The advantage of an SSR is isolation.
If it's an inductive load, a diode across the MOSFET will do nothing to protect it, unless it has a lower breakdown voltage, than the MOSFET's maximum voltage rating i.e. is a TVS, or zener diode. The diode needs to go in parallel with the load.
The RFP30N06LE has an on resistance of 0.047Ohms, with a gate voltage of 5V. Presumably the MCU is powered off 5V, otherwise it's no good.
The on resistance given on the data sheet is specified at a junction temperature of 25
oC and increases with temperature. The data sheet doesn't give the on resistance at higher temperatures, but the article linked to below suggests it will be around 50% higher at 120
oC and 80% higher at the MOSFET's maximum rating of 175
oC. If you aim for a maximum junction temperature of 120
oC, the on resistance will be 0.07Ohms and power dissipation P = I
2R = 20
2*0.07 = 28W.
https://toshiba.semicon-storage.com/eu/semiconductor/knowledge/faq/mosfet/is-the-on-state-resistance-of-a-mosfet-dependent-on-temperature.htmlThe thermal resistance of the die to package is 1.55
oC/W and if the maximum ambient is 40
oC, the permissible temperature rise is 80
oC. The maximum acceptable thermal resistance between the MOSFET junction and case is T
RMAX = T
RISE/Power = 80/28 = 2.86
oC/W.
Subtract the thermal resistance of the die to package and the total thermal resistance of the heat sink plus any thermal pads is 2.86-1.55 = 1.31
oC/W.
I advise using a MOSFET with a lower on resistance, or lower thermal resistance between the junction and case.
Another question: does the induction heater have an enable signal? If so, keep it powered and connect the enable signal to the MCU output.