Voltage exceed source voltage over a coil when opening a switch
There is nothing wrong.

Inductor will store some energy during the time time the switch is closed, which is given by

\$E_L = \frac{1}{2}LI^2\$

=380μJ in your case.

When the switch opens, inductor will develop some voltage on its terminals in order to keep the current constant. Obviously, over time the stored energy will dissipate on the resistors.

Remember that 


Capacitors cannot change their voltage instantly, but can change their current instantly
The exactly opposite is true for inductors. Their current cannot change instantly, but their voltage can


We take advantage of this effect in Boost Converters (google it) in order to step up a DC voltage