Motional emf with rod We know that $\mathcal{E} = -N \frac{\mathrm{d}\phi}{\mathrm{d}t}$. When we have a rod such as the one on the left moving through a constant magnetic field, how is it the case that the flux is changing? For the loop in the right, the area and field are constant, so $\mathcal{E} = 0$. I don't immediately see why this is different for the rod.

 A: I'm currently taking intro E&M and we covered this fairly recently, so I'm going by memory and my textbook here. 
I believe it has to do with the frame of reference you consider. It seems to me that you are analyzing the situation in a lab frame at rest with respect to the wire. In that case, we see the rod move upwards in the +y direction. We also know through the Biot-Savart law that
$$ \vec{B} = \frac{\mu_0}{4\pi} \frac{q\vec{v} \times \hat{r}}{r^2} $$
This can be modified to cater to a wire 
$$ \Delta \vec{B} = \frac{\mu_0}{4\pi} \frac{I \Delta t(\vec{v} \times \hat{r})}{r^2} $$
However, we also know that $ \vec{v} = \frac{\Delta \vec{l}}{\Delta t} $, where $ \Delta \vec{l} $ is simply some vector length inside the wire, in the direction of the conventional current flow. This gives finally
$$ \Delta \vec{B} = \frac{\mu_0}{4\pi} \frac{I \Delta t \frac{\vec{l}}{\Delta t} \times \hat{r}}{r^2} = \frac{\mu_0}{4\pi} \frac{I (\Delta \vec{l} \times \hat{r})}{r^2} $$
Integrating this equation for a long straight wire shows that the magnitude of the electric field some distance, r, away from the wire is given as
$$ |\vec{B}| = \frac{\mu_0}{4\pi} \frac{2I}{r} $$
Using the right hand rule quickly shows that the field is into the plane of the page at the location of the bar. Over the length, L, of the bar it's safe to assume the field is more or less constant since it's not a great distance and we will chose the coordinates for the center of the bar for our r, since this provides a compromise of sorts between the higher B on the left and lower B on the right, but all in all it should be pretty negligible. 
The general idea is that mobile electrons inside the metal bar will experience a Lorentz force (this is the force at time t=0 assuming I is stable at that time, as time progresses, the bar is polatrized and there is an additional force within the bar which can be attributed to the electric field induced due to this polarization.) equal to
$$ \vec{F} = -e(\vec{v} \times \vec{B}) = -ev {\boldsymbol{\hat{\textbf{j}}}} \times \bigg(-\frac{\mu_0}{4\pi} \frac{2I}{\frac{L}{2}+d}\bigg) {\boldsymbol{\hat{\textbf{k}}}} = \frac{\mu_0}{4\pi} \frac{2Ive}{\frac{L}{2}+d}{\boldsymbol{\hat{\textbf{i}}}}$$ 
This result implies that the bar is polarized such that there is an excess of electrons on the right and deficiency of electrons on the left until the electric field until the resulting field, multiplied by the fundamental charge is numerically equal to our above formula for the magnetic force. Note that the electric field due to the polarization sets up so as to halt the change that moved the charges from their rest position.
In order to deduce the emf, we simply wait for the steady state, then take a line integral of the electric field from one end of the bar to the other. I don't have time to type out anymore, but I hope this helps. If you found this helpful I could provide more detail if necessary once I have some more time.
But, in closing, our textbook indicates that, applying a relativistic analysis, if you take your frame of reference to be an observer moving with the same velocity as the bar, then in this frame you will see an electric field evident throughout your reference frame, due to the transformations applied. Since you would be moving upwards in the +y direction, the E field (The E field you see in your frame!) would be directed in the negative x direction such that
$$ E^*_x = -vB $$
So, it all boils down to different mechanisms depending on your point of view.
Unfortunately we haven't discussed the derivation in class, nor do they do it in the book, but I'm sure someone more well versed in relativistic concerns can help you there.
As for the second scenario, it is true that there is no non-Coulomb electric field generated, as that would require a time dependent magnetic field. However, if we apply the Biot-Savart law to each straight segment of wire. It becomes apparent that the vertical wires are horizontally polarized and the top and bottom wires are too horizontally polarized.
This means that there is an electric field within the left wire pointing in the +x direction and in the right wire, there is an electric field within that segment of wire also pointing in the +x direction. However,
$$ EMF = -\int_{L} \vec{E} \cdot d\vec{r} $$ 
where L is simply some path through the wire. If we define L such that it describes a path through the wire from the bottom to the top of the left wire, we find that the integral yields zero since the E field vector is orthogonal to the path vector at each point. The same is true if we follow this procedure for the right wire. 
The last step is to apply this treatment to the top and bottom wires. By the Biot-Savart law, they too are horizontally polarized, and this time mobile electrons are pushed to and build up on the right hand side of the wire in the case of the both the top and bottom wires. It is then clear that the electric field in each wire is constant, uniform and points in the -x direction in the steady state.
Knowing this,
$$ \mathcal{E}_{net} = \int^{x_{f}}_{x_{i}} E_x \space dx + \bigg (-\int^{x_{f}}_{x_{i}} E_x \space dx \bigg) = \int^{x_{f}}_{x_{i}} E_x \space dx + \int^{x_{i}}_{x_{f}} E_x \space dx $$
This reflects the fact that we move in the +x direction in the top wire, down in the right wire (in which the integral is zero), and to the left in the bottom wire, then back up the left wire (in which the field is zero). We also know by our approximations state initially that the magnetic field is practically uniform across the loop, hence it follows that the magnitudes of the electric fields within the loop wire is equal at every point. Thus we find
$$ \mathcal{E}_{net} = E_x (x_f - x_i) + E_x (x_i - x_f) = 0 $$
It is clear that the EMF induced in the top and bottom wires respectively is due to a Lorentz force rather than a time varying magnetic field. Had it been due to a time varying magnetic field, we would have found that the line integral inside the wire, around the whole loop would have been numerically equal to the rate of change of the magnetic flux through the loop with respect to time (assuming one winding). 
P.S. Please let me know if I mixed up my signs/directions :)
Addenum:
Rereading the question, I believe I should add my thoughts on what you describe. It is my impression that Faraday's law does not imply that a time dependent magnetic flux is the only way an EMF can be induced. Instead, it seems to me to be a relation describing quantitatively what must happen IF there is a time varying flux through some area, rather than implying that there must be a time varying magnetic flux in order for any EMF to be induced, in any situation. As shown above, we can still have an EMF induced without requiring a time dependent flux.
