Induction

Lesson D.4 Electromagnetic Induction

IB Physics • Higher Level • Final lesson in Topic D: Fields

Electromagnetic induction explains how changing magnetic flux produces an emf and, in a closed conducting circuit, a current. It underpins generators, transformers, induction cookers and guitar pickups. This lesson connects magnetic flux, Faraday’s law, Lenz’s law, motional emf and alternating-current generators using the IB formulas and algebra, without calculus.

Learning goals

Define magnetic flux and flux linkage, and calculate them for a uniform magnetic field.
Use Faraday’s law to determine induced emf from a change in flux or a flux–time graph.
Apply Lenz’s law to determine current direction and explain its connection to energy conservation.
Derive and apply the motional-emf formula for a straight conductor.
Explain induction as a rectangular coil enters or leaves a magnetic field.
Describe sinusoidal flux and emf in an AC generator and the effect of rotation frequency.
Explain self-induction qualitatively and connect induction to practical applications.

D.4 is an HL topic. The explanations connect to the magnetic forces and conservation of energy studied in D.3.

Notation and assumptions

$\Phi$ is magnetic flux through one turn, $N$ is the number of turns, $N\Phi$ is flux linkage and $\varepsilon$ is induced emf. $B$ is magnetic field strength, $A$ is area and $\theta$ is the angle between the field and the normal to the surface. $L$ denotes the active conductor length in motional-emf questions.

For rotating coils, $\omega$ is angular speed, $f$ is rotation frequency and $T$ is the period. Use SI units, including square metres for area. Unless stated otherwise, assume uniform fields and ideal conductors apart from the stated circuit resistance. Neglect self-induction when calculating current directly from emf and resistance.

$1\,\mathrm{Wb}=1\,\mathrm{T\,m^2}=1\,\mathrm{V\,s},\qquad \omega=2\pi f,\qquad T=\frac{1}{f}$

1. Key ideas: emf, magnetic flux and flux linkage

What is an emf?

An electromotive force, or emf, is energy transferred per unit charge by a source. It is measured in volts and is not a mechanical force, despite its name. A conductor can have an induced emf even when the circuit is open; a sustained circulating current requires a complete conducting path.

A changing magnetic field can produce an electric field that drives charges around a stationary loop. A conductor moving through a magnetic field can also develop an emf because magnetic forces separate its charges. These are two ways to produce electromagnetic induction.

Magnetic flux through a flat surface

Magnetic flux measures the magnetic field passing through a surface. For a uniform field and a flat surface:

Magnetic Flux

$\Phi=BA\cos\theta$

The normal is an imaginary line perpendicular to the surface. The angle in the formula is measured between this normal and the magnetic field. Flux has a sign relative to the chosen normal: it is positive when the field has a component along that normal and negative when the component is opposite.

Why the cosine factor appears

Only the component of the magnetic field perpendicular to the surface contributes to the flux. This component is $B_{\perp}=B\cos\theta$ , so:

$\Phi=B_{\perp}A=(B\cos\theta)A$

Orientation	Angle to the normal	Flux
Field perpendicular to the surface, along the normal	$0^\circ$	$+BA$
Field parallel to the surface	$90^\circ$	$0$
Field perpendicular to the surface, opposite to the normal	$180^\circ$	$-BA$

IB angle and area checks

If a question gives the angle between the field and the plane of the loop, convert it to the complementary angle to the normal. A field at $30^\circ$ to the plane is at $60^\circ$ to the normal. Also remember that $1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2}$ .

Flux linkage

If each turn of an $N$ -turn coil experiences the same flux, the flux linkage is:

$\text{flux linkage}=N\Phi=NBA\cos\theta$

Magnetic flux through one turn and flux linkage through the whole coil are different quantities. Flux linkage is commonly labelled in weber-turns; the turn count is dimensionless.

What can change the flux?

Change the magnetic field strength, for example by moving a magnet or varying the current in an electromagnet.
Change the area exposed to the field, for example by moving a loop into or out of a field region.
Change the orientation of the loop, for example by rotating it.

A rigid loop moving through a completely uniform field at fixed orientation and with constant area has constant flux. Motion alone does not guarantee a net induced emf around the loop.

Worked example 1 — Flux and flux linkage

A coil has $200$ turns, each of area $0.020\,\mathrm{m^2}$ . It is in a uniform $0.40\,\mathrm{T}$ field, with the field at $60^\circ$ to the coil’s normal. Calculate (a) the flux through one turn and (b) the flux linkage.

Show answer and working

$\Phi=BA\cos\theta=(0.40)(0.020)\cos60^\circ=4.0\times10^{-3}\,\mathrm{Wb}$

$N\Phi=(200)(4.0\times10^{-3})=0.80\,\mathrm{Wb\,turn}$

The flux through one turn is $4.0\times10^{-3}\,\mathrm{Wb}$ . The total flux linkage is $0.80\,\mathrm{Wb\,turn}$ .

2. Faraday’s law: calculating induced emf

Faraday’s law states that induced emf is proportional to the rate of change of magnetic flux linkage. For a fixed number of turns, the IB relationship is:

$\varepsilon=-N\frac{\Delta\Phi}{\Delta t}$

Here $\Delta\Phi=\Phi_{\mathrm{final}}-\Phi_{\mathrm{initial}}$ is the signed change in flux through one turn. If the question already gives a change in total flux linkage, use:

$\varepsilon=-\frac{\Delta(N\Phi)}{\Delta t}$

For a finite interval, these equations give the average emf. They also give the emf throughout an interval when flux changes at a constant rate. For a curved flux–time graph, the emf at a particular instant is found from the negative gradient of the tangent, multiplied by $N$ .

The negative sign represents Lenz’s law. To calculate a magnitude, use:

$|\varepsilon|=N\frac{|\Delta\Phi|}{\Delta t}$

What increases the induced emf?

A greater flux change in the same time.
The same flux change in a shorter time.
More turns experiencing the same flux change per turn.

A strong field with constant flux produces no net induced emf around a stationary loop. A weaker field changing rapidly can produce a substantial emf.

Worked example 2 — A changing flux through a coil

The flux through each turn of a $250$ -turn coil decreases uniformly from $3.0\times10^{-3}\,\mathrm{Wb}$ to $1.0\times10^{-3}\,\mathrm{Wb}$ in $0.050\,\mathrm{s}$ . Calculate the induced emf using the stated flux sign convention.

Show answer and working

$\Delta\Phi=1.0\times10^{-3}-3.0\times10^{-3}=-2.0\times10^{-3}\,\mathrm{Wb}$

$\varepsilon=-250\left(\frac{-2.0\times10^{-3}}{0.050}\right)=+10\,\mathrm{V}$

The emf magnitude is $10\,\mathrm{V}$ . Its positive sign refers to the chosen positive direction around the coil. A diagram and viewing direction are needed to translate this into clockwise or anticlockwise current.

3. Lenz’s law and conservation of energy

Lenz’s law states that the induced effect acts in a direction that opposes the change in magnetic flux that produces it. The induced magnetic field may oppose an increasing external field or support a decreasing external field.

A reliable method for determining current direction

Identify the direction of the external flux through the loop.
Decide whether that flux is increasing or decreasing.
Choose the induced magnetic field that opposes the change.
Use the right-hand grip rule: curled fingers give conventional current direction, and the thumb gives the induced field through the loop.

Viewed face-on, an anticlockwise current produces a field out of the page, towards you. A clockwise current produces a field into the page, away from you. Always state the viewing direction.

Why this is an energy-conservation rule

When a magnet approaches a closed coil, the induced effect resists the approach. An external agent must do work to maintain the motion. That work supplies electrical energy, which may be dissipated as thermal energy in the circuit. When the magnet is withdrawn, the induced effect resists the withdrawal.

If the induced effect reinforced the motion that caused it, electrical energy and increasing mechanical energy could be produced without a corresponding energy input. Lenz’s law prevents this violation of energy conservation.

Worked example 3 — An approaching and retreating magnet

The north pole of a bar magnet moves towards a closed conducting coil. View the coil from the magnet’s side. (a) Determine the induced current direction. (b) State what happens when the magnet is held stationary. (c) Determine the current direction when the magnet is pulled away.

Show answer and working

(a) The approaching north pole produces increasing flux directed away from the viewer through the coil. The induced field points towards the viewer to oppose that increase. The induced current is anticlockwise, making the near face of the coil a north pole that repels the approaching magnet.

(b) When the magnet is held stationary and the flux has stopped changing, the induced emf is zero. Any transient current dies away in a resistive circuit.

(c) As the north pole retreats, the away-directed flux decreases. The induced field points away from the viewer to oppose that decrease. The current is clockwise, making the near face a south pole that attracts the retreating magnet.

4. Motional emf in a straight conductor

Consider a straight conductor of length $L$ moving at speed $v$ through a uniform magnetic field $B$ . For the IB perpendicular arrangement, the conductor, its velocity and the magnetic field are mutually perpendicular. The induced emf magnitude is:

$\varepsilon=BvL$

Derivation using magnetic and electric forces

Charges moving with the rod experience magnetic forces. Positive and negative charges separate, creating an electric field inside the rod. In an open-circuit rod, separation continues until the electric and magnetic forces on a carrier balance in magnitude:

$|q|E=|q|vB\quad\Rightarrow\quad E=vB$

The potential difference between the ends is $EL$ , giving:

$\varepsilon=EL=BvL$

To identify the positive end, use the magnetic-force direction for a positive charge. In a metal, electrons move towards the opposite end. If the rod is connected to a closed circuit, an induced current can flow.

Mechanical work supplies electrical energy

A current-carrying rod in the field experiences a magnetic force that opposes its motion. For the perpendicular arrangement:

$F_{\mathrm{mag}}=BIL$

An external force must balance this magnetic force to maintain constant speed. For a resistive circuit, neglecting self-induction:

$I=\frac{\varepsilon}{R},\qquad P_{\mathrm{mechanical}}=Fv=(BIL)v=\varepsilon I=I^2R$

The mechanical power input becomes electrical power dissipated in the resistance. The magnetic field directs the forces; the external agent supplies the energy.

Worked example 4 — A sliding rod and its energy transfer

A $0.30\,\mathrm{m}$ rod slides at $4.0\,\mathrm{m\,s^{-1}}$ through a uniform $0.50\,\mathrm{T}$ field in the perpendicular arrangement. It completes a circuit of total resistance $3.0\,\Omega$ . Calculate the emf, current, opposing magnetic force and mechanical power needed to maintain the speed.

Show answer and working

$\varepsilon=BvL=(0.50)(4.0)(0.30)=0.60\,\mathrm{V}$

$I=\frac{0.60}{3.0}=0.20\,\mathrm{A}$

$F=BIL=(0.50)(0.20)(0.30)=0.030\,\mathrm{N}$

$P=Fv=(0.030)(4.0)=0.12\,\mathrm{W}$

The required mechanical power is $0.12\,\mathrm{W}$ . This equals $I^2R=(0.20)^2(3.0)=0.12\,\mathrm{W}$ , confirming the energy balance.

5. A rectangular coil entering and leaving a field

Consider a rectangular coil moving at constant speed into a uniform field perpendicular to its plane. Let $L$ be the side length perpendicular to the motion and $x$ the distance that has entered the field. The area inside the field is $A=Lx$ .

During entry, the area increases by $\Delta A=L\Delta x$ . Since $\Delta x/\Delta t=v$ :

$|\varepsilon|=N\frac{B\Delta A}{\Delta t}=NB L\frac{\Delta x}{\Delta t}=NB Lv$

Only the area overlapping the field changes. The physical area of the rigid coil remains constant. Assuming a sharp field boundary and negligible fringing, entry and exit each produce constant-emf pulses at constant speed.

Stage	Flux through the coil	Induced effect
Completely outside	Zero and constant.	No induced emf.
Entering	Flux magnitude increases.	Emf appears; induced field opposes the increase.
Fully inside, translating without rotation	Flux is constant.	Net emf around the loop is zero.
Leaving	Flux magnitude decreases.	Emf reverses direction; induced field opposes the decrease.
Completely outside again	Zero and constant.	No induced emf.

When the whole loop is translating within a uniform field, motional effects on opposite sides cancel around the circuit. This explains why individual moving conductors can develop charge separation while the complete loop has no net circulating emf.

Worked example 5 — Emf while a coil enters a field

A $50$ -turn rectangular coil has an active side length of $0.040\,\mathrm{m}$ and a length along its motion of $0.060\,\mathrm{m}$ . It enters a perpendicular $0.30\,\mathrm{T}$ field at $0.20\,\mathrm{m\,s^{-1}}$ . Calculate (a) the emf magnitude during entry and (b) the entry duration.

Show answer and working

$|\varepsilon|=NBLv=(50)(0.30)(0.040)(0.20)=0.12\,\mathrm{V}$

$\Delta t=\frac{0.060}{0.20}=0.30\,\mathrm{s}$

The entry pulse has magnitude $0.12\,\mathrm{V}$ and lasts $0.30\,\mathrm{s}$ . When the coil is fully inside a sufficiently wide uniform field, its net induced emf is zero. On leaving at the same speed, the emf has the same magnitude and opposite sign.

6. AC generators: sinusoidal flux and emf

How an AC generator works

An AC generator rotates a coil in a uniform magnetic field. The changing angle between the coil’s normal and the field changes the flux through the coil. Slip rings and brushes connect the rotating coil to the external circuit while allowing the polarity of the output to alternate.

Choose $t=0$ when the coil’s normal points along the field. For constant angular speed:

$\theta=\omega t,\qquad \Phi(t)=BA\cos(\omega t)$

$N\Phi(t)=NBA\cos(\omega t)$

The flux through each turn varies as a cosine. The peak flux magnitude is $BA$ , and the peak flux-linkage magnitude is $NBA$ .

Obtaining the emf without calculus

Take a rectangular coil rotating about an axis in its plane and perpendicular to the field. Let each active side have length $l$ and let the separation between these sides be $w$ , so $A=lw$ . Each side moves with speed $v=\omega w/2$ . The component that produces emf varies as $v\sin\theta$ .

The emfs from the two opposite active sides add around each turn. Applying the motional-emf relationship gives:

$\varepsilon(t)=2NBl\left(\frac{\omega w}{2}\right)\sin\theta=NBA\omega\sin(\omega t)$

This is a sine wave for the chosen starting orientation and output-polarity convention. Reversing the leads reverses the sign of the measured emf.

$\varepsilon(t)=\varepsilon_{\mathrm{max}}\sin(\omega t),\qquad \varepsilon_{\mathrm{max}}=NBA\omega=2\pi fNBA$

Phase relationship

Flux and emf differ in phase by $90^\circ$ , or one-quarter of a cycle. The emf follows the negative slope of the flux–time graph:

Position in the cycle	Flux per turn	Emf
At $t=0$	$+BA$ ; maximum positive flux.	$0$
At $t=T/4$	$0$ ; flux decreases most rapidly.	$+\varepsilon_{\mathrm{max}}$
At $t=T/2$	$-BA$ ; maximum negative flux.	$0$
At $t=3T/4$	$0$ ; flux increases most rapidly.	$-\varepsilon_{\mathrm{max}}$
At $t=T$	The pattern repeats.	The pattern repeats.

Maximum flux magnitude does not mean maximum emf. At a flux maximum or minimum, the graph has zero slope and the emf is zero. The emf magnitude is greatest when the flux crosses zero.

Effect of changing the rotation frequency

Increasing $f$ increases the output frequency and reduces the period.
The peak emf is proportional to $f$ when $N$ , $B$ and $A$ are fixed.
Doubling the rotation frequency doubles the peak emf and halves the period.
The peak flux $BA$ and peak flux linkage $NBA$ remain unchanged.

More turns, a stronger field or a greater coil area also increases the peak emf for the same rotation frequency. A generator supplies electrical energy by transferring mechanical energy from the system driving its rotation.

Worked example 6 — AC generator output

A coil with $200$ turns and area $0.015\,\mathrm{m^2}$ rotates at $25\,\mathrm{Hz}$ in a uniform $0.30\,\mathrm{T}$ field. Determine (a) its peak emf, (b) its period and (c) its emf at $t=0.010\,\mathrm{s}$ , taking maximum positive flux at $t=0$ .

Show answer and working

$\omega=2\pi f=2\pi(25)=157\,\mathrm{rad\,s^{-1}}$

$\varepsilon_{\mathrm{max}}=NBA\omega=(200)(0.30)(0.015)(157)=141\,\mathrm{V}$

$T=\frac{1}{25}=0.040\,\mathrm{s}$

$\varepsilon(0.010)=141\sin\left[2\pi(25)(0.010)\right]=+141\,\mathrm{V}$

The time $0.010\,\mathrm{s}$ is one-quarter of a cycle, so the emf is at its positive peak. The peak flux through one turn is $BA=4.5\times10^{-3}\,\mathrm{Wb}$ .

7. Interpreting flux–time and emf–time graphs

For a flux–time graph, multiply its slope by $-N$ to obtain the emf. A steeper slope gives a greater emf magnitude. A horizontal section gives zero emf. An increasing positive flux gives a negative emf, while decreasing positive flux gives a positive emf, using a consistent sign convention.

A zero average emf over a complete cycle does not imply that the emf was zero throughout the cycle. An AC generator has positive and negative emf during different parts of each cycle.

You can also recover a flux change from a constant-emf interval using algebra:

$\Delta\Phi=-\frac{\varepsilon\Delta t}{N}$

For several constant-emf intervals, calculate each flux change and add them. The signed rectangular areas of an emf–time graph therefore correspond to $-N\Delta\Phi$ , without needing an integral.

Worked example 7 — Piecewise flux changes

A $100$ -turn coil experiences the following flux through each turn: it rises linearly from zero to $4.0\times10^{-3}\,\mathrm{Wb}$ between $0$ and $0.040\,\mathrm{s}$ , stays constant until $0.100\,\mathrm{s}$ , then falls linearly to $-2.0\times10^{-3}\,\mathrm{Wb}$ at $0.120\,\mathrm{s}$ . Find the emf in each interval.

Show answer and working

$\varepsilon_1=-100\frac{4.0\times10^{-3}-0}{0.040}=-10\,\mathrm{V}$

$\varepsilon_2=0\,\mathrm{V}$

$\varepsilon_3=-100\frac{-2.0\times10^{-3}-4.0\times10^{-3}}{0.020}=+30\,\mathrm{V}$

The emf–time graph has three horizontal sections at $-10\,\mathrm{V}$ , $0\,\mathrm{V}$ and $+30\,\mathrm{V}$ . The signs reflect the direction of each flux change. The idealized sharp transitions would be smoothed in a real apparatus.

8. Self-induction and practical applications

Self-induction: a qualitative explanation

A current in a coil produces magnetic flux through that coil. If the current changes, its own flux linkage changes and an emf is induced in the same coil. This self-induced emf opposes the change in current: it resists a rise in current and acts to sustain a current that is falling.

Self-induction does not oppose a steady, unchanging current. Energy can be stored in the coil’s magnetic field while the current is established and returned as the current decreases. For this lesson, a qualitative understanding is sufficient.

Applications

Electricity generation: turbines driven by wind, water, steam or other energy sources rotate generators. Mechanical energy is transferred to electrical energy.
Transformers: an alternating current in a primary coil produces changing magnetic flux that induces an emf in a secondary coil. Energy is transferred through the changing electromagnetic field.
Induction cooking: a changing magnetic field induces currents in suitable cookware, transferring energy that heats the pan.
Guitar pickups: a vibrating magnetic string changes the flux through a pickup coil, inducing a varying electrical signal.
Eddy-current braking: motion through a field can induce circulating currents in a conductor. Their magnetic effects oppose the motion, converting mechanical energy into thermal energy.

For a transformer with a steady direct current, there is no sustained secondary emf after the switching transient, because the flux is then constant. Similarly, holding a magnet stationary near a stationary pickup does not produce a continuously changing signal.

9. Practice questions with expandable answers

Question 1 — Angle to the plane of a loop

A loop of area $0.012\,\mathrm{m^2}$ is in a uniform $0.25\,\mathrm{T}$ field. The field makes an angle of $30^\circ$ with the plane of the loop. Calculate the magnetic flux magnitude.

Show answer and working

The angle to the normal is $60^\circ$ , not $30^\circ$ .

$|\Phi|=BA\cos60^\circ=(0.25)(0.012)(0.50)=1.5\times10^{-3}\,\mathrm{Wb}$

Question 2 — Turning a coil through half a revolution

A $120$ -turn coil has area $8.0\,\mathrm{cm^2}$ per turn. In a uniform $0.40\,\mathrm{T}$ field, its normal turns from parallel to antiparallel to the field in $0.020\,\mathrm{s}$ . Calculate the average emf magnitude.

Show answer and working

$A=8.0\times10^{-4}\,\mathrm{m^2}$

$\Delta\Phi=(-BA)-(+BA)=-2BA=-6.4\times10^{-4}\,\mathrm{Wb}$

$|\varepsilon_{\mathrm{average}}|=120\frac{6.4\times10^{-4}}{0.020}=3.84\,\mathrm{V}\approx3.8\,\mathrm{V}$

The flux reverses sign, so the flux change magnitude is $2BA$ . The result is an average emf, not the peak emf during the rotation.

Question 3 — Increasing and decreasing field

A closed loop lies in the plane of the page. An external magnetic field points into the page. State the induced conventional current direction when (a) the field increases, (b) the field decreases and (c) the field remains constant.

Show answer and working

(a) The induced field points out of the page to oppose increasing into-page flux. The current is anticlockwise. (b) The induced field points into the page to oppose decreasing into-page flux. The current is clockwise. (c) The flux is constant, so there is no induced emf or sustained induced current.

Question 4 — Rod polarity and current

A vertical rod of length $0.50\,\mathrm{m}$ moves to the right at $6.0\,\mathrm{m\,s^{-1}}$ through a $0.40\,\mathrm{T}$ field into the page. (a) Identify the positive end. (b) Calculate the emf. (c) If it completes a circuit of total resistance $4.0\,\Omega$ , calculate the current and the opposing magnetic force.

Show answer and working

(a) The force on a positive charge is upwards, so the top end is positive. Electrons accumulate towards the bottom end.

$\varepsilon=BvL=(0.40)(6.0)(0.50)=1.2\,\mathrm{V}$

$I=\frac{1.2}{4.0}=0.30\,\mathrm{A}$

$F=BIL=(0.40)(0.30)(0.50)=0.060\,\mathrm{N}$

Conventional current flows upwards through the moving rod. Its magnetic force points to the left, opposing the motion.

Question 5 — A loop crossing a field boundary

A single rectangular loop is $0.080\,\mathrm{m}$ wide perpendicular to its motion and $0.12\,\mathrm{m}$ long along the motion. It moves at $0.40\,\mathrm{m\,s^{-1}}$ into and later out of a wide $0.50\,\mathrm{T}$ field directed into the page. Its resistance is $2.0\,\Omega$ . Find (a) the emf and current magnitudes during entry, (b) the duration of the entry pulse and (c) the current directions during entry and exit.

Show answer and working

$|\varepsilon|=BLv=(0.50)(0.080)(0.40)=0.016\,\mathrm{V}$

$I=\frac{0.016}{2.0}=8.0\times10^{-3}\,\mathrm{A}$

$\Delta t=\frac{0.12}{0.40}=0.30\,\mathrm{s}$

During entry, the current is anticlockwise to oppose increasing into-page flux. During exit, it is clockwise to oppose decreasing into-page flux. While the loop is fully inside the uniform field at fixed orientation, the net induced emf is zero.

Question 6 — Changing generator frequency

A generator has $150$ turns, coil area $0.020\,\mathrm{m^2}$ and magnetic field $0.40\,\mathrm{T}$ . It rotates at $50\,\mathrm{Hz}$ . (a) Calculate its peak emf and period. (b) State what happens to the peak emf, period and peak flux if the rotation frequency doubles.

Show answer and working

$\varepsilon_{\mathrm{max}}=2\pi fNBA=2\pi(50)(150)(0.40)(0.020)=377\,\mathrm{V}$

$T=\frac{1}{50}=0.020\,\mathrm{s}$

At $100\,\mathrm{Hz}$ , the peak emf doubles to approximately $754\,\mathrm{V}$ and the period halves to $0.010\,\mathrm{s}$ . The peak flux per turn remains $BA=8.0\times10^{-3}\,\mathrm{Wb}$ .

Question 7 — Flux linkage rather than flux per turn

A coil has $400$ turns. Its total flux linkage decreases uniformly from $0.60\,\mathrm{Wb\,turn}$ to $0.20\,\mathrm{Wb\,turn}$ in $0.080\,\mathrm{s}$ . Calculate the induced emf magnitude and explain whether the turn count should be used again.

Show answer and working

$|\varepsilon|=\frac{|\Delta(N\Phi)|}{\Delta t}=\frac{|0.20-0.60|}{0.080}=5.0\,\mathrm{V}$

Do not multiply by $400$ again. The stated flux linkage already includes all the turns.

Question 8 — Lab-style investigation of motional emf

A $0.20\,\mathrm{m}$ rod moves through a uniform magnetic field in the perpendicular arrangement. A high-resistance voltmeter measures its open-circuit emf. The readings are shown below.

Speed / $\mathrm{m\,s^{-1}}$	Emf / V
0.50	0.040
1.00	0.080
1.50	0.120
2.00	0.160

(a) State a suitable graph. (b) Explain how its gradient gives the magnetic field strength. (c) Calculate the field strength. (d) State one control variable and one way to improve the measurements.

Show answer and working

(a) Plot emf on the vertical axis against speed on the horizontal axis. The relationship $\varepsilon=BLv$ predicts a straight line through the origin. (b) The gradient $G$ equals $BL$ .

$G=\frac{0.160-0.040}{2.00-0.50}=0.080\,\mathrm{V\,s\,m^{-1}}$

$B=\frac{G}{L}=\frac{0.080}{0.20}=0.40\,\mathrm{T}$

(c) The field strength is $0.40\,\mathrm{T}$ . (d) Keep the active rod length and the perpendicular orientation fixed. Measure speed with a motion sensor or light gates and repeat readings at each speed. A best-fit line using all measurements, with uncertainty estimates, is preferable to relying on only two points.

10. Formula summary and exam readiness

Core relationships

$\Phi=BA\cos\theta,\qquad \text{flux linkage}=N\Phi$

$\varepsilon=-N\frac{\Delta\Phi}{\Delta t}=-\frac{\Delta(N\Phi)}{\Delta t}$

$\varepsilon=BvL\qquad\text{(mutually perpendicular arrangement)}$

$|\varepsilon|=NBLv\qquad\text{(rectangular coil entering or leaving)}$

$\Phi(t)=BA\cos(\omega t),\qquad \varepsilon(t)=NBA\omega\sin(\omega t)$

$\varepsilon_{\mathrm{max}}=NBA\omega=2\pi fNBA,\qquad T=\frac{1}{f}$

For current and power in a resistive circuit, neglecting self-induction:

$I=\frac{\varepsilon}{R},\qquad P=\varepsilon I=I^2R$

Use the angle between the field and the normal, rather than the plane.
Convert areas to square metres before substitution.
Distinguish flux through one turn from flux linkage through the coil.
Use signed flux changes, especially when a coil flips through half a revolution.
State whether a calculated emf is an average, instantaneous or peak value.
Apply Lenz’s law to the change in flux and state the viewing direction.
Explain where the electrical energy comes from.
Recognize that fully translating a closed loop inside a uniform field gives constant flux.
Remember that maximum flux magnitude corresponds to zero generator emf.
Explain self-induction as opposition to a change in current, rather than opposition to a steady current.

11. Topic D review — Connecting the four lessons

Lesson	Central idea	What you should now be able to do
D.1 — Gravitational fields	Mass produces a gravitational field.	Connect force, field strength, potential and energy in gravitational systems.
D.2 — Electric and magnetic fields	Charge and current produce electric and magnetic fields.	Interpret field patterns and distinguish force, field strength and potential.
D.3 — Motion in electromagnetic fields	Fields exert forces on charged particles and currents.	Predict acceleration, circular paths, velocity selection and forces on wires.
D.4 — Induction	Changing magnetic flux linkage produces an emf.	Calculate induced emf, determine its direction and explain generator energy transfers.

Across Topic D, draw clear diagrams, identify the relevant field and state your sign convention. Use conservation of energy to connect the mathematics to the physical process. Practise explaining why a formula applies before substituting numbers.

12. Faith connection — God’s strength and a life of service

Scripture and closing reflection for Topic D

“God is our refuge and strength, a very present help in trouble.” — Psalm 46:1 (KJV)

Creation: Christians understand the dependable order of nature as part of God’s good creation. The study of fields helps us appreciate that order and use it to develop technologies that support human life, from electricity generation to medical equipment.

The Fall: human ability does not guarantee wisdom or goodness. We can misuse knowledge, exploit resources or place personal success above the needs of others. Honest study and responsible use of technology call us to humility and care for our neighbours.

Redemption: God’s answer to human sin is Jesus Christ, who died for our sins and rose again. He offers forgiveness and reconciliation with God. You are invited to turn to him, trust him as your personal God, Saviour and Lord, and receive the new life he gives.

Restoration: Christians look forward to God’s renewal of creation. Until then, we can use our learning to serve others, care for the world and bring hope through faithful action. The ability to understand and apply physics is an opportunity to contribute to that service.

As Topic D closes: seek more than a good examination result. Grow in truthfulness, perseverance and gratitude. Read and study the Bible, spend time in prayer and connect with a church or community of believers who can help you know God more deeply. Let your knowledge become part of a life shaped by the love of Jesus Christ.

Open reflection questions

How could you use what you have learned about fields and induction to serve other people? What does responsible use of energy look like in your school or home? How can you seek God’s strength when learning is difficult, and encourage a classmate who is struggling?

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Induction

Lesson D.4 Electromagnetic Induction

Learning goals

Notation and assumptions

1. Key ideas: emf, magnetic flux and flux linkage

What is an emf?

Magnetic flux through a flat surface

Why the cosine factor appears

IB angle and area checks

Flux linkage

What can change the flux?

Worked example 1 — Flux and flux linkage

2. Faraday’s law: calculating induced emf

What increases the induced emf?

Worked example 2 — A changing flux through a coil

3. Lenz’s law and conservation of energy

A reliable method for determining current direction

Why this is an energy-conservation rule

Worked example 3 — An approaching and retreating magnet

4. Motional emf in a straight conductor

Derivation using magnetic and electric forces

Mechanical work supplies electrical energy

Worked example 4 — A sliding rod and its energy transfer

5. A rectangular coil entering and leaving a field

Worked example 5 — Emf while a coil enters a field

6. AC generators: sinusoidal flux and emf

How an AC generator works

Obtaining the emf without calculus

Phase relationship

Effect of changing the rotation frequency

Worked example 6 — AC generator output

7. Interpreting flux–time and emf–time graphs

Worked example 7 — Piecewise flux changes

8. Self-induction and practical applications

Self-induction: a qualitative explanation

Applications

9. Practice questions with expandable answers

Question 1 — Angle to the plane of a loop

Question 2 — Turning a coil through half a revolution

Question 3 — Increasing and decreasing field

Question 4 — Rod polarity and current

Question 5 — A loop crossing a field boundary

Question 6 — Changing generator frequency

Question 7 — Flux linkage rather than flux per turn

Question 8 — Lab-style investigation of motional emf

10. Formula summary and exam readiness

Core relationships

11. Topic D review — Connecting the four lessons

12. Faith connection — God’s strength and a life of service

Scripture and closing reflection for Topic D

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