Just to show:
The inductance is here 10mH and the input voltage V1 is 10V for easy calculations.
The timed switch SW1 starts to conduct at t=10ms and stops at t=50ms. The inductor has zero resistance, so its current grows to 1A.
At t=50ms the inductor current bulldozes its way through R2 because it's the only available route. The inductor voltage jumps to 10V, it jumps as high as needed for continuing the current. The polarity of the node 3 voltage goes negative because the inductor pulls current through R2 upwards (=from GND towards node 3).
From t=10ms to t=50ms inductor current is fed by voltage divider which as unloaded outputs 5V. Inductor current pulls V(3) onto it knees, the inductor current grows and voltage V(3) decays with time constant L/R. Resistance R=5 ohms (=Thevenin equivalent resistance of the voltage divider).
When the switch opens at t=50ms the inductor current and V(3) decay again with time constant L/R, but this time R=10 ohms, the decay happens with double speed.
I guess your problem is the attempt to approach the practical behaviour of an inductor via a circuit theoretical equation Emf=-L(di/dt). The minus is inserted to make the unmeasurable imagined quantity "electromotive force" to be compliant with the measurable quantities such as voltages between circuit nodes. Check this old discussion of inductor's practical operation: How does the inductor ''really'' induce voltage?