Ideal transformer principle and emf equation

Operating principle

A transformer works on mutual induction. Two windings (primary of $N_1$ turns, secondary of $N_2$ turns) are wound on a common laminated core. An alternating voltage on the primary sends a current that sets up an alternating flux $\phi$ in the core. This flux links the secondary and, by Faraday's law, induces an emf in it. Energy is transferred from primary to secondary at the same frequency, with voltage and current changed in the ratio of turns.

An ideal transformer has no winding resistance, no leakage flux, no core loss, and a core of infinite permeability.

        core (laminated)
     +--------------------+
  ~  |  N1            N2  |  load
 V1  |  primary   secondary|  Z
     +--------------------+
        <----- flux ----->

emf equation

Let the flux be $\phi = \phi_m\sin\omega t$, where $\phi_m$ is the maximum flux (Wb).

$$\begin{aligned} e_1 &= -N_1\frac{d\phi}{dt} = -N_1\omega\phi_m\cos\omega t \\ E_{m1} &= N_1\omega\phi_m = 2\pi fN_1\phi_m \end{aligned}$$

The rms value is

$$E_1 = \frac{E_{m1}}{\sqrt2} = \frac{2\pi}{\sqrt2}fN_1\phi_m = 4.44\,fN_1\phi_m$$

Similarly,

$$E_2 = 4.44\,fN_2\phi_m$$ $$\boxed{E_1 = 4.44fN_1\phi_m = 4.44fN_1B_mA\ \text{V},\qquad \frac{E_1}{E_2} = \frac{N_1}{N_2}}$$

where $B_m$ is the maximum flux density and $A$ the core cross-section. The induced emf lags the flux by 90°. For the ideal case $V_1 = E_1$ and $V_2 = E_2$, hence $V_1/V_2 = N_1/N_2 = a$.