= Solution
Only the signed enclosed <electric current> changes. Between the shells it remains $I$, but outside both it is $I-I=0$. The same circular-loop application of <Ampère's law> gives
$$
\boxed{\mathbf B(r)=\begin{cases}0,&r<a,\\\dfrac{\mu_0I}{2\pi r}\mathbf e_\phi,&a<r<b,\\0,&r>b.\end{cases}}
$$
\b[Opposite currents confine the generated magnetic field to the annulus], as in an ideal infinitely long coaxial cable. The two shell <current sheets> account for the field jumps.
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