= Solution
{c}
Let $x$ cross the barrier from electrode 1 to 2, let $z$ run along its width, and take $\mathbf B=B\widehat{\mathbf y}$. In the ideal barrier-only model choose $A_x=Bz$, which has $(\nabla\times\mathbf A)_y=B$. Then the crossing integral at fixed $z$ is $Btz$. With the sign convention used in the source,
$$
\boxed{\varphi(z)=\varphi(0)-\frac{2\pi Bt}{\Phi_0}z=\varphi(0)-\frac{2\pi\Phi}{W\Phi_0}z,\qquad\Phi=BtW.}
$$
Reversing the field or the phase orientation reverses the slope and does not change the interference envelope. The same result follows by comparing two paths across the junction and using the enclosed <magnetic flux>.
If field penetration into thick identical electrodes is included, the effective magnetic thickness is approximately $t_{\mathrm{eff}}=t+2\lambda_L$, where $\lambda_L$ is the <London penetration depth>. Replace $t$ and $\Phi$ consistently by $t_{\mathrm{eff}}$ and $Bt_{\mathrm{eff}}W$. The linear phase profile assumes a uniform applied field and negligible magnetic self-field of the junction current.
Back to article page
