Solution

In an ideal transformer (ideal means no losses, no leakage, and infinite magnetic permeability)

the voltage induced in each winding is proportional to the number of turns

$V_1 = N_1 \frac{d\Phi}{dt}, V_2 = N_2 \frac{d\Phi}{dt}$

With the secondary winding short‑circuited, the voltage $V_2 = 0$
Since$V_2 = N_2 \frac{d\Phi}{dt}$
this implies that the magnetic flux $\Phi$in the core must be constant (or zero, if starting from zero initial conditions). For the net flux to be zero, the total magnetomotive force must be zero

$N_1 I_1 + N_2 I_2 = 0$

From this, the ratio of magnitudes is :

$\boxed{\frac{I_1}{I_2} = \frac{N_2}{N_1}}$

The negative sign indicates that the currents are in opposite phase, but the ratio of amplitudes is as given.