Electromanetics question
There is no magnetic flux density or magnetic field inside the conductor since the current only flows through the surface. Consider a closed circular path (L) concentric with the wire with radius \$r, r&lt;b\$, by Ampere Circuital Law:
$$\int_L \vec{B}.\vec{dl} = \mu I,$$
Here \$I\$ is the current which crosses the surface bounded by path L. Since there is no current inside the conductor \$I = 0\$.
By symmetry,
$$\int_L \vec{B}.\vec{dl} = B(2\pi r) = 0 \implies B = 0.$$