(1) According to x=4t-2(m), the final uniform motion speed of the conductor bar is vm=4m/s, and the force of the conductor bar is balanced at this time, which is obtained from the balance condition.

F amp = mgsin 30 ①

And f =B2L2vmR? ②

Simultaneous ① ②, m=0. 1kg.

(2) The quantity of electricity passing through the bulb Q =. i△ T =△φ R = BLSR。

What is the displacement of the conductor bar sliding down? s=8m

According to the law of conservation of energy

mgsin30 =Q+ 12mv2

Solution, the heat generated by the light bulb Q = 3.8J 。

(3) Incorrect. Because mgsinθ=B2L2vmR shows that the vm in the formula decreases with the increase of b, and the power of the bulb cannot be improved. Because P=I2R, in order to increase the maximum power of the small bulb when the ab rod slides down, I or resistor R needs to be increased. When moving at a uniform speed, BIL=mgsinθ, so B, L can be reduced, or M, R, θ can be increased.

Answer: (1) The mass m of the conductor rod is 0.1kg; ;

(2) When the speed of the conductor bar is v=2m/s, the heat q generated by the bulb is 3.8J；;

(3) Incorrect. Because mgsinθ=B2L2vmR shows that the vm in the formula decreases with the increase of b, and the power of the bulb cannot be improved. Because P=I2R, in order to increase the maximum power of the small bulb when the ab rod slides down, I or resistor R needs to be increased. When moving at a uniform speed, BIL=mgsinθ, so B, L can be reduced, or M, R, θ can be increased.