Electromagnetic Induction

By the end of this workbook you should be able to:

Calculate magnetic flux using Φ = BA cos θ.
Apply Faraday's law of induction, ε = −N ΔΦ/Δt, to find an induced emf.
Calculate the emf induced in a straight conductor moving perpendicularly through a uniform magnetic field, ε = BvL.
Use Lenz's law and energy conservation to determine the direction of an induced emf.
Describe how a uniform magnetic field induces a sinusoidal emf in a rotating coil, and explain the effect of changing the rotation frequency.
Explain how a transformer works, as a practical application of electromagnetic induction.

Name

1. Motional emf: ε = BvL

Whenever a conductor moves across a magnetic field, or a magnetic field moves across a conductor, an emf is induced. This is called electromagnetic induction. We start with the simplest case: a straight conductor moving perpendicularly through a uniform magnetic field.

Electromagnetic induction is the process in which an emf is produced across a conductor that is experiencing a changing magnetic field. Reminder from Topic B.5: the electromotive force (emf) of any source is the energy transferred per unit charge passing through it.

Why a moving conductor develops an emf

A conductor moving across a field carries free electrons along with it. Each free electron is a moving charge in a magnetic field, so it experiences a magnetic force F = qvB sinθ (Topic D.3). This force pushes electrons towards one end of the conductor, leaving the other end short of electrons — one end becomes negatively charged, the other positively charged. This charge separation is exactly what an emf is: a potential difference between the two ends of the conductor.

Fig. 1.1 — A conductor of length L moves at speed v perpendicular to a uniform field (into the page). Free electrons are pushed towards one end, so that end becomes negative and the other becomes positive: an emf is induced across the ends.

Deriving ε = BvL

The maximum induced potential difference occurs when the magnetic force on each free electron, F_B = qvB, is balanced by the electric force from the charge separation itself, F_E = Eq = (ε/L)q. Setting these equal:

εq/L = qvB
ε = BvL

The emf induced when a straight conductor moves perpendicularly across a uniform magnetic field:

ε = BvL

where B is the field strength, v is the conductor's speed, and L is the length of the conductor inside the field.

Direction matters: if the motion or the field is reversed, the emf reverses too. If both are reversed together, the emf is unchanged. For an emf to be induced, there must be relative motion between the conductor and the field — if both move together at the same velocity, no emf is induced.

If the conductor is wound into a coil of N turns (with one side of the coil inside the field), each turn adds an emf of the same value in series — exactly like adding more cells to a battery:

ε = NBvL

Worked example 1.1 (Hodder D4.1)

Calculate the induced emf produced across a 23.0 cm long conductor moving at 98.0 cm s⁻¹ perpendicularly across a magnetic field of strength 120 μT.

Answer: ε = BvL = (120×10⁻⁶) × 0.98 × 0.23 = 2.7×10⁻⁵ V.

1A straight conductor of length 1.4 m moves perpendicularly across a uniform magnetic field of strength 5.0×10⁻⁴ T at a speed of 3.2 m s⁻¹. Calculate the induced emf.

Hint: substitute directly into ε = BvL. You should get roughly 2.2×10⁻³ V.

2A conductor is moved across a uniform magnetic field, but no emf is induced. Suggest two different reasons why this might be the case.

Either the conductor is moving parallel to the field lines (rather than across them), so it is not "cutting" any field lines; or the conductor and the field are moving together at the same velocity, so there is no relative motion between them (electromagnetic induction requires relative motion).

3A coil of 40 turns, each of length 18 cm, moves perpendicularly across a uniform field of 2.6×10⁻³ T at 0.75 m s⁻¹. Calculate the induced emf.

Hint: use ε = NBvL rather than ε = BvL, since there are 40 turns in series. You should get roughly 0.014 V.

4An aircraft flies horizontally at 250 m s⁻¹ where the vertical component of the Earth's magnetic field is 15 μT. Its wingspan is 42 m. Calculate the emf induced across its wing tips.

Hint: the wings act as a single moving conductor of length L = 42 m, moving at v = 250 m s⁻¹ perpendicular to the vertical field component B = 15 μT. Use ε = BvL. You should get roughly 0.16 V.

2. Magnetic flux and flux linkage

To describe electromagnetic induction in general — not just a conductor sliding across a field — physicists use the idea of magnetic flux. It combines field strength, area, and orientation into a single quantity.

Φ = BA cos θ

Φ = magnetic flux (Wb) · B = magnetic field strength (T) · A = area of the loop or coil (m²) · θ = angle between the field direction and the normal to the area (not the angle to the surface itself!).

Common mistake: θ is measured from the normal (perpendicular) to the surface, not from the surface itself. If the field is perpendicular to the loop, θ = 0°, cos θ = 1, and the equation simplifies to Φ = BA (its maximum possible value for that B and A).

Fig. 2.1 — Magnetic flux depends on the field strength, the area of the loop, and the angle θ between the field and the normal to the loop.

The SI unit of magnetic flux is the weber, Wb (1 Wb = 1 T m²). Rearranging the equation for B perpendicular to A shows why field strength is also called magnetic flux density: B = Φ/A, i.e. flux per unit area.

Magnetic flux linkage

So far we have considered a single loop of wire. If the wire is wound into a coil of N turns, each turn contributes the same flux, so the overall induced emf ends up multiplied by N. This is captured by the idea of magnetic flux linkage:

magnetic flux linkage = NΦ

The units of flux linkage are the same as flux (Wb), though "Wb-turns" is sometimes used to be explicit about the N.

Worked example 2.1 (Hodder D4.3)

(a) Calculate the magnetic flux in a square loop of wire of side 6.2 cm when placed at 45° to a magnetic flux density of 4.3×10⁻⁴ T. (b) Calculate how many turns would be needed on a coil of the same dimensions to create a flux linkage of 8.4×10⁻⁴ Wb.

Answer: (a) Φ = BA cos θ = (4.3×10⁻⁴) × (6.2×10⁻²)² × cos45° = 1.2×10⁻⁶ Wb. (b) N = NΦ/Φ = (8.4×10⁻⁴)/(1.2×10⁻⁶) = 7.2×10² turns.

5A flat coil of area 52 cm² is placed in a field of magnetic flux density 6.0×10⁻³ T, at an angle of 35° to the plane of the coil. Calculate the magnetic flux through the coil.

Hint: the angle given (35°) is between the field and the plane of the coil — you need the angle between the field and the normal to the coil for Φ = BAcosθ, which is 90° − 35° = 55°. You should get roughly 1.8×10⁻⁵ Wb.

6A magnetic field of strength 4.0×10⁻² T passes perpendicularly through a flat coil of 350 turns and area 3.6×10⁻⁵ m². Determine the flux linkage.

Hint: first find Φ = BA (θ = 0° since the field is perpendicular), then multiply by N to get the flux linkage NΦ. You should get roughly 5.0×10⁻⁴ Wb.

7A flat coil of 500 turns and area 7.5 cm² is placed where the magnetic flux density is 8.4×10⁻³ T. The axis of the coil is originally parallel to the field, but the coil is then rotated by 20°. Calculate the change of flux linkage through the coil.

Hint: the coil's axis is the normal direction. If the axis starts parallel to the field, θ (measured from the normal) starts at 0°, so the initial flux linkage is at its maximum, NBA. After rotating by 20°, θ becomes 20°, so the new flux linkage is NBAcos20°. Subtract to find the change — you should get a decrease of roughly 1.9×10⁻⁴ Wb.

3. Faraday's law of electromagnetic induction

Magnetic flux lets us write one general equation that predicts the induced emf in any situation — whatever is actually moving or changing.

ε = N ΔΦ/Δt

Faraday's law: if a coil of N turns experiences a change in magnetic flux ΔΦ in a time Δt, the induced emf is ε = NΔΦ/Δt. In words: the induced emf equals the rate of change of magnetic flux linkage. (Section 4 adds a direction to this equation, via Lenz's law.)

Faraday's law applies to every method of induction — a conductor moving through a field, a field moving past a conductor, or a changing current inducing an emf in a nearby circuit. Let's see how it reduces to familiar equations in each case.

Case 1 — a conductor moving across a uniform field

Picture a rod of length L sliding at speed v along two parallel rails, in a uniform field B. The field is constant, but the area enclosed by the circuit changes as the rod moves. Since there is only one loop, Faraday's law becomes:

ε = ΔΦ/Δt = Δ(BA)/Δt = B(ΔA/Δt)

In time Δt the rod sweeps out an area ΔA = LvΔt (the "area swept out"), so ΔA/Δt = vL, and:

ε = B(ΔA/Δt) = BvL

— exactly the equation from Section 1, now derived from Faraday's law rather than from balancing forces on individual electrons.

Worked example 3.1 (Hodder D4.4)

Two parallel horizontal conducting rails, 44 cm apart, sit in a uniform field of 8.7×10⁻⁴ T acting vertically downwards. A rod moves along the rails at 48 cm s⁻¹. (a) Determine the emf induced across the loop. (b) Determine how much extra magnetic flux passes through the circuit when the rod moves 25 cm.

Answer: (a) ε = BvL = (8.7×10⁻⁴)(0.48)(0.44) = 1.8×10⁻⁴ V. (b) Increase in area = 0.25 × 0.44 = 0.11 m²; increase in flux = area × B = 0.11 × (8.7×10⁻⁴) = 9.6×10⁻⁵ Wb.

Case 2 — a coil moving into or out of a field (or a changing field through a stationary coil)

Here the coil's area A is constant, but the field passing through it changes. Faraday's law becomes:

ε = N(ΔΦ/Δt) = NA(ΔB/Δt)

Worked example 3.2 (Hodder D4.5)

A coil of 40 turns and area 5.0 cm² is moved from completely outside to completely inside a uniform magnetic field of strength 0.34 T in 0.56 s. Determine the average magnitude of the induced emf.

Answer: ε = NA(ΔB/Δt) = 40 × (5.0×10⁻⁴) × (0.34/0.56) = 1.2×10⁻² V.

Case 3 — mutual induction between two separate circuits

A changing current in one circuit (A) creates a changing magnetic field, which passes through a second, completely separate circuit (B) and induces an emf in it. This is called mutual induction. For a fixed arrangement of the two circuits, ΔΦ/Δt (and therefore the induced emf, ε) in circuit B is proportional to the rate of change of current, ΔI/Δt, in circuit A. You will not be expected to answer detailed quantitative questions on mutual induction — but it is the basic principle behind the transformer, which you will meet in Section 6.

8A coil of 220 turns moves from outside a strong uniform field of 0.095 T to a position fully inside it, in a time of 0.90 s. The coil has dimensions 4.5 cm × 7.0 cm. (a) Calculate the change of magnetic flux in the coil. (b) State any assumption made in part (a). (c) Determine the change of magnetic flux linkage. (d) Calculate the average induced emf.

Hint: (a) Φ = BA, using the coil's area in m². (b) assume the field is perpendicular to the coil throughout (θ = 0°) and that the coil moves at a constant rate. (c) multiply by N. (d) divide the change in flux linkage by the time taken.

9The magnetic flux through a coil of 850 turns increases from zero to 3.6×10⁻⁵ Wb in 3.4 ms. Calculate the magnitude of the average induced emf.

Hint: use ε = NΔΦ/Δt directly, converting the time to seconds first. You should get roughly 9.0 V.

10A train travels at 42 m s⁻¹ through a region where the Earth's magnetic field is 46 μT, acting at 60° to the horizontal. An axle on the train has a length of 1.5 m. (a) Calculate the area swept out by the axle every second. (b) Determine the component of the field acting perpendicular to the axle's motion. (c) Calculate the magnitude of the induced emf across the axle.

Hint: (a) area swept per second = speed × length = vL. (b) only the vertical component of the Earth's field induces an emf in a horizontal axle moving horizontally — use B sin60° (the vertical component) if 60° is measured from the horizontal. (c) then use ε = BvL with that perpendicular component.

4. Lenz's law and energy conservation

Faraday's law tells us the size of an induced emf. The negative sign in ε = −NΔΦ/Δt tells us about its direction — and that direction is not arbitrary. It is a direct consequence of the conservation of energy.

Lenz's law: the direction of an induced emf (and current) is always such that it opposes the change that produced it. It was the physicist Heinrich Lenz who added this direction to Faraday's law, giving the complete, signed equation:

ε = −N ΔΦ/Δt

Why energy conservation demands this

If a current is generated by electromagnetic induction, energy must be transferred to that current from somewhere — energy cannot be created from nothing. That energy usually comes from the kinetic energy of whatever is moving (the conductor, or the magnet). For that energy to be transferred, the moving object must do work against an opposing force — so the induced current's magnetic effect must oppose the motion that created it, requiring a force (and therefore work) to keep the motion going.

If Lenz's law were reversed (i.e. if the induced current assisted the motion instead), the moving object would speed up, generating an even larger current, which would speed it up further — energy would be created from nothing. This would violate the conservation of energy, so it cannot happen.

Consider a magnet approaching a coil (Topic D.4, Fig. D4.9): the induced current turns the coil into an electromagnet. By Lenz's law, the coil's near end must become the same pole as the approaching magnet, so that the two repel — opposing the magnet's approach. Work must be done to push the magnet closer, and that work is transferred to the induced current. If the magnet is instead pulled away, the induced current reverses, creating an attractive force that opposes the magnet leaving.

Worked example 4.1

A bar magnet is dropped, north pole down, through a horizontal conducting loop. Describe, using Lenz's law, the direction of the force the loop exerts on the magnet as it approaches and as it leaves.

Answer: As the magnet approaches, the loop's induced current makes its upper face a north pole, repelling the approaching north pole and opposing its fall. As the magnet moves away (below the loop), the induced current reverses, making the loop's lower face a south pole, which attracts the departing magnet — again opposing its motion. In both cases, the magnetic force does negative work on the falling magnet, so it falls more slowly than it would in free fall.

Eddy currents

Lenz's law also applies inside solid conductors, not just wire loops. When a solid piece of metal experiences a changing magnetic flux, circulating currents called eddy currents are induced within it. By Lenz's law, these currents oppose the change that caused them — which is why, for example, a magnet falling through a copper or aluminium tube falls surprisingly slowly: the eddy currents induced in the tube wall create a magnetic field that opposes the magnet's motion. The electrical energy in eddy currents is transferred to internal (thermal) energy via resistive heating (P = I²R), which is exactly how an induction cooker heats a metal pan.

11A bar magnet, north pole facing down, is moved away from a solenoid that is part of a complete circuit. State the polarity induced at the top of the solenoid (nearest the departing magnet), and explain your reasoning using Lenz's law.

Hint: Lenz's law says the induced effect must oppose the change — here, the magnet is leaving, so the induced current must try to attract it back. An attractive force between a departing north pole and the solenoid's top requires the solenoid's top to become a south pole.

12Explain why a magnet falls more slowly through a copper tube than through an identical plastic tube, even though copper is not magnetic.

As the magnet falls, the changing magnetic flux through the copper (a conductor) induces eddy currents in it, even though copper itself is not magnetic. By Lenz's law, these eddy currents create their own magnetic field that opposes the magnet's motion, producing a retarding force. In a plastic tube (an insulator), no eddy currents can flow, so there is no such opposing force and the magnet falls freely.

13Explain, in terms of energy transfers, why a conductor being pushed across a magnetic field to generate a current requires a continuous external force to keep it moving at constant speed.

By Lenz's law, the induced current creates a magnetic force on the conductor that opposes its motion. To keep the conductor moving at constant speed against this opposing force, an external force must do work on it. That work is the source of the electrical energy delivered to the circuit — energy is not created from nothing, it is converted from the work done by whatever is pushing the conductor.

5. AC generators and the effect of frequency

Electromagnetic induction generates most of the world's electrical energy, using coils that rotate inside a magnetic field.

A uniform magnetic field induces a sinusoidally varying emf in a coil that rotates within it, at constant angular speed.

Fig. 5.1 — A coil rotating about a fixed central axis between the poles of a fixed magnet (the faded rectangle shows an earlier position in the rotation, with the red arrows showing the direction of rotation). The induced emf is greatest when the coil's plane is parallel to the field (cutting field lines fastest), and zero when its plane is perpendicular to the field (momentarily moving parallel to the field lines).

Carbon "brushes" press against rotating "slip rings" to connect the spinning coil to the external circuit without the wires twisting. As the coil turns at constant angular speed in a uniform field, the emf induced varies smoothly between a maximum in one direction and the same maximum in the other — a sine wave.

The effect of changing the rotation frequency

If the coil rotates faster (a higher frequency, f), the magnetic flux through it changes more rapidly, so — by Faraday's law — a larger peak emf is induced, and the whole cycle repeats in a shorter time (period T = 1/f). Halving the frequency halves the rate of change of flux linkage and therefore halves the peak induced emf — while doubling the time period.

Key relationship: peak emf ∝ frequency of rotation, while the time period T = 1/f. Faster rotation means a bigger, faster-oscillating signal; slower rotation means a smaller, slower one.

Explore it yourself: drag the slider to change the coil's rotation frequency.

Rotation frequency

Frequency: 20 Hz Peak emf: 6.0 V Period: 50.0 ms

reference (20 Hz) current frequency

Worked example 5.1

A generator produces a peak emf of 340 V when its coil rotates at 50 Hz. Calculate the peak emf if the frequency were reduced to 20 Hz, assuming everything else stays the same.

Answer: Peak emf is proportional to frequency, so ε_peak,new = 340 × (20/50) = 136 V. The period would also increase, from 1/50 = 0.020 s to 1/20 = 0.050 s.

Real generators use turbines to provide the rotation — driven by high-pressure steam, falling water, or wind — and many-turned coils wound on high-permeability cores to maximise the induced emf. Mains electricity in most of the world is generated and delivered as alternating current (ac) rated at 230 V, 50 Hz (120 V, 60 Hz in North America). The quoted "230 V" is the RMS voltage — the steady dc voltage that would deliver the same power — not the peak voltage, which actually reaches about ±325 V. The difference between peak and RMS voltage is not part of our course, but interesting to look at if you are considering studying physics or engineering.

14A generator's coil rotates at 60 Hz and produces a peak emf of 170 V. (a) Calculate the period of the output. (b) Predict the new peak emf and period if the frequency is increased to 90 Hz.

Hint: (a) T = 1/f. (b) peak emf scales in direct proportion to frequency, while the period scales inversely with frequency. You should find the period roughly halves-to-thirds and the peak emf increases to roughly 255 V.

15Using the interactive widget above: set the frequency to 40 Hz, then to 10 Hz. Record the peak emf and period the widget displays at each setting, and confirm that quadrupling the frequency quadruples the peak emf and reduces the period to a quarter of its previous value.

Hint: at 10 Hz you should see a small, slow wave (long period); at 40 Hz a large, fast wave (short period) — exactly four times the peak emf and a quarter of the period of the 10 Hz case, since both quantities scale linearly (directly or inversely) with frequency.

16Explain, in terms of Faraday's law, why increasing the rotation frequency of a generator's coil increases the peak induced emf.

Faraday's law states that the induced emf equals the rate of change of magnetic flux linkage, ε = NΔΦ/Δt. A faster rotation means the coil sweeps through the same change in orientation (and flux) in less time, so ΔΦ/Δt is larger at every point in the cycle — including at the peak. A larger rate of change of flux linkage means a larger induced emf.

6. Application: the transformer

A transformer uses electromagnetic induction to step an alternating voltage up or down. Two separate coils — the primary (connected to the input supply) and the secondary (connected to the output) — are wound around a shared iron core, but are not electrically connected to each other at all. This is exactly the mutual induction described in Section 3, Case 3: the alternating current in the primary coil creates a constantly changing magnetic flux, the iron core channels that flux efficiently through the secondary coil, and the changing flux linkage induces an alternating emf in the secondary — with no wires touching.

Fig. 6.1 — A transformer: primary coil (Nₚ turns, blue) and secondary coil (Nₛ turns, red) wound directly around a common iron core, each connected to its own circuit by a pair of leads. The alternating flux produced by the primary links the secondary, inducing an alternating emf in it.

Deriving the transformer equation from Faraday's law

Both coils are wound on the same iron core, so the same changing flux, and therefore the same rate of change of flux, ΔΦ/Δt, passes through every turn of both coils at once. Applying Faraday's law (Section 3) to each coil separately:

εₚ = Nₚ(ΔΦ/Δt)
εₛ = Nₛ(ΔΦ/Δt)

Dividing each equation by its own number of turns isolates the shared quantity, ΔΦ/Δt, which must therefore be the same for both:

εₚ/Nₚ = ΔΦ/Δt
εₛ/Nₛ = ΔΦ/Δt
so εₚ/Nₚ = εₛ/Nₛ

The primary and secondary coils are each just one part of a complete circuit, so it makes more sense to talk about the potential difference, V, across each coil rather than an "emf" belonging to neither circuit. Replacing ε with V:

Vₚ/Nₚ = Vₛ/Nₛ

which rearranges to the usual form of the transformer equation:

Vₛ / Vₚ = Nₛ / Nₚ

Vₚ, Vₛ = primary and secondary (r.m.s.) voltages (V) · Nₚ, Nₛ = number of turns on the primary and secondary coils. The same changing flux links every turn of both coils, so the induced emf per turn is identical — the total emf on each side simply scales with how many turns that side has.

If Nₛ > Nₚ, the secondary voltage is larger than the primary — a step-up transformer. If Nₛ < Nₚ, the secondary voltage is smaller — a step-down transformer. A real transformer is not perfectly efficient: some energy is lost as heat, mostly through eddy currents induced in the iron core itself (see Section 4) and through electrical resistance in the windings. Laminating the core — building it from thin, electrically-insulated sheets rather than one solid block — breaks up the paths available to eddy currents and greatly reduces this loss, which is why real transformer cores are always laminated rather than solid.

Worked example 6.1

A transformer has 500 turns on its primary coil and is connected to a 230 V a.c. supply. How many turns are needed on the secondary coil to produce an output of 12 V?

Answer: Rearranging Vₛ/Vₚ = Nₛ/Nₚ gives Nₛ = Nₚ × (Vₛ/Vₚ) = 500 × (12/230) = 26 turns (to the nearest whole turn). Since Nₛ < Nₚ, this is a step-down transformer.

Try it yourself: the interactive transformer simulation

The simulation below lets you build your own transformer: set the number of turns on each coil, the primary voltage, and the efficiency, then read the secondary voltage straight off the output meter. Try predicting Vₛ from Vₚ × (Nₛ/Nₚ) before you change the sliders, then check your prediction against the simulation.

Drag your mouse across the 3D transformer to rotate the view. Use the sliders to set Nₚ, Vₚ and Nₛ, and watch Vₛₛ update in real time. Try switching to DC Mode and note what happens to the output — then switch off Laminated Core and watch the core glow red as eddy-current heating increases.

Interactive simulation by Dr Jones Physics — drjonesphysics.com/transformer. Used with permission.

Interactive 3D transformer simulation — available in the online version of this workbook. See drjonesphysics.com/transformer.

17Using the simulation, set Nₚ = 10, Nₛ = 25 and Vₚ = 12 V, with efficiency at 100% and the laminated core switched on. (a) Predict Vₛ using Vₛ = Vₚ × (Nₛ/Nₚ). (b) Read off the simulation's actual output voltage. Explain any small difference between your prediction and the simulation's reading.

Hint: your prediction should be Vₛ = 12 × (25/10) = 30 V. Even at "100% efficiency" in the simulation, the reading may be a little different because the simulation still applies the efficiency slider value as a multiplying factor on top of the ideal turns-ratio calculation — check that the efficiency slider really is set fully across before comparing.

18With the same turns (Nₚ = 10, Nₛ = 25) and Vₚ = 12 V, reduce the efficiency slider to around 50%, then switch the laminated core toggle off. (a) State what happens to the output voltage as efficiency decreases. (b) Explain, using ideas from Section 4, why switching off the laminated core makes the core visibly heat up in the simulation.

(a) The output voltage falls, since a lower efficiency means a smaller fraction of the ideal turns-ratio voltage is actually delivered to the secondary. (b) A solid (non-laminated) iron core provides a continuous conducting path for eddy currents to circulate within the core itself. By Lenz's law these eddy currents oppose the changing flux that induces them, and the electrical resistance of the iron dissipates their energy as heat (P = I²R) — exactly the same eddy-current heating effect discussed for induction cookers in Section 4. Laminating the core breaks up these circulation paths, choking off the eddy currents and keeping the core cool.

19A step-down transformer is needed to convert a 24 V a.c. supply to 8.0 V. If the primary coil has 30 turns, calculate the number of turns needed on the secondary. Then set the simulation as close as possible to these values (within its slider limits) and check that the output voltage is close to 8.0 V.

Hint: rearrange Vₛ/Vₚ = Nₛ/Nₚ to make Nₛ the subject: Nₛ = Nₚ × (Vₛ/Vₚ) = 30 × (8.0/24) = 10 turns. Set Nₚ = 30, Nₛ = 10 and Vₚ = 24 V in the simulation (all within the slider ranges) and confirm the output reads close to 8.0 V at full efficiency.

7. Equation summary

Quantity / situation	Equation
Magnetic flux	Φ = BA cosθ
Magnetic flux density (from flux)	B = Φ/A
Magnetic flux linkage (N-turn coil)	NΦ
Motional emf (straight conductor)	ε = BvL
Faraday’s law of induction, with Lenz’s addition	ε = −NΔΦ/Δt
Force on a current-carrying conductor	F = BIL sinθ

8. Glossary

Electromagnetic induction: The generation of an emf (and, in a closed circuit, a current) as a result of a changing magnetic flux linking a conductor or coil.
Magnetic flux, Φ: A measure of the total magnetic field passing through a given area, Φ = BA cosθ, where θ is measured from the normal to the area. Measured in webers (Wb).
Magnetic flux linkage: The total flux linking all N turns of a coil, NΦ; this is the quantity that actually determines the induced emf for a multi-turn coil.
Faraday’s law of induction: The magnitude of the induced emf equals the rate of change of magnetic flux linkage: ε = −NΔΦ/Δt. This single law explains motional emf, a coil moving in or out of a field, and mutual induction.
Lenz’s law: The direction of an induced emf (and any resulting current) always opposes the change that produced it. This is a direct consequence of the conservation of energy — if the induced effect instead reinforced the change, energy could be created from nothing.
Eddy currents: Induced currents that circulate within the bulk of a solid conductor (rather than around a single defined loop), always in a direction that opposes the change causing them (Lenz’s law). They dissipate energy as heat, which is useful in induction cookers but wasteful in transformer cores, where lamination is used to suppress them.
Mutual induction: The induction of an emf in one circuit as a result of a changing current (and hence changing flux) in a separate, nearby circuit.
Alternator / a.c. generator: A device that induces a sinusoidally varying emf by rotating a coil at constant angular speed within a uniform magnetic field; slip rings and brushes maintain electrical contact with the rotating coil.
Transformer: A device that uses mutual induction between two coils wound on a shared iron core to step an alternating voltage up or down, according to Vₛ/Vₚ = Nₛ/Nₚ.
Step-up / step-down transformer: A transformer with more turns on the secondary than the primary (step-up, increases voltage) or fewer turns on the secondary (step-down, decreases voltage).