= Solution
For the field generated by the infinitely long, axially symmetric shell currents, <axisymmetry> and the right-hand rule give $\mathbf B=B_\phi(r)\mathbf e_\phi$. A circular loop has $\oint\mathbf B\cdot d\boldsymbol\ell=2\pi rB_\phi(r)$. Its enclosed <electric current> is zero inside the inner shell, $I$ between the shells and $2I$ outside both. <Ampère's law> therefore gives
$$
\boxed{\mathbf B(r)=\begin{cases}0,&r<a,\\\dfrac{\mu_0I}{2\pi r}\mathbf e_\phi,&a<r<b,\\\dfrac{\mu_0I}{\pi r}\mathbf e_\phi,&r>b.\end{cases}}
$$
The positive $\mathbf e_\phi$ direction is counterclockwise viewed from positive $z$. The ideal <current sheets> make the tangential <magnetic field> discontinuous at each shell; these formulas specify the three open regions, with no additional imposed background field.
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