What is the current induced in a wire connected to an ideal transformer using Faraday's law?

Question

Imagine an ideal transformer interfacing an ideal current source with a ramp function \$i_1(t)=a t\$ connected to the primary winding and a resistor \$R\$ connected to the second winding. We want to know the current in the second loop \$i_2\$.

We could proceed in two ways,

1. We compute the current in the secondary loop, using ideal transformer's theory we obtain $$i_2 = \frac{N_1}{N_2} i_1 \tag 1 $$.

2. We could also reason that in an ideal transformer, the magnetic flux in the secondary is the same as in the primary \$\phi_2 = \phi_1 \$, with $$ \phi_1 = B_1 A_1 = \mu n_1 i_1 A_1 \tag 2 $$ with \$n_1 = N_1 / l_1\$. We can then use Ohm's and Faraday's laws to derive \$i_2\$, $$ i_2 = \frac{v_2}{R} = - \frac{\mathcal{E}_2}{R} = - \frac{1}{R} N_2 \frac{d \phi_2}{d t} = \frac{\mu n_1 N_2}{R} \frac{d i_1}{d t} \tag 3 $$ which is clearly different from the expression obtained in equation (1).

Can somebody tell me what is wrong in this reasoning?

Answer 1

An ideal transformer has infinite magnetizing inductance. Therefore with finite voltage applied, if the secondary is open circuit, no current flows. You can't apply I =a.t to an ideal transformer.

When the secondary is loaded, current does flow. N1.i1==N2.i2 is true, but the net flux in the core is 0.

Thus your (1) is the correct behaviour; (2) doesn't apply