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Consider an infinitely long solenoid. The field at some interior
point is B (=u0*n*I as usual). Say the point is on the axis. Slice
the solenoid in half at that point. By symmetry, both the left
semi-infinite half and the right semi-infinite half must make equal
contributions to B. Hence, the field on the axis at the end of a
semi-infinite tightly-wound solenoid is B/2.