How can one write $\varepsilon=I R$ for a loop moving in a constant magnetic field? A loop is moving down with some part of it in a constant magnetic field pointing into the screen as shown.
We know the emf due to magnetic force is given by
$\mathcal{E}=-\frac{d \Phi}{d t}$.
If the loop has a total resistance of $R$ then why is it that the current in the loop will be
$I=\mathcal{E}/R$.
 A: Good question. In my experience, most introductions to E&M don't give a great explanation of the exact similarities and differences between voltage and electromotive force (emf), and when you can and can't use the concepts interchangeably.
Voltage and emf are both formally defined the exact same way, as the (negative) line integal of the electric field over some path: $-\int_A^B {\bf E} \cdot d{\bf l}$. But the terms are used in different contexts. Voltage is used in the electrostatic context, where no magnetic fields change significantly over time and the electric field is generated by fixed electric charges (or chemical potentials, etc.). emf is used in the context of Faraday's law, where the electric field is induced by a time-varying magnetic field. In situations where both source charges are time-varying magnetic fields play a role, the distinction can get a little blurry.
Another difference is that in a magnetostatic situation, the value of the line integral is path-independent and only depends on the endpoints. Therefore, voltage is usually thought of a property relating two points in space, while emf explicitly depends on the path that the wire takes between the endpoints.
But both of them have the effect of pushing current through conductors. You can use emf in the place of voltage in pretty much all formulas regarding DC electric currents through wires, as long as the situation is quasistatic enough that you can assume that the electric field is uniform along the wire (which in practice is almost always the case).
A: No proof exists for the combined relation:
$I=-\frac{d \Phi}{d t}/R$
...because it is not true in all cases.
This happens because the current induced in the loop generates it own internal magnetic flux that opposes the external magnetic flux according to the Lenz Law. That reaction modifies the variable $\Phi$  in the equation above.
Consider the limiting case when the resistance of the shorted loop is zero (this can really happen in superconducting loops).
According to that Ohm's Law relation: $I=\mathcal{E}/R$ 
...any induced $\mathcal{E}$ would generate infinite current in the loop. That would be absurd.
What happens in reality, in such case, is that the SUM of the external flux + the flux generated by the current induced in the loop = CONSTANT.
See this simulation of a R=0 loop moving through the field of a permanent magnet
The simulation above depicts the loop moving over the field of a magnet instead of across a uniform field, but the principle of changing the flux encompassed by the loop is the same.

The reason for this is that any change in net flux requires a nonzero  $\mathcal{E}$ around the loop, which requires an infinite current, so the net magnetic flux through the loop cannot change. The flux from the self-inductance $L$ of the loop must be equal and opposite to the external flux, thus we conclude that in a shorted coil without resistance, the max induced current is INDEPENDENT from $\frac{d \Phi}{d t}$ and simply equal to $I=-\frac{\Phi}{L}$.
A: In this situation, there will be a magnetic force acting on the free electrons in the upper segment of the loop, pushing them to the left and producing a potential difference, vBl, between the two ends of that segment. On the other hand, a closed conducting loop totally within a uniform magnetic field which is changing with time will experience an emf which pushes electrons around the loop, but voltage differences will be dissipated from point to point as the current moves through the resistance of the conductor.
A: If the loop is going down at a velocity $v_y$, for  a magnetic field B, loop width $L$, and portion of loop height in the field $= y$, $$\frac{\partial \phi}{\partial t} = \frac{B \partial (Ly)}{\partial t} = \frac{LB \partial y}{\partial t} = LBv_y$$
As you said, an electric field is generated in the conductor:$$E = -\frac{\partial \phi}{\partial t}$$ If the velocity is constant, $E$ is also constant, and we have a static situation: $V = \oint{Edl} = RI$.
If it is not constant, the changing magnetic field generated by the current in the conductor generates an E-field that opposes that current. So, strictly speaking, $V \neq RI$ in this case. In practice that second effect is very small, except for a big indutance and small resistance of the conductor.
Both effects must be evaluated together, what mathematically means that the 2 Maxwell equations:
$$\nabla \times E = -\frac{\partial B}{\partial t}$$
$$\nabla \times B = \mu_0j + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}$$
have to be fulfilled simultaneously for the total fields.
A: Ohm's law in the form
$$
I = \frac{V}{R}
$$
where $V$ is voltage between two points, is merely a special case of general Ohm's law applicable only to situations where all electromotive forces active in the metal conductor are due to electrostatic field. This is not the case here, as the current is not due to electrostatic field, but due to special electromotive force of the conductor itself acting on the mobile charges forming the current, called motional EMF. It appears whenever conductor moves in magnetic field.
General Ohm's law (ignoring magnetic forces) is
$$
\mathbf E + \mathbf E^* = \sigma \mathbf j
$$
where $\mathbf E$ is total electric field, $\mathbf E^*$ is total non-electric force per unit charge (in this case, force due to the wire moving in magnetic field).
We can integrate both sides of the general Ohm's law for some segment of the circuit in the following way:
$$
\int_{start~of~s}^{end~of~s} (\mathbf E + \mathbf E^*) \cdot d\mathbf s = \int_{start~of~s}^{end~of~s} \sigma \mathbf j \cdot d\mathbf s
$$
The left-hand side is called total electromotive force for segment $s$ and we can denote it $\mathscr{E}_s$.
If we imagine the circuit wire is torus of cross-section area $A$ and length $L$, we can do the integration over the whole closed circuit, and then we get
$$
\mathscr{E}_{circuit} = \frac{\sigma A}{L} I.
$$
We know from high school that ohmic resistance of circuit is
$$
R = \frac{\sigma A}{L}
$$
so we finally arrive at the equation
$$
\mathscr{E}_{circuit} = RI.
$$
This is the other (integral) form of the general Ohm's law.
