At this time, the ampere force on the conductor is F =BIL, I=ER+r, and E=BLv.

Get f amp = b2l2vr+r。

According to the equilibrium condition, f = f a+μ g.

V=5m/s is obtained by substituting into the solution.

(2) Let the distance that the conductor moves along the frame from rest be x 。

By e = △φ△t

q=I？ △t，get

q =△φR+R = BLxR+R

X=q(R+r)BL=2×(2+2)0.8×0.5=20m。

According to the law of conservation of energy, draw a conclusion.

Fx=μmgx+Q+ 12mv2

Substitute the solution to get q =1.5j.

So the electric heat generated by the metal bar is QR =12q = 0.75j.

Answer: (1) The speed of the conductor bar is 5m/s when it moves at a uniform speed;

(2) The electric heat generated by the conductor bar is 0.75 J from the beginning to the end of uniform motion.