A. Assume that the falling height of the wire frame is h, and the wire frame is in free fall. The speed of the wire frame when entering the magnetic field is v2=2gh, v=2gh. The speed of the two wire frames when entering the magnetic field is equal.

The Ampere force F=BIL=BBLvRL=B2L2vR experienced by the wire frame when entering the magnetic field. When A enters the magnetic field, it moves in a straight line at a uniform speed, then B2L2vR=mg. The side length of A is twice that of B. The resistance is 4 times that of B.

The speeds of the two wire frames are equal. From B2L2vR, we can know that the ampere force on B is equal to the ampere force on A; the cross-sectional area of ??the wires is the same, and the side length of A is twice that of B. times, the density of A is 12 times that of B, then the masses of A and B are equal,

The gravity experienced by B is equal to the gravity of A. When B enters the magnetic field, the Napoleon experienced by B is equal to the gravity, and the gravity experienced by B is The resultant force is zero, and B moves in a straight line at a uniform speed, so A is correct;

B. The induced current when the wire frame enters the magnetic field I=ER=BLvR, the side length of A is twice that of B, and the resistance of A It is 4 times that of B. After entering the magnetic field, the ratio of the induced current intensity in A and B is 1:2, so B is wrong;

C. The amount of induced charge Q=It=BLvR×Lv=BL2R, The side length of A is twice that of B, and the resistance of A is four times that of B. When both frames enter the magnetic field, the electric charge passing through the cross section is equal, so C is correct;

D. Electric energy consumed W=I2Rt=(BLvR)2×R×Lv=B2L3vR, v is the same as B, the side length of A is twice that of B, and the resistance of A is 4 times that of B. When both frames enter the magnetic field, the The ratio of electric energy is 2:1, so D is wrong;

So AC is chosen.