In a circuit when the switch is closed the magnetic field produced by the solenoid is very strong

what would happen to the copper ring located next to the circuit (which is free to move) when the switch is closed? when the switch is opened? and how would the answers change if the charge were reversed...
sooo confused please help lol

The copper ring will be pushed away. This happens both when the switch is closed and when it is opened; the direction of the current makes no difference.

The solenoid's changing magnetic field (at the time of switching on or switching off) induces a current in the copper (electromagnetic induction). The direction of the induced current is such that the magnetic field it produces repels the field causes by the current in the solenoid (Lenz's Law). Note there must be a changing field from the solenoid for this to happLen's with DC you only see the effect at the momnet of switching on or off.

If you want to know more detail, just do a search on 'jumping ring experiment' and/or Lenz's Law.