The circuit you show is a current regulator, not a fuse. Once the output is shorted, VT2 is controlled to limit the current at about 5-6A, and it will dissipate accordingly (>60W), as long as the short circuit is present. It will quickly overheat and get destroyed.
Instead, you'd probably want a circuit that latches into the OFF state. To reset it, the power must be removed then reapplied, after the short circuit was removed of course.
simulate this circuit – Schematic created using CircuitLab
R4 is the current sense resistor, R3-VT1 detect the excessive current, VT3 latches VT1 once it starts turning on. R2 limits the SCR ON current. C1-R5 add rudimentary immunity against transients. D1 protects against negative spikes generated by the load inductance.
Instead of a permanent latch, we can turn the load off for some fixed duration, then turn it back on - and off again if the short persists. This can be done with a monostable.