Inductor held with hands
When you close the switch, current through the inductor will build up linearly at a rate of di = V*dt/L = 98A per second. 

When you open the switch the final current will initially keep flowing (in the same direction) through the inductor. Since the switch is open the only possible path is now through the body. If the body's resistance was 20k&ohm; and you opened the switch after 1 second, then the current would reach 98A and voltage would spike to 98A*20k&ohm; = 1.96MV. 

Since the current is flowing in the same direction through the inductor it must go in reverse through the body (ie. upwards in your diagram) compared to the current that was flowing through it when the switch was closed.  

As the magnetic energy in the inductor is drained out the current (and voltage) will decay exponentially with a time constant of L/R, so the body will only receive that 98A at 1.96MV for a short time. As to what would happen to it, how should I know? - I'm an engineer not a doctor!