(1) Before the conductor bars are locked, the area of the closed loop remains unchanged, as follows: △ b△ t = k.

According to Faraday's law of electromagnetic induction: E=△φ△t=△BS△t=3 16kL2.

By closed-circuit ohm's law: I=ER sum =3kL28R?

According to Lenz's law, the direction of induced current is clockwise or B → A.

(2) When the conductor bar just leaves the guide rail, the stress is shown in the figure.

E= 12B0Lv

Me = eh?

F= 12B0IL，

Get: F=B0L2v4R？

According to Newton's second law: mg-F=ma?

So: a = g-B02V2V4MR?

(3) obtained from the conservation of energy: mgh =12mv2+q.

h=34L

Solution: Q=3mgL4- 12mv2.

Answer: (1) The loop current before unlocking is 3kL28R；;

(2) The acceleration when sliding to the end of the guide rail is G-B02V2V4MR;

(3) Joule heat generated during exercise is 3mgl4- 12mv2.