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(1) According to the formula of free fall: v2=2gH,

We can get the speed of the box just entering the magnetic field: v0=2gH,

Let the speed of the box reach V' when it is stable in the magnetic field, then there are:

mg = bi×2L = 2 blv′Rb×2L

Available: v'=mgR4B2L2,

If v0 ≥ v ′, the box will slow down after entering the magnetic field, and its maximum speed is v0 = 2gH；;

If v0 < v ′, the box will accelerate after entering the magnetic field, and its maximum speed is: v ′ = mgr4b2l2;

(2) When the box falls at a speed? v？ Power of induced current in the box: p = i2r = 4bl2v2r;

(3) According to the energy conversion relation, it can be concluded that mg(H+h)= 12mvt2+I02Rt.

So the equivalent current: I0=mg(H+h)? 12mv2tRt。

Answer: (1) When v0 ≥ v ′, the maximum speed of falling box vm = 2gH；; When v0 < v', the maximum speed of the box falling vm═mgR4B2L2; ;

(2) When the falling speed of the box is V, the power of the induced current in the box at this time is p = 4bl2v2r;

(3) The expression i0 = mg(h+h) for constant current l0? 12mv2tRt。