Will an emf be induced across the ends of a stationary metal rod placed in a time-varying magnetic field?

This is more of an expansion of leongz's answer.

TL;DR: The situation is incomplete. There may be an emf, and there may be a deflection. Existence of emf and deflection are independent. We cannot calculate the value of the emf from the given data (i.e., from a given time-varying $\bf B$ field)

Your fundamental issue is that Maxwell's equations (of which Faraday's law is one) are not "cause and effect". You cannot "plug in" a value of magnetic field and get a corresponding value of $\bf E$ field induced by the $\bf B$ field. All Maxwell's equations tell you is "which kinds of $\bf E$ and $\bf B$ fields can coexist given so-and-so conditions".

Trying to solve the situation via Maxwell's equations

I remember solving a similar situation via Maxwell's equations and being surprised by the answer.

The "initial conditions" were $\mathbf {B}=\beta t\hat k$, $\rho=0$ (no charge), $\mathbf{J}=0$ (no current).

Solving{*} for $\mathbf{E}$, using the differential+microscopic form of Maxwell's equations(since the integral form can only get you the value of $\bf E$ at certain positions at many times), I got:

$$\mathbf{E}=\hat i (lx + \frac{\beta}{2}y+az+c_1)+\hat j(-\frac\beta{2}x+my+bz+c_2)+\hat k(ax+by+nz+c_3)$$

where $a,b,l,m,n,c_1,c_2,c_3$ are arbitrary constants subject to $l+m+n=0$

Note that this is a family of electric fields (Setting certain constants to zero, you get concentric ellipses IIRC). All this means is that any $\bf E$ field of this type can coexist with a $\bf B$ field.

Implication for your problem

This means that your initial conditions are insufficient/inconsistent. Along with such a magnetic field, any type of electric field satisfying the above equations can exist--and must exist.

So, in addition to knowing how your magnetic field is changing with time, you need to know:

• Which one of these bajillion electric fields is present
• Where is the rod in relation to this electric field?

These can usually be determined if you know the boundary conditions for the system. In a physical situation, these can be extracted from the setup.

Some more analysis

Let's choose a simple solution and analyse it. I'm taking the case where the coexisting electric field is just concentric circles.

In this diagram, the blue stuff is the $\bf B$ field, and the green stuff is the $\bf E$ field. Being lazy, I haven't added arrows to it (I also haven't spaced the circles properly. There should be more space between the inner ones and less space between the outer ones). The other things are just rods and wire loops.

To avoid confusion, when I refer to "emf", I mean "the energy gained/lost in moving a unit test charge along a given path". Mathematically, the path integral $\int_{path}\mathbf{E}\cdot \mathrm{d} \vec l$. I'll come to voltmeters and the like later.

Let's first look at the rods. The yellow rod $AB$ will have no emf across its ends, since the $\bf E$ field is perpendicular to its length at all points. On the other hand, the magenta rod $CD$ has an emf across its ends. This emf can be easily calculated via some tricks--avoiding integration--but let's not get into that right now.

You now can probably see why the second point "Where is the rod in relation to this electric field?" matters.

On the other hand, this second point is not necessary for a loop. Indeed, neither is the first point.

Going around the loop, both loops (cyan and red in diagram) will have an emf $-A\frac{\partial\mathbf{B}}{\partial t}$. It's an interesting exercise to try and verify this without resorting to Faraday's law--take an electric field $\mathbf{E}=kr\hat\tau$ and do $\int \mathbf{E}\cdot\mathrm{d}\vec l$ around different loops of the same area. You should get the same answer.

But, you cannot divide this emf by four and say that each constituent "rod" of the loop has that emf. For example, in the cyan loop $EFGH$, $EF$ has no emf, and the rest have different emfs. "dividing by four" only works (in this case) if the loop is centered at the origin.

Voltmeters

Voltmeters are an entirely different matter here. The issue with voltmeters is that, even for so-called "ideal" voltmeters, the p.d. measured depends upon the orientation of the voltmeter.

Reusing the same situation:

Here, the black wires are part of the current loop (and peripherals). The yellow/magenta wires are to be swapped in and out for using the voltmeters. Only one set of colored wires will be present at a given time.

Here, the "yellow" voltmeter will measure a pd three times that of the "magenta" one. This is due to the fact that it spans thrice the area and consequently has thrice the flux.

Remember, induced $\bf E$ fields are nonconservative, so voltmeters complicate things. They don't really tell you anything tangible, either.

If the voltmeter were an everyday, non-ideal galvanometer based voltmeter, there would be extra complications due to there being a second loop.

One more thing about rods

A rod can additionally cause the extra complication of being polarizable/magnetizable. Then, you have to consider the macroscopic Maxwell equations, with the $\bf D,P,M,H$ fields and bound/free charges/currents. But then you need to know about the material of the rod. Or, just find a hypothetical rod with $\mu_r=\varepsilon_r=1$ and use it.

Also, the charges in a rod will tend to redistribute, nullifying the electric field and thus the emf in the rod.

Conclusion

The given data is incomplete. There is a truckload of different $\mathbf E$ fields that you can use here, and you're not sure which one it is. Additionally, even if we knew which field it was, the orientation of the rod comes into the picture.

So, the rod will have a motional emf, but this emf may be zero. The exact value of this emf cannot be calculated if you only know $\bf B$.

An ideal voltmeter, again, may show deflection. Not necessarily, though.

_{*Solving simultaneous PDEs in four variables is not too fun, so I did make some assumptions regarding the symmetry of the situation to simplify stuff. So the given family of solutions is a subset of the actual solution set. That doesn't hamper this discussion though.}

PLEASE NOTE: This answer is provided only as a student's reference. Following the principle of theoretical classical electromagnetism, solution might be much more complex and possibly exceed the ambitions of the question.

The answer can be found using Lorenz force

$$\vec{F} = q \vec{E} + q \vec{v} \times \vec{B}.$$

Of course, there are always "free" electrons within metal rod. However, if there is no bar movement and $v = 0$, there is also no magnetic force on electrons and there will be no induced emf.

If rod, however, moves within constant magnetic field, so do all the electrons within it, magnetic force push them in one direction, concentrating electrons on one side of the rod. This creates electric field within the rod and consequently measureable emf.

(Back to your question: If you connected an ideal voltmeter (with infinite resistance) across the ends of the rod, then you do create a loop. But let's suppose for the sake of argument that we have some kind of loopless voltmeter based on some entirely new principle.)

According to the Maxwell's equation $$\nabla\times\mathbf{E}=-\frac{1}{c}\frac{\partial \mathbf{B}}{\partial t},$$ a time-dependent magnetic field $\mathbf{B}$ must co-exist with an electric field $\mathbf{E}$. So, even when the rod is stationary, the electric field will cause charges in the rod to move and gather near the ends of the rod, giving a potential difference across the rod.

Specifically, since the electric field inside a perfect conductor must be zero, the charges will move in order to exactly cancel out the electric field that comes with the magnetic field. This happens regardless of the presence of a voltmeter.