I assume what is meant by Faraday's law of induction is what Griffiths refers to as the "universal flux rule", the statement of which can be found in this question. This covers both cases 1) and 2), even though in 1) it is justified by the third Maxwell equation1 and in 2) by the Lorentz force law.
The universal flux rule is a consequence of the third Maxwell equation, the Lorentz force law, and Gauss's law for magnetism (the second Maxwell equation). To the extent that those three laws are fundamental, the universal flux rule is not.
I won't comment on whether the universal flux rule is intuitively true. But the real relationship is given by the derivation of the universal flux rule from the Maxwell equations and the Lorentz force law. You can derive it yourself, but it requires you to either:
- know the form of the Leibniz integral rule for integration over an oriented surface in three dimensions
- be able to derive #1 from the more general statement using differential geometry
- be able to come up with an intuitive sort of argument involving infinitesimal deformations of the loop, like what is shown here.
If you look at the formula for (1), and set $\mathbf{F} = \mathbf{B}$, you see that
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{a} &= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} + \iint_{\Sigma} \mathbf{v}(\nabla \cdot \mathbf{B}) \cdot \mathrm{d}\mathbf{a} - \int_{\partial \Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= - \iint_{\Sigma} \nabla \times \mathbf{E} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= -\int_{\partial \Sigma} \mathbf{E} + \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell
\end{align*}
where we have used the third Maxwell equation, Gauss's law for magnetism, and the Kelvin--Stokes theorem. The final expression on the right hand side is of course the negative emf in the loop, and we recover the universal flux rule.
Observe that the first term, $\iint_\Sigma \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a}$, becomes the electric part of the emf, so if the loop is stationary and the magnetic field changes, then the resulting emf is entirely due to the induced electric field. In contrast, the third term, $-\int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell$, becomes the magnetic part of the emf, so if the magnetic field is constant and the loop moves, then the resulting emf is entirely due to the Lorentz force. In general, when the magnetic field may change and the loop may also move simultaneously, the total emf is the sum of these two contributions.
If you are an undergrad taking a first course in electromagnetism, you should know the statement of the universal flux rule, and you should be able to justify it by working out specific cases using the third Maxwell equation, the Lorentz force law, or some combination thereof, but I can't imagine you would be asked for the proof of the general case from scratch, as given above.
The universal flux rule only applies to the case of an idealized wire, modelled as a continuous one-dimensional closed curve in which current is constrained to flow, that possibly undergoes a continuous deformation. It cannot be used for cases like the Faraday disc. In such cases you will need to go back to the first principles, that is, the third Maxwell equation and the Lorentz force law. There is no shortcut or generalization of the flux rule that you can apply. You should be able to do this on an exam.
1 This equation is also often referred to as "Faraday's law" (which I try to avoid) or the "Maxwell--Faraday equation/law" (which I will also avoid here because of the potential to cause confusion).
Both of these follow from desirable properties of this hypothetical magnetic charge, namely:
- Magnetic charge is conserved.
- Magnetic field lines radiate outwards from positive magnetic charges.
- The net force between two magnetic charges moving at constant speed along parallel tracks is less than that between two stationary charges.
All three of these properties hold for electric charges. The last one may not be as familiar, but it basically works as follows: if we have a positive electric charge moving at constant velocity, it generates a magnetic field in addition to its electric field. A second positive electric charge moving parallel to the first one will therefore experience a magnetic force, and if you work out the directions, this force works out to be attractive. Thus, the net force between the two charges (electric and magnetic together) is less than the magnitude of the force they would exert on each other if they were at rest. [ASIDE: This can also be thought of in terms of the transformation properties of forces between different reference frames in special relativity, if you prefer to think of it that way.]
Now, the conservation of electric charge can be written in terms of the continuity equation:
$$
\vec{\nabla} \cdot \vec{j}_e + \frac{ \partial \rho_e}{\partial t} = 0
$$
Note that this can be derived from Ampère's Law and Gauss's Law ($\epsilon_0 \vec{\nabla} \cdot \vec{E} = \rho_e$), using the fact that the divergence of a curl is always zero:
$$
0 = \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) = \mu_0 \vec{\nabla} \cdot \vec{j}_e + \mu_0 \frac{\partial ( \epsilon_0 \vec{\nabla} \cdot \vec{E})}{\partial t} = \mu_0 \left( \vec{\nabla} \cdot \vec{j}_e + \frac{\partial \rho_e}{\partial t} \right)
$$
If we want to extend Maxwell's equations to magnetic charges, we need to have a magnetic version of Gauss's Law and add in a magnetic current term to Faraday's Law:
$$
\vec{\nabla} \cdot \vec{B} = \alpha \rho_m \qquad \vec{\nabla} \times \vec{E} = \beta \vec{j}_m - \frac{\partial \vec{B}}{\partial t} ,
$$
where $\alpha$ and $\beta$ are arbitrary proportionality factors. But if we try to derive a continuity equation for magnetic charge from these two facts (as we did above for electric charge), we get
$$
\beta \vec{\nabla} \cdot \vec{j}_m - \alpha \frac{\partial \rho_m}{\partial t} = 0,
$$
and this is equivalent to the continuity equation if and only if $\alpha = - \beta$. Beyond this, the choice of $\alpha$ is to some degree arbitrary; different values correspond to different choices of which type of magnetic charge we call "positive", and what units we use to measure it. If we want to have magnetic field lines radiating away from "positive" magnetic charges, then we will want $\alpha > 0$; the usual choice in MKS units is to pick $\alpha = \mu_0$ (and $\beta = -\mu_0$), as you have in your equations above.
This negative sign in the magnetic current term Faraday's Law then implies that the electric field lines created by a moving magnetic charge will obey a "left-hand rule" instead of a "right-hand rule". In other words, the direction of $\vec{E}$ created by a moving magnetic charge would be opposite the direction of $\vec{B}$ created by a moving electric charge. If we still want two magnetic charges moving along parallel tracks to exhibit a lesser force than what they feel when at rest, then we must also flip the sign of the $\vec{v} \times \vec{E}$ term in the Lorentz force law to compensate for this flip.
Best Answer
Maxwell's equations in vacuum are symmetric bar the problem with units that you have identified. In SI units $$ \nabla \cdot {\bf E} = 0\ \ \ \ \ \ \nabla \cdot {\bf B} =0$$ $$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}\ \ \ \ \ \ \nabla \times {\bf B} = \mu_0 \epsilon_0 \frac{\partial {\bf E}}{\partial t}$$
If we let $\mu_0=1$, $\epsilon_0 =1$ (effectively saying we are adopting a system of units where $c=1$, then these equations become completely symmetric to the exchange of ${\bf E}$ and ${\bf B}$ except for the minus sign in Faraday's law. They are symmetric to a rotation (see below).
If the source terms are introduced then this breaks the symmetry, but only because we apparently inhabit a universe where magnetic monopoles do not exist. If they did, then Maxwell's equations can be written symmetrically. We suppose a magnetic charge density $\rho_m$ and a magnetic current density ${\bf J_{m}}$, then we write $$ \nabla \cdot {\bf E} = \rho\ \ \ \ \ \ \nabla \cdot {\bf B} = \rho_m$$ $$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} - {\bf J_m}\ \ \ \ \ \ \nabla \times {\bf B} = \frac{\partial {\bf E}}{\partial t} + {\bf J}$$
With these definitions, Maxwell's equations acquire symmetry to duality transformations. If you put $\rho$ and $\rho_m$; ${\bf J}$ and ${\bf J_m}$; ${\bf E}$ and ${\bf H}$; ${\bf D}$ and ${\bf B}$ into column matrices and operate on them all with a rotation matrix of the form $$ \left( \begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array} \right),$$ where $\phi$ is some rotation angle, then the resulting transformed sources and fields also obey the same Maxwell's equations. For instance if $\phi=\pi/2$ then the E- and B-fields swap identities; electrons would have a magnetic charge, not an electric charge and so on.
Whilst one can argue then about what we define as electric and magnetic charges, it is an empirical fact at present that whatever the ratio of electric to magnetic charge (because any ratio can be made to satisfy the symmetric Maxwell's equations) all particles appear to have the same ratio, so we choose to fix it that one of the charge types is always zero - i.e. no magnetic monopoles.
I mention all this really as a curiosity. It seems to me that the real symmetries of Maxwell's equations only emerge when one considers the electromagnetic potentials.
e.g. if we insert $B = \nabla \times {\bf A}$ and $E= -{\bf \nabla V} - \partial {\bf A}/\partial t$ into our Ampere's law $$\nabla \times (\nabla \times {\bf A}) = \frac{\partial}{\partial t} \left({\bf -\nabla V} - \frac{\partial {\bf A}}{\partial t}\right) +{\bf J}, $$ $$-\nabla^2 {\bf A} +\nabla(\nabla \cdot {\bf A}) = -\nabla \frac{\partial V}{\partial t} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J}.$$ Then using the Lorenz gauge $$\nabla \cdot {\bf A} + \frac{\partial V}{\partial t} = 0$$ we can get $$ \nabla^2 {\bf A} - \frac{\partial^2 {\bf A}}{\partial t^2} + {\bf J} = 0$$ A so-called inhomogeneous wave equation. A similar set of operations on Gauss's law yields $$ \nabla^2 V - \frac{\partial^2 V}{\partial t^2} + \rho= 0$$
These remarkably symmetric equations betray the close connection between relativity and electromagnetism and that electric and magnetic fields are actually part of the electromagnetic field. Whether one observes $\rho$ or ${\bf J}$; ${\bf E}$ or ${\bf B}$, is entirely dependent on frame of reference.