Electromagnetic induction: Year 5 test

Electromagnetic induction opens the Year 5 physics syllabus: in the science-oriented liceo scientifico the test falls between September and November, often before maths has even finished the chapter on limits. It is the first grade of the final year, and for many students it is also the first moment when physics stops being “apply the formula” and asks you to reason about a direction, a sign, a cause.

The most widespread misunderstanding is to think that induction is about the magnetic field. It is not: it is about the flux, that is, how much field passes through a given surface, and above all about its changes. An enormous but constant magnetic field induces absolutely nothing, and a very weak field that changes quickly can light a bulb. Anyone who does not get this distinction into focus in the first week never will, and pays for it again at the maturità (Italy’s school-leaving exam).

Here you will find flux, the Faraday-Neumann-Lenz law with its minus sign properly explained, the three set-ups that come up in the test, three worked exercises and the five mistakes that cost the most marks. The map of the whole electromagnetism block with a view to the exam is instead in electromagnetism for the maturità.

Flux: the quantity everything depends on

Φ (B) = B \cdot S = B S cos θ

It is measured in webers ( $1 Wb = 1 T \cdot m^{2}$ ) and read like this: how many field lines pass through the surface. The angle $θ$ is the one between $B$ and the normal to the surface, not between $B$ and the plane of the loop: it is worth a clean mark in the test, because mixing them up shifts the cosine by $90°$ and turns a maximum into a zero.

From here follows the only classification you need. Flux can change in three ways, and every exercise in the test is one of these three:

What changes	Typical example	Formula you will use
the field $B$	magnet brought closer, changing current in a nearby circuit	$ΔΦ = S Δ B cos θ$
the area $S$	rod sliding on two rails	$ΔΦ = B Δ S$
the angle $θ$	loop rotating → alternator	$Φ (t) = B S cos (ω t)$

When you open the text of a problem, the first question to ask yourself is not “which formula do I use” but which of the three is changing. The formula comes afterwards, by itself.

Faraday-Neumann-Lenz

E = - N \frac{ΔΦ ( B )}{Δ t}

where $N$ is the number of turns of the winding and $E$ is the electromotive force (emf). If the circuit has resistance $R$ , the induced current follows from Ohm’s law:

i = \frac{∣ E ∣}{R}

Two points that are worth marks. The emf depends on the rate of the change, not on its size: halving the time doubles the emf. And $N$ is systematically forgotten — if the text speaks of a coil of 200 turns, that 200 multiplies everything.

The minus sign: Lenz’s law

The minus sign is not an accounting convention and does not mean that the emf is negative. It tells you the direction of the induced current, with this rule:

The induced current flows in the direction that opposes the change that produced it.

Mind the word change: the current does not oppose the field, it opposes its change. In practice:

if the linked flux increases, the induced current generates a field opposite to the external one, to brake the increase;
if the flux decreases, the induced current generates a field in the same direction, to sustain the flux that is fading.

The deeper reason is conservation of energy: if the induced current went along with the change instead of opposing it, the phenomenon would feed itself and you would get energy from nothing. This sentence, written in the theory part of the test, is always worth a mark.

In practice, with a magnet approaching a loop: the flux increases, the loop reacts by generating a field that counters it, and the macroscopic result is that the loop repels the magnet. If you move it away, it attracts it. In both cases it opposes what you are doing — and that is why you have to do work to move it.

The most frequently asked set-up: the sliding rod

A rod of length $l$ slides with speed $v$ on two conducting rails, immersed in a field $B$ perpendicular to the plane. It is the most frequent problem in the chapter, and the formulas can all be derived from the flux.

E = B l v, i = \frac{B l v}{R}, F_{braking} = B i l = \frac{B ^{2} l ^{2} v}{R}

The third formula is the one that makes the exercise interesting: the induced current, immersed in the field, experiences a magnetic force that brakes the rod — Lenz again. To keep it moving at constant speed you have to apply an equal and opposite force, and the mechanical power you spend is exactly the power dissipated by the Joule effect:

P_{mechanical} = F v = \frac{B ^{2} l ^{2} v ^{2}}{R} = R i^{2} = P_{dissipated}

This equality is the nicest consistency check in the chapter, and the first one I get students to do in lessons: if the two numbers do not match, there is an error somewhere.

Self-induction, alternator, transformer

Self-induction. A changing current in a circuit generates a changing flux in the circuit itself, and therefore an emf that opposes the change:

Φ = L i, E = - L \frac{Δ i}{Δ t}, W = \frac{1}{2} L i^{2}

The inductance $L$ is measured in henries. The last formula says that an inductor carrying a current stores energy in the magnetic field, exactly as a capacitor stores it in the electric field.

Alternator. A loop rotating with angular velocity $ω$ in a uniform field:

Φ (t) = N B S cos (ω t) ⟹ E (t) = N B S ω sin (ω t)

The emf is maximum when the flux is zero (loop parallel to the field) and zero when the flux is maximum. It seems counter-intuitive until you remember that what counts is the change, not the value: it is the point where the flux changes fastest.

The effective (r.m.s.) values, which are the ones read by instruments:

V_{eff} = \frac{V _{max}}{2}, i_{eff} = \frac{i _{max}}{2}

Transformer. Two windings on the same core:

\frac{V _{2}}{V _{1}} = \frac{N _{2}}{N _{1}}, \frac{i _{2}}{i _{1}} = \frac{N _{1}}{N _{2}}

The two proportions are inverted, and it is the favourite trap of tests. The reason is conservation of energy: in an ideal transformer $V_{1} i_{1} = V_{2} i_{2}$ , so if the voltage goes up the current goes down in the same proportion. And a transformer works only with alternating current: with direct current the flux is constant, so it induces nothing.

Three worked exercises

1. A loop in a changing field

A square loop of side $l = 0.20$ m and resistance $R = 2.0 Ω$ is immersed in a field perpendicular to its plane, which goes from $0.50$ T to $0.10$ T in $Δ t = 0.20$ s. Find the emf and the induced current.

The area is $S = (0.20)^{2} = 0.040 m^{2}$ , and the field is perpendicular to the plane, so parallel to the normal: $cos θ = 1$ .

ΔΦ = S Δ B = 0.040 \cdot (0.10 - 0.50) = - 0.016 Wb

∣ E ∣ = \frac{∣ΔΦ∣}{Δ t} = \frac{0.016}{0.20} = 0.080 V

i = \frac{0.080}{2.0} = 0.040 A

The direction: the flux is decreasing, so the induced current flows in such a way as to generate a field in the same direction as the external one, to sustain it. This sentence must be written down, because it is half the marks of the exercise.

2. Rod on rails, with the energy balance

A rod of length $l = 0.50$ m slides at $v = 4.0$ m/s on rails in a perpendicular field $B = 0.30$ T. The circuit has resistance $R = 1.5 Ω$ . Find the emf, the current, the braking force, and check the energy balance.

E = B l v = 0.30 \cdot 0.50 \cdot 4.0 = 0.60 V

i = \frac{0.60}{1.5} = 0.40 A

F = B i l = 0.30 \cdot 0.40 \cdot 0.50 = 0.060 N

The force points opposite to the velocity: it brakes the rod. The mechanical power needed to keep it at constant speed is

P_{mech} = F v = 0.060 \cdot 4.0 = 0.24 W

and the power dissipated in the circuit

P_{diss} = R i^{2} = 1.5 \cdot (0.40)^{2} = 0.24 W

They coincide, as they must: all the work you do to drag the rod ends up as heat in the resistor. If in a test these two numbers do not coincide, you have gone wrong earlier.

3. Transformer

The primary of an ideal transformer has $N_{1} = 1000$ turns and voltage $V_{1} = 230$ V; the secondary has $N_{2} = 50$ turns. Find $V_{2}$ , and the current in the primary if $i_{2} = 2.0$ A flows in the secondary.

V_{2} = V_{1} \frac{N _{2}}{N _{1}} = 230 \cdot \frac{50}{1000} = 11.5 V

For the current the proportion is inverted:

i_{1} = i_{2} \frac{N _{2}}{N _{1}} = 2.0 \cdot \frac{50}{1000} = 0.10 A

The check is immediate with the powers: $230 \cdot 0.10 = 23$ W in the primary and $11.5 \cdot 2.0 = 23$ W in the secondary. An ideal transformer does not create energy, it redistributes it between voltage and current.

The five mistakes that cost the most marks

1. Getting the angle wrong in the flux. $θ$ is between the field and the normal to the surface. If the text says “the field makes $30°$ with the plane of the loop”, the angle to use is $60°$ .

2. Treating the minus sign as an algebraic sign. Lenz’s law indicates a direction, it does not make the emf negative. In the calculations you work with magnitudes and the direction is discussed in words, as in exercise 1.

3. Forgetting the number of turns. $E = - N ΔΦ/Δ t$ . If the text says “coil of 200 turns”, that factor is there.

4. Inverting the transformer proportions. Voltages are directly proportional to the turns, currents inversely. The check with the powers, $V_{1} i_{1} = V_{2} i_{2}$ , catches it in five seconds.

5. Thinking that a constant field induces something. It induces nothing, however intense. You need a change — of $B$ , of $S$ or of $θ$ .

The set-up mistakes that recur throughout high-school physics — units, signs, reading the text — I have collected in typical physics mistakes in high school.

The two weeks before the test

Every day, for a week, write on a blank sheet the definition of flux and the Faraday-Neumann-Lenz law, and next to them the three ways in which flux can change. Two minutes.
Take ten problem texts and write down only which of the three is changing ( $B$ , $S$ or $θ$ ), without solving them. It is the exercise that pays off most on this chapter.
Do five exercises on the rod, each time closing with the comparison between mechanical power and dissipated power. If you pick up this habit, you correct yourself.
Train Lenz’s direction in words, on five different situations: magnet approaching, magnet moving away, loop entering the field, loop leaving, a nearby current increasing. The test almost always asks for the direction, and it is not a calculation: it is reasoning that has to be said in one sentence.

How we work on this in lessons

On induction I spend almost the whole first lesson on flux and on Lenz, with no alternator or transformer formulas. Anyone who can say in one sentence why the loop repels an approaching magnet will then solve all the exercises in the chapter; anyone who has learned the formulas but not the why gets stuck at the first problem written differently from those in the book.

The other thing I insist on is the energy balance of the rod: making it come out by hand, every time, ties induction to what was studied in Year 3 on work and energy, and turns four unconnected formulas into a single story. Lessons are online and one-to-one: rates and packages are on the pricing page, and the 10-hour package is worth it if you want to cover all of electromagnetism with a view to the maturità and not only the October test.

The quickest way to see where you stand is to try the simulated test on electromagnetic induction: fifteen questions in fifty minutes, with a breakdown of weak points by topic at the end. The correct answer to every question is free and needs no registration; the guided step-by-step solution is unlocked with the first lesson. If after the test you see that the calculations hold but Lenz’s direction does not (or vice versa), write to me for a free introductory call: we look at the syllabus of your test and I tell you honestly how many lessons you need. The full picture of how I set up the path is on the page for physics tutoring.

Electromagnetic induction: preparing for the Year 5 test