HL only Fields BETA D.4

Induction

This topic explores how relative motion between conductors and magnetic fields, or changes in magnetic environments, generate electromotive force (emf). It details the mathematical and directional principles that govern modern electric generators, transformers, and induction systems.

Part of IB Physics (2025-2030 syllabus) — Standard and Higher Level.

Key points

Magnetic flux ( $\Phi$ ) measures the total magnetic field passing through a given area, accounting for the angle between the field vectors and the surface normal.
Magnetic flux linkage ( $N\Phi$ ) scales the flux by the number of turns ( $N$ ) in a coil, representing the cumulative electromagnetic interaction.
Faraday's law states that the magnitude of an induced electromotive force ( $\text{emf}$ ) is proportional to the rate of change of magnetic flux linkage.
Lenz's law is an expression of the law of conservation of energy, stating that the induced current will flow in a direction that creates a magnetic field opposing the original change in flux.
A straight conductor of length $L$ moving at a constant speed $v$ perpendicular to a magnetic field $B$ generates a steady motional $\text{emf}$ given by $\varepsilon = BvL$ .
A coil rotating with a constant angular velocity $\omega$ in a uniform magnetic field produces a sinusoidally alternating $\text{emf}$ due to the continuously changing projected area.

Subtopic by subtopic

Magnetic flux and flux linkage

Magnetic flux measures how much magnetic field passes through a surface. For a uniform field of strength $B$ crossing a flat area $A$ , the flux is given by:

$\Phi = BA\cos\theta$

Here $\theta$ is measured between the field direction and the normal to the surface (the line perpendicular to it), never the surface itself. Flux is a maximum when the field passes straight through the loop ( $\theta = 0^\circ$ ) and zero when the field lines skim along the plane of the loop ( $\theta = 90^\circ$ ). Flux is measured in webers, where $1\ \text{Wb} = 1\ \text{T m}^2$ .

A coil with $N$ turns threads the same flux $N$ times over, so the useful quantity for coils is the flux linkage:

$N\Phi = NBA\cos\theta$

For example, a $500$ -turn coil of area $1.0 \times 10^{-3}\ \text{m}^2$ with its plane perpendicular to a $0.20\ \text{T}$ field has a flux linkage of $N\Phi = 500 \times 0.20 \times 1.0 \times 10^{-3} = 0.10\ \text{Wb}$ .

You must be able to compute flux and flux linkage at any orientation, and recognise the three ways flux can change:

altering the field strength $B$
altering the area $A$
altering the angle $\theta$

Faraday's law of electromagnetic induction

Whenever the flux linkage through a circuit changes, an emf is induced. Faraday's law gives its size: the induced emf equals the rate of change of flux linkage, given by:

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

The minus sign records the direction and is explained by Lenz's law. The faster the change, the larger the emf: the same magnet dropped through a coil more quickly produces a taller, narrower voltage pulse, even though the total flux change is identical.

Because $\Phi = BA\cos\theta$ , there are three distinct ways to induce an emf:

change the field strength (switching an electromagnet on or off)
change the area of the circuit in the field (a loop being stretched, or a rod sliding along rails)
change the orientation (a rotating coil)

On a graph of flux linkage against time, the induced emf at any instant is the negative of the gradient, so steep sections of the flux graph correspond to large emfs and flat sections to zero emf.

Induction cooktops use this directly: a rapidly alternating field beneath the pan continually changes the flux through the metal base, inducing currents that heat it. You must be able to calculate an average emf from $\frac{\Delta(N\Phi)}{\Delta t}$ and translate between flux-time and emf-time graphs.

Lenz's law

Lenz's law fixes the direction of induced effects: the induced emf, and any resulting current, acts to oppose the change in flux that created it. It supplies the minus sign in Faraday's law.

To apply it, follow three steps:

identify how the external flux is changing (growing or shrinking, and in which direction)
state that the induced current must oppose that change
use the right-hand grip rule to find the current direction that produces the opposing field

Pushing the north pole of a magnet towards a coil makes the flux into the coil grow, so the induced current circulates to make the near face of the coil a north pole, repelling the approaching magnet. Pulling the magnet away reverses everything: the coil now attracts it, resisting the separation.

Lenz's law is a fundamental manifestation of the conservation of energy. If the induced current assisted the change in flux instead of opposing it, pushing a magnet into a coil would induce a current that attracted the magnet further, accelerating it without limit while generating electrical energy from nothing.

Because of the opposing force required by Lenz's law, mechanical work must be done on the system to push the magnet against the opposing field, and this work is the exact source of the electrical energy generated in the coil. Eddy-current brakes exploit this: the opposing forces on induced currents in a moving metal disc convert kinetic energy directly into heat.

Motional emf (a conductor moving in a field)

When a straight conductor moves through a magnetic field, its free electrons move with it, so each feels a magnetic force $F = qvB$ . This force pushes electrons towards one end of the conductor, leaving the other end positive, and the separated charge builds an internal electric field $E$ .

Charge stops accumulating when the electric and magnetic forces on the charges balance, $qE = qvB$ , giving $E = vB$ and hence a potential difference across a conductor of length $L$ given by:

$\varepsilon = BvL$

This is the motional emf. The formula assumes $B$ , $v$ and the conductor are mutually perpendicular; only the velocity component perpendicular to the field counts, and a conductor moving parallel to the field cuts no flux and generates no emf.

If the moving conductor forms part of a complete circuit, as in the classic rod sliding along conducting rails joined through a resistor, the emf drives a current $I = \frac{\varepsilon}{R}$ , and the field then exerts a force $F = BIL$ on the rod that opposes its motion (Lenz's law again). Keeping the rod at constant speed requires an applied force doing work at exactly the rate electrical energy is dissipated in the resistor.

You must be able to:

calculate motional emfs
identify which end of the conductor becomes positive by considering the magnetic force on the charges
explain the energy transfer in rail-and-rod circuits

emf from a coil rotating in a magnetic field

Rotating a flat coil at a steady angular speed $\omega$ in a uniform field changes the angle $\theta$ continuously, so the flux linkage varies as:

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

The induced emf is the negative rate of change of this, giving the sinusoidal output:

$\varepsilon = NBA\omega\sin(\omega t)$

The peak value is $\varepsilon_0 = NBA\omega$ . This is the principle of the alternating-current generator: slip rings and brushes connect the spinning coil to the external circuit so the alternating emf can drive a current.

The emf and the flux linkage are a quarter-cycle out of phase. When the coil's plane is perpendicular to the field, the flux linkage is at its maximum but momentarily not changing, so the emf is zero; when the plane lies parallel to the field, the flux is zero but changing at its fastest rate, so the emf is at its peak.

Spinning the coil twice as fast doubles the peak emf (since $\varepsilon_0 \propto \omega$ ) and halves the period, so the output graph becomes both taller and more compressed.

You must be able to:

sketch the flux-linkage and emf graphs one above the other with the correct phase relationship
calculate the peak emf from $\varepsilon_0 = NBA\omega$
describe how the output changes when $B$ , $A$ , $N$ or the rotation rate is altered

Formulae

\Phi = B A \cos\theta

To calculate the magnetic flux through an area $A$ when the magnetic field $B$ is uniform and oriented at an angle $\theta$ relative to the normal of the surface area.

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

To calculate the average induced electromotive force in a coil of $N$ turns over a time interval $\Delta t$ when the magnetic flux changes.

\varepsilon = BvL

To calculate the motional electromotive force induced across the ends of a straight conductor of length $L$ moving at speed $v$ perpendicular to a uniform magnetic field $B$ .

\varepsilon = N B A \omega \sin(\omega t)

To calculate the instantaneous alternating electromotive force induced in a flat coil of $N$ turns and area $A$ rotating at a constant angular frequency $\omega$ in a uniform magnetic field $B$ .

\varepsilon_0 = N B A \omega

To calculate the peak (maximum) emf of a coil of $N$ turns and area $A$ rotating at angular frequency $\omega$ in a uniform field $B$ ; the instantaneous emf oscillates between $+\varepsilon_0$ and $-\varepsilon_0$ .

Definitions

Magnetic flux ( $\Phi$ ): The product of the area of a surface and the component of the magnetic field strength normal to that surface, measured in webers ( $\text{Wb}$ ).
Magnetic flux linkage ( $N\Phi$ ): The product of the magnetic flux passing through a single loop and the total number of turns ( $N$ ) in the wire coil.
Faraday's law of induction: A fundamental law stating that the magnitude of the induced electromotive force in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit.
Lenz's law: The rule stating that the direction of any induced electromotive force and resulting current is such that it opposes the change in magnetic flux that produced it.
Motional emf: The electromotive force induced across a conductor moving through a magnetic field, caused by the magnetic force pushing the conductor's free charges towards opposite ends.

Worked examples

A flat, rectangular coil consisting of $120$ turns and enclosing an area of $4.0 \times 10^{-3}\ \text{m}^2$ is placed perpendicular to a uniform magnetic field of $0.45\ \text{T}$ . The coil is rotated by $90^\circ$ about an axis in its plane in a time interval of $0.15\ \text{s}$ , ending up parallel to the magnetic field. Calculate the average electromotive force ( $\text{emf}$ ) induced in the coil during this rotation.

1
Identify the initial state: The coil is perpendicular to the field, so the angle $\theta$ between the normal to the area and the magnetic field is $0^\circ$ . The initial flux is $\Phi_{\text{initial}} = B A \cos(0^\circ) = 0.45\ \text{T} \times 4.0 \times 10^{-3}\ \text{m}^2 = 1.8 \times 10^{-3}\ \text{Wb}$ .
2
Calculate the initial flux linkage: $N\Phi_{\text{initial}} = 120 \times 1.8 \times 10^{-3}\ \text{Wb} = 0.216\ \text{Wb}$ .
3
Identify the final state: The plane of the coil is parallel to the field, meaning the normal to the area is perpendicular to the field ( $\theta = 90^\circ$ ). Thus, $\Phi_{\text{final}} = B A \cos(90^\circ) = 0\ \text{Wb}$ , and the final flux linkage is $0\ \text{Wb}$ .
4
Apply Faraday's law to calculate the magnitude of the average induced $\text{emf}$ : $\varepsilon_{\text{avg}} = \left| \frac{\Delta(N\Phi)}{\Delta t} \right| = \frac{0.216\ \text{Wb} - 0\ \text{Wb}}{0.15\ \text{s}}$ .
5
Compute the numerical value: $\varepsilon_{\text{avg}} = 1.44\ \text{V}$ .

Answer: $1.4\ \text{V}$

A conducting rod of length $0.60\ \text{m}$ slides at a constant speed of $4.5\ \text{m s}^{-1}$ along two frictionless horizontal rails, perpendicular to a uniform vertical magnetic field of $0.25\ \text{T}$ . The rails are joined through a resistor of resistance $1.5\ \Omega$ ; the rod and rails have negligible resistance. Calculate the motional emf, the induced current, and the force required to keep the rod moving at constant speed.

1
The rod, the field and the velocity are mutually perpendicular, so the motional emf is $\varepsilon = BvL = 0.25\ \text{T} \times 4.5\ \text{m s}^{-1} \times 0.60\ \text{m} = 0.675\ \text{V}$ .
2
The emf drives a current around the circuit of $I = \frac{\varepsilon}{R} = \frac{0.675\ \text{V}}{1.5\ \Omega} = 0.45\ \text{A}$ .
3
The field exerts a force on the current-carrying rod of $F = BIL = 0.25\ \text{T} \times 0.45\ \text{A} \times 0.60\ \text{m} = 0.0675\ \text{N}$ , directed against the motion by Lenz's law.
4
For the rod to move at constant speed, the applied force must balance this opposing force, so $F_{\text{applied}} = 6.8 \times 10^{-2}\ \text{N}$ .
5
Check with energy conservation: the mechanical power $P = Fv = 0.0675\ \text{N} \times 4.5\ \text{m s}^{-1} = 0.30\ \text{W}$ equals the electrical power $P = I^2R = (0.45\ \text{A})^2 \times 1.5\ \Omega = 0.30\ \text{W}$ .

Answer: $\varepsilon = 0.68\ \text{V}$ , $I = 0.45\ \text{A}$ , applied force $= 6.8 \times 10^{-2}\ \text{N}$

The flat coil of a model generator has $200$ turns, each of area $2.5 \times 10^{-3}\ \text{m}^2$ , and rotates at a frequency of $50\ \text{Hz}$ in a uniform magnetic field of $0.080\ \text{T}$ . Taking $t = 0$ at the instant when the plane of the coil is perpendicular to the field, calculate the peak emf produced and the instantaneous emf at $t = 2.0\ \text{ms}$ .

1
Convert the rotation frequency to angular frequency: $\omega = 2\pi f = 2\pi \times 50\ \text{Hz} = 314\ \text{rad s}^{-1}$ .
2
Calculate the peak emf: $\varepsilon_0 = NBA\omega = 200 \times 0.080\ \text{T} \times 2.5 \times 10^{-3}\ \text{m}^2 \times 314\ \text{rad s}^{-1} = 12.6\ \text{V}$ .
3
Since the flux linkage is at its maximum at $t = 0$ , the emf follows $\varepsilon = \varepsilon_0 \sin(\omega t)$ , and at $t = 2.0\ \text{ms}$ the phase is $\omega t = 314\ \text{rad s}^{-1} \times 2.0 \times 10^{-3}\ \text{s} = 0.628\ \text{rad}$ .
4
Evaluate the sine of the phase: $\sin(0.628\ \text{rad}) = 0.588$ .
5
Calculate the instantaneous emf: $\varepsilon = 12.6\ \text{V} \times 0.588 = 7.4\ \text{V}$ .

Answer: $\varepsilon_0 = 12.6\ \text{V}$ ; at $t = 2.0\ \text{ms}$ , $\varepsilon = 7.4\ \text{V}$

Common mistakes

×Using the incorrect angle in the magnetic flux formula. Students often use the angle between the magnetic field vector and the *plane of the loop* rather than the angle relative to the *normal to the loop's surface*.
×Neglecting to multiply the magnetic flux by the number of turns ( $N$ ) in the coil when calculating total electromotive force or flux linkage.
×Forgetting that if a conductor moves parallel to the magnetic field lines, no magnetic flux is cut, and therefore no motional $\text{emf}$ is induced.
×Confusing the graphs of flux linkage versus time and induced $\text{emf}$ versus time for a rotating coil. The induced $\text{emf}$ is the negative derivative of the flux linkage, which introduces a $90^\circ$ (or $\frac{\pi}{2}\ \text{rad}$ ) phase shift.

Exam tips

✓When asked to **explain** Lenz's law in a specific scenario, always structure your answer in three distinct steps: 1) Identify how the external magnetic flux is changing, 2) State that the induced current must oppose this change, and 3) Describe the direction of the magnetic field and current needed to create that opposition.
✓When you are asked to **derive** the motional $\text{emf}$ equation $\varepsilon = BvL$ for a moving conductor, start by equating the magnetic force ( $F_B = qvB$ ) and the electric force ( $F_E = qE$ ) acting on free charges within the conductor under equilibrium, and substitute the relation $E = \frac{V}{L}$ .
✓Pay close attention to whether the question asks for the 'magnitude' of the induced $\text{emf}$ or its directional value. If direction is crucial, ensure the negative sign from Faraday-Lenz's law is accounted for qualitatively or quantitatively.

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