $\Phi_C=-\Phi$ for that diagram, and its convention gives $\boxed{aL=e\Phi}$. If instead $\Phi$ is defined with the normal associated to $+x$, the relation is $aL=-e\Phi$.
There is no requirement that the <magnetic field> be nonzero on the ring: it can be confined to a <solenoid> inside it. The nonzero circulation of the <magnetic vector potential> still affects the <Electron> through the <Aharonov-Bohm effect>.
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