The Figures https://www.dropbox.com/s/jyyw28u4cgavn8y/Photo%201-8-14%2011%2020%2045%20pm.jpg
The arrows towards A and towards B show direction of induced currents, not force.

Fig. 22.4a:
Magnetic field: from N to S
Motion of straight conductor in field: upwards
Hence, by applying Fleming's Right hand rule, induced current flows in the straight conductor from B to A.

Alternatively, as described in the textbook, we can apply Lenz's law. The induced current will flow to oppose the motion of the straight conductor. Since the conductor moves up, there would be an opposing force downwards on the conductor.
Magnetic field: N to S
Force: downwards (the "result" of the field and current)
Applying Fleming's left hand rule, the induced current must flow from B towards A in the metal conductor.

1. Fleming's Left Hand Rule (also known as Motor Rule):
- we want to predict the direction of the force (e.g. to turn the coil in a motor)
- given magnetic field and electrical current (in conductor) -> deduce the direction of the force (on the conductor)

2. Fleming's Right Hand Rule (also know as Dynamo Rule)
- we want to predict the direction of the induced e.m.f. or induced current (e.g. that flows out from a generator)
- given magnetic field and direction of motion (in conductor relative to field) -> deduce the direction of the induced e.m.f. (across a conductor) or induced current (that flows in the conductor)

Q3: the magnetic field is N to S, current is along XY, so the motion must be perpendicular to both, by Fleming's right hand rule.

Q4: Rotating a wire loop within a magnetic field is essentially similar to an a.c. generator coil rotation. Referring to the voltage (e.m.f.) vs time graph of th generator (in notes or textbook), the maximum voltage (& current) occurs when the plane of the coil (or loop in this qn) is parallel to the magnetic field direction.

1. magnitude of voltage (induced emf) changes with orientation of the plane of the coil.
Refer to the slides in the pdf file posted above.

2. EM Induction Exercises MCQ Q10:
Following the question, answer should be (2) only, so B.

Alternately, if we modify the question option (1) "An a.c. generator ... replacing the two slip rings with a commutator",
then this is correct, and answer would be (1) and (2), so C.

5) for Q10d exercises, why is it for figure a, we adjust rheostat to maximum current for higher voltage? I thought current is inversely proportional to current?

1) 18.2 cii: the carbon brushes are in contact with the slip rings of the generator which acts like a "battery". So polarity of the carbon brushes mean which brush is positive "+" (where the current flows out to the external circuit) and negative "-" where the current flows back into the generator coil, just like the terminals of a battery.

2) E.g. 18.3b: we are looking at the "change" from when there is current to zero current (switch just opened). This corresponds to a magnetic field in P decreasing to zero, so there is a "rate of decrease of magnetic flux linked to the coil Q" hence a "momentary" or "short-lived" induced emf and induced current in the coil Q, by Faraday's law. The galvanometer would show a momentary deflection (deflect then go to zero).

3) E.g. 18.3e iv:
PQRS section is a series circuit. The secondary coil between P and Q (1000 turns) acts like a "battery" driving a current through this circuit. Using sum of emf = sum of p.d. in this circuit,
e.m.f. across secondary coil PQ = p.d. across cable PR + p.d. across cable QS (identical to PR) + p.d. across the coil across RS.
100 V = 1 V + 1 V + required p.d. across coil RS

4) MCQ 1: yes answer should be A.

If a coil A (primary coil) has a current flowing through it, the magnetic field set up within it is not strong.
If another coil B (secondary coil) is placed next to coil A, the magnetic field linked to B from A is weak.
If a soft iron core is now inserted into A, the magnetic field in A would be much stronger.
See http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/imgmag/icore.gif
If the soft iron core is extended into coil B, B would experience a much stronger field linked to it from A.
Any change in A's magnetic field (due to a change in current), would thus induce a larger current in B.

Yes, in (e) you could also use P = IV to derive the same answer, since V = 8.325 V from part (d) is the p.d. across the long leads.
In other questions, V across the cables or leads is not calculated or given, so P = I^2 R gives a direct calculation of P.

Soft iron loses its magnetism easily when the current is switched off, unlike steel which would keep its magnetism for some time when current is switched.
Hence, the magnetic field linked by the soft iron core from the primary coil to the secondary coil is able to change rapidly and continuously with the changing current (a.c.) in the primary coil to induce a changing current in the secondary coil.