Voltage in an inductor - DC circuit
When the switch goes open-circuit, the inductor tries to keep the same current flowing AND in the same direction. The only way it can achieve this is by making its upper terminal go negative. So, at the instant in time that the switch opens, the current is forced through the 10 kohm resistor by a large negative voltage at its upper terminal.
Your upper multisim experiment is incorrect because you are using the formula for back emf (i.e. it has a negative sign) and you have no resistance in your circuit to develop the voltage across when the inductor has charged to its peak current.
In other words, the two &quot;experiments&quot; in your question are not the same. Just think about what your upper multisim &quot;experiment&quot; is telling you; when the switch is closed, how can the inductor voltage be anything else other than the applied supply voltage? Your analysis is faulty.