Corrected exercises: electromagnetic induction (203-SN2-RE)

$\Phi=BA\cos\theta=0.400\times(0.120)^{2}\times\cos 30.0°=0.400\times 0.0144\times 0.8660=4.99\times 10^{-3}$ Wb, that is $4.99$ mWb. Had the problem given $30.0$° between the field and the PLANE of the loop, the angle with the normal would be $60.0$° and the flux would drop to $0.400\times 0.0144\times 0.500=2.88\times 10^{-3}$ Wb, nearly half as much. This is the most profitable trap of the whole chapter for an examiner, because it does not show up on rereading: both answers are plausible. The remedy is mechanical, underline the word “normal” or the word “plane” in the problem before writing anything.

The field of the wire is $B(r)=\dfrac{\mu_{0}I}{2\pi r}$: it is NOT uniform over the frame, since it is almost four times stronger at the near edge than at the far edge. The formula $BA$ therefore cannot be used, and the frame has to be cut into vertical strips of width $dr$ and area $L\,dr$, each at constant field. $\Phi=\displaystyle\int_{a}^{b}\dfrac{\mu_{0}I}{2\pi r}L\,dr=\dfrac{\mu_{0}IL}{2\pi}\ln\dfrac{b}{a}$. Numerically: $\dfrac{4\pi\times 10^{-7}\times 15.0\times 0.300}{2\pi}\times\ln 3.50=9.00\times 10^{-7}\times 1.253=1.13\times 10^{-6}$ Wb, that is $1.13$ μWb. The logarithm is the signature of this geometry, and you must learn to recognize it: whenever a field in $\dfrac{1}{r}$ crosses a flat surface, a $\ln\dfrac{b}{a}$ appears.

$B=\mu_{0}nI=4\pi\times 10^{-7}\times 800\times 2.50=2.51\times 10^{-3}$ T. Flux per turn: $\Phi=BA=2.51\times 10^{-3}\times 4.00\times 10^{-4}=1.01\times 10^{-6}$ Wb. Total flux through the $500$ turns: $N\Phi=5.03\times 10^{-4}$ Wb. Two precautions: the area to use is that of the SOLENOID, not that of the outer coil, since there is a field only inside the solenoid; and the total flux, often written $\lambda$ and called flux linkage, is not the flux in the strict sense, which remains defined per turn. It is nevertheless the one that enters Faraday's law.

In none of the three, but for different reasons, and that is what makes the question interesting. Case a): the area quadruples, so the flux quadruples. Case c): the cross-section quadruples too, but be careful, if the length of the solenoid is doubled while keeping the same total number of turns, $n$ is halved, and the flux only doubles; at constant $n$ it quadruples. Case b): this is the only one where the enlargement leaves a partial invariant. Doubling $a$, $b$ AND $L$ multiplies the flux by $\dfrac{2L}{L}=2$, since the ratio $\dfrac{b}{a}$, and therefore the logarithm, does not change. The flux doubles instead of quadrupling. The general lesson is that a field in $\dfrac{1}{r}$ exactly offsets one dimension of the enlargement, which is why logarithmic quantities appear wherever a geometry has no scale of its own.

On approach, the north pole sends toward the loop a field pointing at it, so the flux through it INCREASES. The induced current must oppose this: it creates an opposite field, so it flows COUNTERCLOCKWISE as seen from the magnet, which presents a north face to the magnet. Two norths repel: the loop is pushed away, and work has to be done to bring the magnet closer. On withdrawal, the flux DECREASES; the induced current reverses, it creates a field in the same direction as that of the magnet to try to maintain the flux, presents a south face, and the loop is ATTRACTED. In both cases the effect brakes the motion, and this requirement is no coincidence: if it were the other way round, the magnet would accelerate on its own and we would have perpetual motion.

Outward field that increases: the induced current creates an inward field inside the loop, so it flows CLOCKWISE. As soon as the strength becomes constant, $\dfrac{d\Phi}{dt}=0$ and the induced current vanishes immediately, even though the field stays very strong. This is the most important point of the question: a strong field produces nothing, only a CHANGE produces something. A superconducting magnet at rest induces nothing at all.

Flattening a circle into an ellipse at constant perimeter DECREASES the area, so the inward flux decreases. The induced current opposes this decrease by also creating an inward field: it flows CLOCKWISE, as seen by the reader, since that is the direction that produces an inward field. The resulting force tends to make the loop return to its circular shape, the one of greatest area: once again the system resists what is done to it. A copper ring placed in a strong field and carrying an induced current really does stiffen, and this is a design problem in powerful magnet coils.

While the current builds up, the flux through the second loop goes from zero to its final value: it increases. The current induced in the second loop opposes this and therefore flows in the OPPOSITE direction to that of the first, that is counterclockwise, and the two loops repel each other. Once steady state is reached, the current in the first loop is constant, the flux no longer varies, and the induced current in the second loop becomes EXACTLY ZERO: the two loops stop influencing each other electrically. This is exactly the principle of the transformer, and it is also why a transformer does not work with direct current.

At each alternation, the flux through the ring changes, a current is induced in it and, by Lenz's law, it opposes this change. One might expect the ring to be pushed away during the rising phases and pulled in during the falling phases, but the time average is repulsive: the inductance of the ring makes the induced current lag behind the field, so the repulsive phases outweigh the attractive ones. The ring therefore jumps, and it heats up. If it has a SLIT, the circuit is open: no current can flow, no force arises, and the ring stays in place. This is the clearest experimental demonstration that induction needs a CLOSED circuit to produce a current, even though the emf itself still exists across the slit.

The counterexample is the second half of a), or situation c). When the magnet moves AWAY, the induced current creates a field in the SAME direction as that of the magnet, it reinforces it instead of opposing it. Likewise in c), the induced current creates an inward field when the existing field is already inward. The correct statement is therefore: the induced current opposes the CHANGE in flux, never the flux itself. This one-word nuance is what separates a right answer from an answer that is wrong half the time, and it is the first reflex to install on this chapter.

$\varepsilon=-N A\dfrac{dB}{dt}$ with $\dfrac{dB}{dt}=0.240-0.120\,t$. Hence $\varepsilon(t)=-200\times 2.50\times 10^{-3}\times(0.240-0.120t)=-0.120+0.0600\,t$, in volts. At $t=0$: $\varepsilon=-0.120$ V, of magnitude $120$ mV. At $t=2.00$ s: $\varepsilon=0$ exactly. At $t=4.00$ s: $\varepsilon=+0.120$ V, the same magnitude as at the start but the OPPOSITE sign. Notice that $NA=0.500$ m$^{2}$ here, which simplifies every calculation: the emf is simply half of $\dfrac{dB}{dt}$.

$\dfrac{dB}{dt}=0$ gives $t=\dfrac{0.240}{0.120}=2.00$ s, and $B(2.00)=0.0800+0.480-0.240=0.320$ T. This is the point marked on the curve. The connection is the heart of the exercise: the emf vanishes exactly at the instant when the field is MAXIMUM. This is not paradoxical, it is the definition of a maximum, a horizontal tangent, therefore a zero derivative, therefore a zero emf. The problem could just as well state a perfectly constant field of ten teslas, and the emf would be zero. This is the most concrete formulation of the rule of the chapter: it is never the value of the field that induces, it is its slope.

$I=\dfrac{|\varepsilon|}{R}=\dfrac{0.120}{12.0}=1.00\times 10^{-2}$ A, that is $10.0$ mA at $t=0$. The current decreases linearly in magnitude until it vanishes at $t=2.00$ s, then it grows again but in the OPPOSITE direction, to reach $10.0$ mA again at $t=4.00$ s. Physically: during the first two seconds the flux increases and the induced current opposes this; afterwards the flux decreases and the induced current reverses to try to maintain it. The reversal of the current coincides with the maximum of the field.

$q=\displaystyle\int_{0}^{4}I\,dt=\dfrac{N}{R}\displaystyle\int_{0}^{4}\dfrac{dB}{dt}A\,dt=\dfrac{NA\,\Delta B}{R}$, and $\Delta B=B(4.00)-B(0)=0.0800-0.0800=0$. The net charge is therefore EXACTLY ZERO. That is the surprise: a current did flow for four seconds, it heated the coil, and yet no net charge crossed the cross-section. The explanation is that the current changed direction halfway and the two halves cancel exactly, as the symmetry of the parabola requires. The general lesson is that the induced charge depends ONLY on the initial and final states of the flux, never on the path between the two, whereas the heat depends on the whole path.

$W=\displaystyle\int_{0}^{4}\dfrac{\varepsilon^{2}}{R}dt=\dfrac{1}{12.0}\displaystyle\int_{0}^{4}(-0.120+0.0600t)^{2}dt$. Setting $u=-0.120+0.0600t$, with $du=0.0600\,dt$, the integral becomes $\dfrac{1}{0.0600}\displaystyle\int_{-0.120}^{0.120}u^{2}du=\dfrac{1}{0.0600}\times\dfrac{2(0.120)^{3}}{3}=\dfrac{3.456\times 10^{-3}}{0.180}=1.92\times 10^{-2}$. Divided by $R$: $W=\dfrac{1.92\times 10^{-2}}{12.0}=1.60\times 10^{-3}$ J, that is $1.60$ mJ. This energy does not come from nowhere: it is supplied by the device that makes the field vary, which has to work against the field created by the induced current. This is the energy counterpart of Lenz's law, and the contrast with question d) is the real lesson of the exercise: the charge is zero but the energy bill is not.

The formula $B\ell v$ assumes that ALL the points of the conductor have the same speed, which is true in translation and false in rotation: the foot of the rod is motionless, the tip moves at $\omega\ell$, and each intermediate point at $\omega r$. So we cut the rod into elements $dr$ at distance $r$ from the axis, each contributing $d\varepsilon=B(\omega r)\,dr$, then add up: $\varepsilon=\displaystyle\int_{0}^{\ell}B\omega r\,dr=\dfrac{1}{2}B\omega\ell^{2}$. The factor $\dfrac{1}{2}$ is therefore the average of the speeds along the rod, and writing $B\ell(\omega\ell)$ gives exactly double, which is the most common wrong answer. Alternative check: in one revolution, the rod sweeps the area $\pi\ell^{2}$ in $\dfrac{2\pi}{\omega}$ seconds, so $\dfrac{d\Phi}{dt}=B\dfrac{\pi\ell^{2}\omega}{2\pi}=\dfrac{1}{2}B\omega\ell^{2}$. Both routes give the same result.

$\varepsilon=\dfrac{1}{2}\times 0.350\times 60.0\times(0.250)^{2}=0.656$ V. For the sign, apply the Lorentz force to a positive carrier in the rod: $\vec{F}=q\vec{v}\times\vec{B}$. The velocity is tangential, the field comes out of the page, and the cross product gives a force directed OUTWARD, along the rod. Positive charges therefore accumulate at the free end, which is at the HIGHEST potential, and the foot of the rod is at the lowest potential. The reflex to keep is to always go back to $q\vec{v}\times\vec{B}$ for the sign of a motional emf: Lenz's law gives the direction of the current in a closed circuit, but it says nothing about the polarity of an isolated conductor.

$\omega=\dfrac{1800\times 2\pi}{60}=188.5$ rad/s. The disk behaves like an infinity of radial rods in parallel, all with the same emf: $\varepsilon=\dfrac{1}{2}B\omega R^{2}=\dfrac{1}{2}\times 0.500\times 188.5\times(0.120)^{2}=0.679$ V. Note that a conversion to radians per second is essential before any calculation, and that forgetting the factor $\dfrac{2\pi}{60}$ produces an error of about a factor of ten, which rarely goes unnoticed but costs all the marks of the question.

$I=\dfrac{\varepsilon}{R}=\dfrac{0.679}{0.0200}=33.9$ A. Electrical power: $P=\varepsilon I=0.679\times 33.9=23.0$ W, entirely dissipated in the resistance of the circuit. Torque: in steady state the motor supplies exactly the electrical power produced, so $\Gamma=\dfrac{P}{\omega}=\dfrac{23.0}{188.5}=0.122$ N·m. This torque is the mechanical translation of Lenz's law: the induced current, flowing radially in the field, experiences a tangential Laplace force that BRAKES the disk, and the motor has to overcome it. Without it, the disk would produce energy without consuming any.

The flaw is visible in the emf: $0.679$ V for a disk turning at $1800$ revolutions per minute in half a tesla, less than a single battery cell, while the current reaches $34$ A. The Faraday disk is a source of very low voltage and very high current, which is the worst possible compromise for carrying energy: the line losses are $RI^{2}$, and this enormous current makes them prohibitive. The structural cause is that there is only one “turn” of circuit, the radius, whereas an alternator winds $N$ turns and so multiplies the emf by $N$, several hundred in practice. On top of that, the output of the disk is direct, so it cannot be stepped up by a transformer, whereas the alternating output of an alternator can be transformed and transmitted at high voltage. For these two reasons, and not for lack of ingenuity, the homopolar disk remained a laboratory curiosity.

The field exists only INSIDE the primary solenoid: $B=\mu_{0}n_{1}I_{1}$, and it is zero outside. The flux through one turn of the secondary is therefore limited to the cross-section of the primary, $A_{1}=\pi(0.0200)^{2}=1.257\times 10^{-3}$ m$^{2}$, whatever the actual size of the secondary turns. This is the method point of the question, and it generalizes: the flux common to two circuits is always the one passing through the SMALLER of the two, or more exactly through the region where the field exists. $M=\dfrac{N_{2}\Phi_{2}}{I_{1}}=\mu_{0}n_{1}N_{2}A_{1}=4\pi\times 10^{-7}\times 1200\times 300\times 1.257\times 10^{-3}=5.68\times 10^{-4}$ H, that is $0.568$ mH.

$|\varepsilon_{2}|=M\left|\dfrac{dI_{1}}{dt}\right|=5.68\times 10^{-4}\times 250=0.142$ V, that is $142$ mV. Notice that the emf does NOT depend on the value of the primary current, only on its rate of change: a perfectly steady current of a thousand amperes would induce nothing at all, whereas a current of one milliampere cut off in one microsecond would induce hundreds of volts. This is the principle of the ignition coil of an engine, which produces tens of kilovolts from a twelve-volt battery.

Primary: $L_{1}=\mu_{0}n_{1}^{2}A_{1}\ell_{1}=4\pi\times 10^{-7}\times(1200)^{2}\times 1.257\times 10^{-3}\times 0.400=9.10\times 10^{-4}$ H, that is $0.910$ mH. Secondary: its turn density is $n_{2}=\dfrac{300}{0.0500}=6000$ turns per metre, five times that of the primary. $L_{2}=\mu_{0}n_{2}^{2}A_{1}\ell_{2}=4\pi\times 10^{-7}\times(6000)^{2}\times 1.257\times 10^{-3}\times 0.0500=2.84\times 10^{-3}$ H, that is $2.84$ mH. The secondary, although eight times shorter, has an inductance three times LARGER, because inductance grows as the square of the turn density.

$k=\dfrac{M}{\sqrt{L_{1}L_{2}}}=\dfrac{5.68\times 10^{-4}}{\sqrt{9.10\times 10^{-4}\times 2.84\times 10^{-3}}}=\dfrac{5.68\times 10^{-4}}{1.61\times 10^{-3}}=0.354$. And $\sqrt{\dfrac{\ell_{2}}{\ell_{1}}}=\sqrt{\dfrac{0.0500}{0.400}}=\sqrt{0.125}=0.354$: identical. The statement in the problem is a trap and has to be corrected. It is true that all the flux OF THE PRIMARY passes through the secondary, but coupling is measured in both directions, and the converse is false: the short secondary creates a field that fills only an eighth of the length of the primary, so the other seven eighths of the primary turns see almost nothing. It is this geometric asymmetry that drops $k$ to $0.354$. To approach $k=1$, the two windings must occupy the same volume, which in practice is achieved by winding them one over the other along the whole length of a closed ferromagnetic core.

$M_{12}=M_{21}$ is a general theorem, a consequence of the symmetry of the magnetic energy of a system of coupled circuits. Its practical strength is precisely this: you are free to calculate $M$ in the most CONVENIENT direction. Here, the easy direction is primary to secondary, because the field of a long solenoid is uniform, known and simple, $\mu_{0}n_{1}I_{1}$. In the other direction, you would have to calculate the field created by a SHORT coil, which is neither uniform nor given by an elementary formula, and then integrate that field over the $480$ turns of the primary, most of which lie in the fringe region where the field flares out. The calculation is feasible but requires elliptic integrals, whereas the result is the same to the last decimal. Recognizing the easy direction is what the reciprocity theorem really gives a student.

We apply the integral form of Faraday's law, $\displaystyle\oint\vec{E}\cdot d\vec{\ell}=-\dfrac{d\Phi}{dt}$, on a circle of radius $r$ centred on the axis. By symmetry, $\vec{E}$ is tangential and of constant magnitude on this circle, so the left-hand side is $E(2\pi r)$. For $r<R$, the enclosed flux is $B\pi r^{2}$, so $E(2\pi r)=\pi r^{2}\dfrac{dB}{dt}$ and $E=\dfrac{r}{2}\dfrac{dB}{dt}$: the field grows LINEARLY with $r$. For $r>R$, the enclosed flux no longer grows, it is capped at $B\pi R^{2}$, so $E(2\pi r)=\pi R^{2}\dfrac{dB}{dt}$ and $E=\dfrac{R^{2}}{2r}\dfrac{dB}{dt}$: the field falls off as $\dfrac{1}{r}$. The structure is exactly that of the magnetic field of a thick wire, and for the same mathematical reason.

At $r=3.00$ cm: $E=\dfrac{0.0300}{2}\times 0.150=2.25\times 10^{-3}$ V/m, that is $2.25$ mV/m, the first marked point. At $r=R=5.00$ cm: $E=\dfrac{0.0500}{2}\times 0.150=3.75$ mV/m, the maximum, reached at the edge and visible as the sharp peak of the curve. At $r=10.0$ cm: $E=\dfrac{(0.0500)^{2}}{2\times 0.100}\times 0.150=1.875$ mV/m, the second point. Check: the two expressions give the same value at $r=R$, which should always be verified on a piecewise law.

No, and this is the deepest content of the chapter. Faraday's law in integral form does not relate $\vec{E}$ to the LOCAL magnetic field, it relates its circulation around a contour to the flux ENCLOSED by that contour. A circle of radius $10.0$ cm surrounds the whole region where $B$ varies, so it encloses a changing flux, so it carries a nonzero circulation, even though at each of its points the magnetic field is strictly zero. This is a nonlocal effect, and it is the same phenomenon that lets a transformer work although its windings sit in a region where the leakage field is weak. The differential version, $\vec{\nabla}\times\vec{E}=-\dfrac{\partial\vec{B}}{\partial t}$, says in fact that outside the curl is zero: the field there is locally irrotational, while having a nonzero circulation around a contour that encircles the region. This is possible because the outside domain is not simply connected.

Circle of radius $3.00$ cm. First method, through the field: $\varepsilon=E\times 2\pi r=2.25\times 10^{-3}\times 2\pi\times 0.0300=4.24\times 10^{-4}$ V. Second method, through the flux: $\varepsilon=\pi r^{2}\dfrac{dB}{dt}=\pi(0.0300)^{2}\times 0.150=4.24\times 10^{-4}$ V. Identical. Circle of radius $10.0$ cm. Through the field: $1.875\times 10^{-3}\times 2\pi\times 0.100=1.18\times 10^{-3}$ V. Through the flux: $\pi R^{2}\dfrac{dB}{dt}=\pi(0.0500)^{2}\times 0.150=1.18\times 10^{-3}$ V, using $R$ and not $r$, since beyond the cylinder there is no more flux to add. This is the error to avoid: the radius of the contour and the radius of the region carrying the flux are two different things.

The two voltmeters really do measure different voltages, and neither is faulty. The reason is that the induced electric field is NOT conservative: its circulation around a closed loop is $-\dfrac{d\Phi}{dt}$ and not zero. The integral $\displaystyle\int\vec{E}\cdot d\vec{\ell}$ between two points then depends on the path, and the two voltmeters, whose leads run on either side of the region, follow two paths that differ by a complete loop: their readings differ by exactly the total emf, $1.18$ mV here. What this reveals is important and unsettling: the very notion of POTENTIAL no longer makes sense in a region where $\dfrac{\partial\vec{B}}{\partial t}\ne 0$. You can only define $V$ if $\displaystyle\oint\vec{E}\cdot d\vec{\ell}=0$ for every loop, which is true in electrostatics and false here. The question “what is the voltage between A and B?” is then ill-posed until the path has been specified, which is why, in power electronics, as much care goes into the LAYOUT of the measurement loops as into the choice of instruments.

$U_{2}=U_{1}\dfrac{N_{2}}{N_{1}}=120\times\dfrac{200}{2400}=10.0$ V. Current in the load: $I_{2}=\dfrac{10.0}{5.00}=2.00$ A. Power: $P=U_{2}I_{2}=20.0$ W. The transformer is a step-down by a factor of twelve: it divides the voltage by twelve and, in return, multiplies the available current by twelve.

First method, through conservation of power, which is the very definition of an ideal transformer: $I_{1}=\dfrac{P}{U_{1}}=\dfrac{20.0}{120}=0.167$ A. Second method, through the turns ratio: $I_{1}=I_{2}\dfrac{N_{2}}{N_{1}}=2.00\times\dfrac{1}{12}=0.167$ A. Identical. The point to watch is the DIRECTION of the ratio: voltages follow $\dfrac{N_{2}}{N_{1}}$ and currents follow the inverse. The check that never fails is to verify that the product $UI$ is the same on both sides.

Reflected impedance: $Z_{1}=\left(\dfrac{N_{1}}{N_{2}}\right)^{2}Z_{2}=(12.0)^{2}\times 5.00=720$ Ω. Direct check: $\dfrac{U_{1}}{I_{1}}=\dfrac{120}{0.167}=720$ Ω. The source therefore does not see a resistance of five ohms at all, it sees seven hundred and twenty ohms. This is the most useful property of the transformer after voltage conversion, and it explains its historic role in audio: an eight-ohm loudspeaker cannot be driven directly by a vacuum tube that wants to see several kilohms, and the output transformer does exactly that matching. The square of the ratio is essential: dividing the voltage by twelve and multiplying the current by twelve divides the impedance by one hundred and forty-four.

$\Phi_{\max}=\dfrac{U_{\text{rms}}}{4.44\,N_{1}f}=\dfrac{120}{4.44\times 2400\times 60.0}=\dfrac{120}{6.394\times 10^{5}}=1.88\times 10^{-4}$ Wb. The coefficient $4.44$ equals $\sqrt{2}\pi$ and comes from converting the maximum value to the rms value of a differentiated sinusoid. The formula says something important for design: at a given voltage, the flux required is inversely proportional to the FREQUENCY. This is why an aircraft transformer at $400$ Hz is several times smaller than a mains transformer at $60$ Hz for the same power, and why modern electronics works at tens of kilohertz.

Efficiency: $\eta=\dfrac{20.0}{20.0+1.20}=\dfrac{20.0}{21.2}=0.943$, that is $94.3$ %. With direct current, the flux in the core is CONSTANT, so $\dfrac{d\Phi}{dt}=0$, so the induced emf in the secondary is zero: the transformer transmits strictly nothing, whatever the applied voltage. Worse, the primary then presents only the ohmic resistance of its winding, a few ohms instead of the $720$ Ω calculated in c), because that impedance came entirely from induction. The current drawn would explode, and the winding would burn out within seconds. This is a classic accident, and it is the best practical demonstration that what limits the current in a transformer is not its resistance but its inductive reaction, that is, Lenz's law.

False, and this is error number one of the chapter. Faraday's law involves $\dfrac{d\Phi}{dt}$, never $\Phi$: a perfectly steady ten-tesla field induces exactly zero, whereas a ten-millitesla field cut off in a millisecond induces far more. Correct statement: it is the RATE OF CHANGE of the flux that produces the emf, and a strong but constant field produces nothing at all. The example to give is the magnetic resonance imaging scanner, whose main magnet, several teslas strong, induces nothing as long as it is steady: it is the much weaker switched gradients that induce currents in the patient.

False, and a single word is missing for it to be true. The induced current opposes the CHANGE in flux, not the flux. When the flux DECREASES, the induced current creates a field in the SAME direction as the applied field, to try to maintain it: it reinforces it instead of opposing it. Correct statement: the induced current flows in the direction such that its own field opposes the CHANGE in the flux that gives rise to it. A magnet pulled away from a coil is attracted by it, and this is the shortest counterexample to give in an exam.

The statement confuses emf and current. The emf exists as soon as the flux changes, independently of any circuit: it is the work per unit charge done by the induced electric field, and it shows up as a measurable voltage across the open terminals of the coil. What requires a closed circuit is the CURRENT, and therefore also the Lenz force and the dissipation. Correct statement: an open coil does develop an emf $-N\dfrac{d\Phi}{dt}$ across its terminals, it simply delivers no current, feels no force and dissipates no energy. This is in fact the operating mode of most inductive sensors, whose open-circuit voltage is read.

False. Lenz's law is nothing other than conservation of energy applied to induction. If the induced current reinforced the change instead of opposing it, every disturbance would amplify itself: a magnet brought toward a coil would be drawn in, accelerate, produce a stronger current that would draw it in even harder, and we would have an inexhaustible source of energy. Correct statement: Lenz's law is the minus sign of Faraday's law, and this sign is imposed by conservation of energy. This is also why it is sometimes presented as the moderation principle, under which every system resists what is imposed on it.

False, and this is the content of exercise 6. The potential can only be defined if the electric field is CONSERVATIVE, that is, if its circulation around every closed loop is zero. But Faraday's law says exactly the opposite as soon as the flux varies: $\displaystyle\oint\vec{E}\cdot d\vec{\ell}=-\dfrac{d\Phi}{dt}\ne 0$. Correct statement: in such a region, the integral of $\vec{E}$ between two points depends on the path followed, two voltmeters connected to the same terminals by different paths display different values, and the notion of potential difference loses its meaning. It only regains it in the static regime, or in practice when the measuring loop encloses no changing flux.