Lenz's and Faraday's Law Induction. Help on resistor and switches

[cy19861126]

Lenz's and Faraday's Law Induction. Help on resistor and switches!

Homework Statement

Find the direction of the current in the resistor shown below a) at the instant the switch is closed b) after the switch has been closed for several minutes and c) at the instant the switch is opened

Homework Equations

V = -N(flux/t)
flux = BA

The Attempt at a Solution

Seriously, I have no idea how to start with this problem. ok, so let's look at (a) first. When switch is closed, the current is moving from the right side of the battery, through the loop and ends at the left side. Okay, and then at (b) after the switch has been closed for several minutes, what difference would it make? Maybe the value of V will decrease because of the above equation, V = flux/t? Okay, and at the instant the switch is opened, there would be no current? Somehow, I think the answer to this question is way too easy. Please check if I did anything wrong. Any help would be appreciated, thanks

 

[cy19861126]

Okay, I read my text and I'm guessing the answer might be as follow but I do not know why:
a) Right to left
b) no current
c) left to right

 

[MarkusNaslund19]

This is all about Lenz's law. Know these laws!

When the switch is closed, we get a current in the left solenoid. This current is counter clockwise. Mind you the current is the direction of the positive charges. So this creates a magnetic field to the right, according to the right hand rule. The law says that this change in magnetic field will tend to be countered. So a magnetic field in the opposite direction is induced by the second solenoid. According to the right hand rule, the current is clockwise. Following the loops of wire of the second solenoid you will see it passes the resister from right to left.

If there is no immediate change in magnetic field of the system, there is no induced current.

Same concept as my big paragraph up. See if you can follow.