If you step-change the pipe-and-wire assembly in that field, at first there will be no induced voltage, and the pipe will have sheared the field lines in that direction. Which is a graphical way of putting it: the pipe's outer wall has an inducted current, which you are seeing superimposed upon the ambient field. This reflected field happens to be just the right magnitude to cancel out the field in the interior, so the induced voltage is zero.
Over time, the field lines seep through the wall. Resistive materials act as a gooey trap for field lines. (Type I superconductors have no resistance, and exclude field lines for all time. Type II superconductors exhibit hysteresis loss -- called flux pinning -- so that, if the magnitude of influence is great enough, some flux will pass through anyway. I don't know if this is a sudden thing or a gradual thing (consider Barkhausen noise in ferromagnetic materials!), but the effect is simply: nonzero AC resistance, and therefore some induced voltage in the wire.)
For the steel pipe, the field lines divert aroundthe wire, but sooner or later, the field lines trapped above and below the wire must move around the pipe, and the only way for that to happen is for the wire to be cut by lines. In this case, there is an induced voltage, merely delayed by the losses of the steel.
For pipes of the same dimensions (dia, thickness), the induced voltage will be greater for steel, but not by a gross amount. The time constant (i.e., how much the step change is slowed down by) is proportional to wall thickness (or rather, a power of it).
Tim