If the walker must constantly approach B, as the puzzle suggests, then my answer seems to be slightly different from some of those proposed.

In that case, construct point A at the origin, and B at 1,0, then construct circle c, radius 1, on B. Construct point C somewhere in the second quadrant, and draw a ray from B through C. Construct a perpendicular through A to BC; these lines cross at D. D traces all the points between which chords can be drawn such that the distance between the walker and B is constantly decreasing. By construction, since angle ADB is a right angle, the figure traced out as D moves is a circle, say d, or more precisely a semicircle, of radius 1/2.

True it is that there are points outside d but inside c that are closer to B than A. But to access those points, the walker must cross d twice, with the implication that at some point during that leg of his walk the walker has receded from B.

For N=2, the distance to be walked is sqrt(2) miles.

For N=3, the distance to be walked is 1.5 miles.

For N=4, the distance to be walked is 2 sqrt (2 -sqrt(2)) miles. etc.

The longest possible route is therefore P/2, where P is the perimeter of the regular polygon having 2N sides, i.e. N * sin(pi/(2*N)), as suggested by Charlie.  

And the walk can never exceed pi/2 miles.