How does a magnetic field produce an electromotive force? As I keep reading through texts and online, I keep getting conflicting views when in magnetic fields and electromotive forces are involved.
From what I can see, the emf is the work done per unit charge integrated along a closed loop by a source that is not electrostatic. If the Lorentz force does no work, how can a magnetic field produce an emf?
 A: The emf is not the work done per unit charge integrated along a closed loop by a source that is not electrostatic, it's just:
$$ \mathscr E =\oint \vec{f}_s \cdot d\vec{\ell},$$
where the integral is around the circuit, and $\vec{f}_s$ is the net force per unit charge on the conduction charges that move about within the circuit element.  This might seem the same, but it isn't.  The integral for the emf is around a loop at some fixed time.  The work done between two points needs to follow the charges in space and time.  In statics there isn't much difference, but in dynamics (changing fields or moving wires) it matters.
The Lorentz force can do work, but I think you are specifically asking about how a magnetic field can produce an emf.  There are two ways a magnetic field can be responsible for an emf.
The first is cheating.  If the magnetic field at a point in space changes, then it is responsible for producing an electric field and that electric field can do work and in fact $\vec{E}$ can be line integrated along the circuit, and the result is equal to the flux of $\vec{\nabla}\times\vec{E}$ through the surface determined by the circuit, so equal to the flux of $-\partial \vec{B}/\partial t$ through the surface determined by the circuit.
The second is if the wire is moving.  If there wire is moving, then you can compute $\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}$ where $\vec{v}=\vec{w}+\vec{u}$ is the velocity of the conduction (mobile) charge, and $\vec{w}$ is the velocity of the wire element itself, and $\vec{u}$ is the relative velocity of the mobile charge through the wire.  This is the magnetic contribution to the emf because $\vec{v}\times\vec{B}$ is the magnetic force per unit charge on the mobile charges.
We can evaluate this:
$$ \oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\oint \vec{B} \cdot (\vec{v}\times d\vec{\ell})=-\oint \vec{B} \cdot ((\vec{w}+\vec{u})\times d\vec{\ell}).$$
Then notice that $\vec{u}$ is along the wire, so parallel to $d\vec{\ell}$, so 
$$ \oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\oint \vec{B} \cdot (\vec{w}\times d\vec{\ell})$$
In a small amount of time, there is a circuit at one place, and then a time $\Delta t$ later the circuit is somewhere else and each part moves $\vec{w}\Delta t$ in space.  Imagine a circuit at one time, and at the later time (so a latter circuit, and a former circuit), and there is a ribbon in between.  Each part of the ribbon has an area $d\vec{a}=(\vec{w}\times d\vec{\ell})\Delta t$.  Make your time interval so small that $\vec {B}$ doesn't change much (in time) from when the circuit is in one place to the other or anywhere inside.  Then for a fixed time in that interval $\Delta t$, $\vec{\nabla}\cdot \vec{B}=0$ so the total flux through the surface $S$ bounded by the latter circuit ($C_2$), the former circuit ($C_1$) and the ribbon ($R$) is zero.  
$0=\oint_S \vec{B}\cdot d\vec{a}=\int_{C_1} \vec{B}\cdot d\vec{a}+\int_{C_2} \vec{B}\cdot d\vec{a}+\int_{R} \vec{B}\cdot d\vec{a}.$
So $$\Delta t\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\Delta t\oint \vec{B} \cdot (\vec{w}\times d\vec{\ell})=\int_{C_1} \vec{B}\cdot d\vec{a}+\int_{C_2} \vec{B}\cdot d\vec{a}.$$
Where $d\vec{a}$ points outwards in both cases.

So if the $\vec{B}$ field is changing we get an induced electric field and a corresponding electric emf of:
$$\oint_{C_1} \vec{E} \cdot d\vec{\ell}=\int_{C_1} (\vec{\nabla}\times\vec{E})\cdot d\vec{a}=\int_{C_1} (\vec{\nabla}\times\vec{E})\cdot d\vec{a}=\int_{C_1} (-\partial \vec{B}/\partial t)\cdot d\vec{a}.$$
So for a small time interval $\Delta t$:
$$\Delta t\oint_{C_1} \vec{E} \cdot d\vec{\ell}=\Delta t\int_{C_1} (-\partial \vec{B}/\partial t)\cdot d\vec{a}=-\int_{C_1} (\vec{B}(t_0+\Delta t)-\vec{B}(t_0))\cdot d\vec{a}.$$
And if the circuit is moving we get:
$$\Delta t\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-\int_{C_1} \vec{B}\cdot d\vec{a}+\int_{C_2} \vec{B}\cdot d\vec{a}.$$
Where this time both the $d\vec{a}$ vectors point in the direction associated with the direction of the oriented circuit.
Now if you compute the magnetic flux $\Phi=\int_{C} \vec{B}\cdot d\vec{a}$, then it's time derivative has two parts (from the product rule), one from the changing $\vec{B}$ (for fixed circuit) and one from a fixed $\vec{B}$ (and changing circuit).
So putting them together, $$\oint \vec{E} \cdot d\vec{\ell}+\oint (\vec{v}\times\vec{B}) \cdot d\vec{\ell}=-d\Phi/dt.$$

Thus the negative of the time rate of change of the magnetic flux is literally equal to the integral of the Lorentz force per unit charge around the circuit, and the electric part of the Lorentz force is due to the parts of the $\vec{B}$ field that are changing at some point, and the magnetic parts of the Lorentz force contribute where the circuit element itself is moving.
So the Lorentz force exactly contributes the emf due to the changing magnetic flux.  The magnetic part because of the moving circuit, the electric part because of the induced electric field from the changing magnetic field.  Both matter in general.
A: The electromotive force induced by a time-varying magnetic field upon a stationary wire is given by the equation
$$ \mathscr E_{induced} = {\int_a^b}_C \vec{E_{rot}} \cdot d\vec{\ell}$$
where

*

*$\vec{E_{rot}}$ is the divergence free (or rotational) component of the $\vec{E}$ field, and is a solution to the equation

$$\vec{\nabla} \times \vec{E_{rot}} = \frac{\partial\vec{B}}{\partial t}$$

*

*the integral is taken from a point $a$ on the wire to a point $b$ on the wire along the wire path $C$
By Helmholtz's decomposition theorem, a well-behaved 3D vector field can be decomposed into two components, one curl-free and the other, divergence-free.
In the case where $a=b$, and $C$ is a simple closed loop, the equation becomes:
$$ \mathscr E_{induced} = {\oint}_C \vec{E_{rot}} \cdot d\vec{\ell}$$
which (in an overall neutrally charged wire) is equal to
$$ \mathscr E_{induced} = {\oint}_C \vec{E} \cdot d\vec{\ell}$$
where $\vec{E}$ is the total electric field, and not merely the divergence-free component.
It is important to draw the distinction between $\vec{E}$ and $\vec{E_{rot}}$ because in a steady state where there is no time varying change in charge density $\rho$, conduction current (the flow of charges) is "conserved" in a circuit. By "conserved" I do not mean over time, but throughout the circuit. That is, Kirchhoff's Current Law holds.
$$\vec{\nabla} \cdot \vec{J} = 0$$
where $\vec{J}$ is current density.
However by the "microscopic" version of Ohm's Law,
$$\vec{J} = \sigma \vec{E}$$
where $\sigma$ is the conductivity of the material under consideration (the wire in this case).
Combining the last two equations we have
$$\vec{\nabla} \cdot \sigma \vec{E} = 0$$
which, implies that where $\sigma$ is invariant (in space),
$$\vec{\nabla} \cdot \vec{E} = 0$$
In any circuit with current, charges quickly re-arrange themselves to ensure that the above equation holds. Consequently, if we used $\vec{E}$ instead of $\vec{E_{rot}}$ to find the induced EMF in a wire, we would arrive an an erroneous result. This "spontaneous" rearrangement of charges to make the divergence of $\vec{E}$ equal to 0 (in large swaths of the circuit), does not affect $\vec{E_{rot}}$.
The formula given for the EMF induced by a time-varying magnetic field gives the correct expression to apply in Kirchhoff's (original) Voltage Law found in his 1845 paper "Ueber den Durchgang eines electrischen stromes etc.".



*when the wires \$1,2,...n\$ form a closed figure,
$$I_1R_1 + I_2R_2 + ... I_nR_n$$
= the sum of all electromotive forces that are on the way:


(my translation).
That is, one should use the formula
$$ \mathscr E_{induced} = {\int_a^b}_C \vec{E_{rot}} \cdot d\vec{\ell}$$
to find the induced EMF, when such EMF's are found "on the way".
