Grainger's Ideal Transformer Doesn't Make Sense In Power Systems Analysis (Grainger/Stevenson), the authors illustrate a two-winding transformer in Figure 2.2 (p43). Using the properties of an ideal transformer (e.g., "core losses and winding resistances are zero"), the authors conclude that "the voltages $e_1$ and $e_2$ induced by the changing flux must equal the terminal voltages $v_1$ and $v_2$, respectively"
If $e_1$ (which I think is the back emf for the primary coil) cancels out $v_1$, how can we expect there to be any current in the primary coil?
A similar question was posed years ago, but the best answer is along the lines of "an ideal transformer isn't a real object" (Physics in the primary coil of an ideal transformer). If $e_1$ = $v_1$ as in Figure 2.2, doesn't $i_1 = 0$?
\begin{align}
v_1&=e_1=N_1\frac{d\Phi}{dt}\\
v_2&=e_2=N_2\frac{d\Phi}{dt}
\end{align}

Figure 2.2
Edit: Thank you all who contributed to this post. I found a clear explanation at https://forum.allaboutcircuits.com/threads/induced-emf-supposed-to-oppose-the-applied-voltage.83821/
"The back EMF in a inductor is equal to the voltage across the inductor. Don't know what you mean by "cancel out the supply voltage"? It's the same as having a voltage across a resistor. The resistor will provide a resistance (back EMF) equal to the supply voltage and the voltage across the resistor is equal to the supply voltage."

 A: 
If $e_1$ (which I think is the back emf for the primary coil) cancels out $v_1$, how can we expect there to be any current in the primary coil?

In the book by Stevenson and Grangier, $e_1$ does not seem to refer to induced electromotive force (emf, back emf), but to voltage assigned to two points on the winding close to the core, i.e. potential drop when going from one point to the other in the chosen positive direction. In other words, they use the usual convention where all symbols, even $e$'s, refer to potential drops in the chosen direction, rather than the induced EMF in that direction.
You are correct that in ideal winding (zero ohmic resistance), electric force due to induced EMF would have the same magnitude but opposite direction to electric force due to potential difference, so they would cancel each other. This is necessary because in ideal conductor (assumption of no resistance), total electric field has to vanish.
Current in the winding is still present though. There is no contradiction, because in ideal conductor there need not be force in order for the current to exist. It's like in mechanics; there need not be force in order for motion to persist, if there is no friction. Back in ideal transformer, the currents are determined by behaviour of the power source on the primary and the load on the secondary.
In a real transformer, the winding has some resistance, which can make the induced EMF weaker than the potential difference $e_1$, and then non-zero total electric force is present in the conductor, to drive the current against the force of resistance.
A: 
If e1 (which I think is the back emf for the primary coil) cancels out v1, how can we expect there to be any current in the primary coil?

The current in the primary of an ideal transformer is not caused by a voltage difference between the voltage applied to the primary and the back emf generated in the primary. Rather, the current in the primary of an ideal transformer is a consequence of current in the secondary.
If there is no current in the secondary of an ideal transformer, there will be no current in the primary. However if there is a voltage applied to the primary, there will be an emf appearing at the secondary. If there is a circuit on the secondary that allows it, the emf on the secondary will cause current to flow through the secondary. This secondary current causes a primary current to flow.
In a transformer, the flux is proportional to the algebraic sum of the currents in the windings times the turns in those windings.
$$\Phi \mathscr{R}=N_pI_p + N_sI_s$$
where $\mathscr{R}$ is the magnetic reluctance of the transformer's magnetic circuit.
However, in an ideal transformer, the magnetic reluctance is treated as approaching zero. So,
$$N_pI_p + N_sI_s =0$$
So, when a current appears in the secondary, an appropriate current must appear in the primary.
A: 
If e1 (which I think is the back emf for the primary coil) cancels out
v1, how can we expect there to be any current in the primary coil?

If the secondary circuit is open, the situation in the primary is an inductor under an AC voltage.
$$V = L\frac{dI}{dt}$$
For example, if $V = V_0 cos{\omega t}$, $$I = \frac{V_0}{L\omega}sin(\omega t) + C$$
There is a AC voltage (same as the source) that can be measured by a voltmeter, and a AC currrent in the coil.
For a transformer $L$ is huge, so that the primary currect is really close to zero when the secondary is open.
In this case, the electro-mechanical analogy is useful because the differential equations are the same, except for the meaning of the variables and constants:
$$F = m\frac{dv}{dt}$$
If we move a mass back and forth, the mass reacts on us with the same force that we apply on it, and nevertheless there is a non zero velocity. The force plays here the role of the voltage, and velocity the role of the current.
