Final answer:
The ratio between the electric currents (I) created by the electromagnetic induction (emf) in the two cases is 2.
Step-by-step explanation:
The ratio between the electric currents (I) created by the electromagnetic induction (emf) in the two cases can be determined by considering the emf induced in each side of the loop.
The emf induced on each side is given by the equation emf = Blv sin(θ), where B is the magnetic field, l is the length of the loop, v is the velocity of the magnet, and θ is the angle between the velocity and the magnetic field.
Since both magnets are identical, the magnetic field (B) and the velocity (v) will be the same for both cases.
The only difference in the two cases is the height (h) from which the magnets are falling. Let's assume that the height (h) is 2 units for the first case and 1 unit for the second case.
For the first case, the emf induced on each side is emf = Blv sin(θ) = B(2l)v sin(θ).
For the second case, the emf induced on each side is emf = Blv sin(θ) = B(l)v sin(θ).
To find the ratio between the electric currents, we divide the emf in the first case by the emf in the second case:
Ratio = (emf1 / emf2) = (B(2l)v sin(θ)) / (B(l)v sin(θ)) = 2