A changing flux is a battery

For a coil of $N$ turns linked by flux $\Phi_B$,

$$\varepsilon = -N\frac{d\Phi_B}{dt}.$$

Every symbol on the right except $N$ is inside a derivative. The magnitude of $\Phi_B$ is not in the formula anywhere, which means these three claims are all wrong for the same reason:

1. "The field is strongest here, so the EMF is biggest here." Strength is not rate.
2. "The magnet is deep inside the coil, so the flux is enormous and so is the EMF." Held still, the derivative is zero and so is the EMF.
3. "The EMF peaks when the flux peaks." At a peak the tangent is flat, so the EMF is exactly zero there.

On a graph of $\Phi_B$ against $t$, the EMF is the slope, flipped in sign and scaled by $N$. Flat stretches give zero however high they sit. The steepest stretch gives the largest EMF however low it sits. A corner in the flux graph is a jump in the EMF graph.

The rotating loop makes the point permanently. With $\Phi_B = BA\cos\omega t$, differentiating gives $\varepsilon = NBA\omega\sin\omega t$: a sine against a cosine, a quarter cycle apart. The EMF is largest exactly where the flux passes through zero.

The minus sign in Faraday's law is a direction, and it points at $d\Phi_B/dt$, not at $\vec{B}$. The induced current always flows so that its own magnetic field fights whatever the flux is doing.

1. Pick a normal for the loop and note which way the external flux points through it.
2. Ask whether that flux is growing or shrinking. This is the step people skip.
3. Growing: the induced current makes field against it. Shrinking: the induced current makes field along it, propping it up.
4. Curl the right hand to convert "field into the page inside the loop" into "clockwise current", and the reverse.

The case that exposes the error is a decreasing into-page flux. A student who has learned "oppose the field" says counterclockwise. The loop actually drives current clockwise, adding its own into-page field to defend the flux it is losing. Half of all Lenz problems on the exam are of this kind: withdrawing magnets, collapsing fields, shrinking loops.

Underneath, this is energy conservation with a compass attached. If the induced current ever helped the change along, the system would accelerate itself and generate free energy.

Since $\Phi_B = BA\cos\theta$, the product rule gives three independent ways to induce an EMF:

$$\frac{d\Phi_B}{dt} = \underbrace{A\cos\theta\,\frac{dB}{dt}}_{\text{changing field}} + \underbrace{B\cos\theta\,\frac{dA}{dt}}_{\text{changing area}} + \underbrace{-BA\sin\theta\,\frac{d\theta}{dt}}_{\text{rotation}}.$$

"The field is constant, so there is no induction" is only true when the other two terms vanish as well. The standard counterexample is a conducting bar of length $L$ sliding at speed $v$ along rails in a constant field $B$. The enclosed area grows at $dA/dt = Lv$, so

$$|\varepsilon| = B\frac{dA}{dt} = BLv,$$

the motional EMF, with no $dB/dt$ anywhere. A generator uses the third lever instead, rotating a coil in a fixed field.

The mirror-image error also exists: a closed loop translating entirely inside a large uniform field induces nothing, because every side sweeps through identical field and the flux never moves. Motion is not the criterion. Changing flux is.

A changing magnetic flux drives current in a ring that contains no battery. What pushes the charges is a genuine electric field, produced by $d\vec{B}/dt$ rather than by any charge, and Faraday's law in its field form says so:

$$\oint \vec{E}\cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}.$$

The left side is the work per unit charge around a closed path, and it is not zero. That single fact separates this field from every electrostatic field met so far:

1. Its field lines close on themselves. They do not start on positive charge and end on negative charge, because there is no charge involved.
2. It is non-conservative. Carry a charge once around the ring and it gains energy $q\varepsilon$, which is what heats the ring.
3. It has no potential function. There is no single-valued $V$ to write down, so "the potential difference around the loop" is not a defined quantity, and two voltmeters attached to the same two points on opposite sides of a ring genuinely read differently.

Note also that the induced $\vec{E}$ exists outside the solenoid producing it, where $\vec{B}$ is essentially zero. What the ring responds to is the flux it encircles, not the field at its own location.