Examine the circuit shown in figure 12. When the switch is closed, assume the magnetic field produced by the solenoid is very strong.
a) Explain what will happen to the copper ring (which is free to move) when the switch is closed
b)Explain what will happen to the copper ring when there is a steady current in the circuit.
c)Explain what will happen to the copper ring when the switch is opened.
d)How would your answers change if the terminals on the power supply are reversed? Explain your answer.
Law of electromagnetic induction
An electric current is induced in a conductor whenever the magnetic field int he region of the conductor changes with time.
When a current is induced in a coil by a changing magnetic field, the electric current is in such a direction that its own magnetic field opposes the change that produced it.
The Attempt at a Solution
a) When the switch is closed current will flow from the positive to the negative. Based on the right hand rule the bottom of the coil will be "N" while the top of the coil will be "S". But I don't know how the copper ring will react. It'll probably be attracted towards the coil but I don't know why.
b) When there is a steady current in the circuit the copper ring will no longer be attracted to the current since the current becomes constant? Again I have no idea why.
c) When the switch is open the current will stop flowing and the copper ring will no longer be affected?
d) I have no idea.
Could anyone explain some of the concepts I need to know to solve this problem? I'm having difficulty understanding what the copper ring and the coil, current, etc do.
Any help will be greatly appreciated.
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