The current the EMF drives, and the drag it pays for

Two stages, in this order, every time:

1. Geometry and motion set the EMF. $\varepsilon = -N\,d\Phi_B/dt$. No resistance appears in this equation, so nothing about the wiring can change the answer.
2. The circuit sets the current. $I = \varepsilon/R$. Double $R$ and the current halves while the EMF sits exactly where it was.

The clean test is an open loop. Cut a gap in a ring and wave a magnet at it: the EMF is completely unchanged, and it now appears as a measurable voltage across the gap, with zero current. "An open loop has no induction" gets this backward, and so does "more resistance means less EMF".

Resistance does eventually reach back into the mechanics, but only through the current. For a bar of length $L$ sliding at $v$ on rails closed by resistance $R$,

$$\varepsilon = BLv, \qquad I = \frac{BLv}{R}, \qquad F = BIL = \frac{B^2L^2v}{R}.$$

So $R$ changes the drag on the bar. It does that by changing $I$, one link further down the chain, and never by changing $\varepsilon$.

A rod moving through a field is a battery, and which end is its positive terminal is not a matter of taste. Every charge carrier in the rod is dragged along at velocity $\vec{v}$, so it feels

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

Do that cross product with the rod's velocity, never with the rod's own direction. The component of $\vec{F}$ along the rod pushes positive carriers to one end, which charges positive; the other end is left negative. Charge piles up until the electrostatic pull balances the magnetic push, at which point the rod's terminals sit at

$$\varepsilon = \int (\vec{v}\times\vec{B})\cdot d\vec{\ell} = BLv$$

for the usual perpendicular case. In the external circuit, conventional current then leaves the positive terminal, exactly as with a chemical cell.

Two habits keep this straight. Order matters: $\vec{B}\times\vec{v}$ is the negative of $\vec{v}\times\vec{B}$ and names the wrong end every single time. And separation is not creation: the rod stays neutral overall while its ends carry equal and opposite charge.

Once the induced current exists, the field pushes on it, and the direction is never in doubt: the force opposes the relative motion that caused the induction.

1. A magnet approaching a ring is repelled: the ring's near face behaves like a matching pole.
2. A magnet receding from a ring is attracted: the ring tries to hold on. Both cases resist.
3. A bar sliding on rails feels a retarding force $B^2L^2v/R$, so keeping it at constant speed requires a steady applied force even on frictionless rails.
4. A magnet dropped down a copper tube reaches a terminal velocity, though copper is not ferromagnetic at all.

Argue it from energy rather than from memory. The resistor dissipates $I^2R$ every second, and that energy comes from the agent doing the moving. For the sliding bar, check the books:

$$P_{\text{agent}} = Fv = \frac{B^2L^2v^2}{R} = I^2R = P_{\text{dissipated}}.$$

If the induced force ever assisted the motion instead, the bar would accelerate on its own and heat the resistor, which is a free-energy machine. Note also that the magnetic force does no net work on the carriers; it is the agent's push that pays, with the field acting only as the go-between.

Put a second coil near a first. What the secondary sees is the flux the primary makes, and what it responds to is the derivative of that flux, so what actually crosses the gap is $dI_1/dt$:

$$\varepsilon_2 = -M\frac{dI_1}{dt}.$$

Three consequences worth committing:

1. A steady primary current, however large, induces nothing. A $200$ A DC bus next to a coil is silent.
2. Closing and opening the primary switch each produce a blip, and the two blips run in opposite directions, since one is a rise and the other a fall.
3. Transformers require AC for exactly this reason. On DC a transformer passes one pulse at switch-on and then nothing.

The same logic explains why a magnet dropped through a coil writes an S-shaped trace rather than a single bump. On the way in the flux climbs and the current runs one way; at the instant the magnet is centered the flux is at a maximum, so its slope, and the current, pass through zero; on the way out the flux collapses and the current reverses. Every passage records both signs, and the second lobe is taller and narrower than the first because the magnet is moving faster by then.