Electronic – eddy current direction

The question asks for the current, not the voltage.

The voltage (emf) generated would follow a straight line (not a curve) as the rate at which the area grows is proportional to the length of the wire - and the length of the wire grows linearly.

The current that flows as a result of this depends on the resistance of the loop. The circumference of the loop grows linearly with the horizontal position of the wire (similar triangles - when the wire has moved twice as far to the right, all the sides are twice as big).

The current is voltage divided by resistance. Both grow linearly; their ratio is constant.

If magnetic field B is coming out of the page and increasing, then the current in the conducting loop FBCE will flow clockwise.

Take a look at segment EF. If the current is flowing clockwise from E up to F, then the right hand rule tells us that the magnetic field generated by that current will create a circular magnetic field around the segment that goes out of the page on the outside of the loop and into the page inside the loop...

...same goes for all the other segments, with the net result being that the current circulating clockwise in the loop will generate a magnetic field that goes into the page inside the loop, which will resist the magnetic flux that is increasing out of the page.

There is another rule of thumb that gives the same answer

With the thumb of your right hand point in the direction of decreasing flux, your fingers will curl around in the direction of the electric field.

Note that in this rule, your thumb points in the direction of decreasing flux. This reflects the negative sign in Faraday's law. Also note that the decreasing flux could mean a field pointing out of the page increasing in time, or a field pointing into the page and decreasing in time.

No current will flow though segment AD.

Since AD does not enclose the changing flux, it does not experience any emf along its length so no current flows though it. All of the forces on its charged particles are perpendicular to its length.

You might think that the current flowing around the outside loop FBCE might create a voltage across AD and that might cause a current to flow though it, and this would be true in a battery-powered circuit because the battery creates an electric field that drives the current around the loop. But in this case, there is no electric field between A and D- they are at the same voltage potential. This is a hard concept to grasp when you are used to looking at battery driven circuits. Imagine this circuit sitting flat on a table. The charges are uniformly distributed around all of the conducting elements so there are no voltage differentials and no electric fields. Now you start increasing the flux. This causes a force on every charge particle that is curly and clockwise. This force pushes the particles around the perimeter loop. They are now moving, but their density has not changed. They are still uniform density so there is still no electric field anywhere. Make sense? (Note that this really is only accurate if the loop is circular, but this does not effect our questions here because the rectangular loop is symmetrical around R1.)

Since you are asking such a great question, I think you'd really enjoy the following resources:

1. Matter and Interactions. Explains this stuff in terms of fields and forces and the movement of charged particles, which is very different from the way electronics is typically presented and helps explains stuff that doesn't make sens when you only think in terms of traditional circuit theory. Completely dissolves the false distinction between static electricity and circuit theory.

2. Electricity and Magnetism (Berkeley Physics Course, Vol. 2). Explains this stuff in terms of special relativity. There is no magnetic field, just the normal electrical field as felt by moving particles due to Lorenz transformation. Mind blowing and makes everything finally come together including the deep reality of what electromagnetic waves actually are. Note you can likely find the first edition of this book for free on the web. The 3rd edition is expensive, but includes more problems with solutions which are great for self-study.

3. Lec 16: Electromagnetic Induction. Covers the above question and ends with a demonstration that will likely keep you awake thinking for many nights. Be prepared to watch this over and over!