Skip to main content
ExamExplained
TAS · Physics
Physics study scene
§-Syllabus dot point
TASPhysicsSyllabus dot point

How does a changing magnetic field create electricity?

Apply Faraday's and Lenz's laws to electromagnetic induction, generators and transformers.

How a changing magnetic flux induces an EMF through Faraday's law, how Lenz's law fixes its direction, and how generators and transformers apply induction.

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

Electromagnetic induction is the reverse of the motor effect: instead of using a current to make motion, you use motion or a changing field to make a current. It is the basis of almost all electrical power generation.

Magnetic flux

Magnetic flux Φ\Phi measures how much magnetic field passes through a loop of area AA:

Φ=BAcosθ\Phi = B A \cos\theta

where θ\theta is the angle between the field and the normal (perpendicular) to the loop. Flux is measured in webers. Flux changes if the field strength changes, the area changes, or the loop rotates relative to the field.

Faraday's law

Faraday found that an EMF (voltage) is induced whenever the flux through a circuit changes. The size of the induced EMF is:

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

where NN is the number of turns in the coil and ΔΦΔt\dfrac{\Delta\Phi}{\Delta t} is the rate of change of flux. The faster the flux changes and the more turns the coil has, the larger the induced EMF. A steady, unchanging flux induces nothing; only change matters.

Motional EMF

A special case is a straight conductor of length LL moving at speed vv perpendicular to a field BB. The free charges in the conductor feel the magnetic force qvBqvB and pile up at the ends until an electric field balances it, giving a motional EMF:

ε=BLv\varepsilon = BLv

This is the same physics as Faraday's law: as the rod moves it sweeps out area, so the flux through the circuit it is part of changes at the rate BLvBLv. It explains the voltage induced across an aircraft wing moving through Earth's field and the output of a simple generator.

Lenz's law

The negative sign in Faraday's law represents Lenz's law: the induced current flows in the direction that opposes the change in flux producing it. If the flux is increasing, the induced current creates a field to oppose the increase; if it is decreasing, the induced current tries to maintain it. Lenz's law is a statement of conservation of energy. If the induced current aided the change, you would get energy for free.

Generators and transformers

A generator is essentially a motor run in reverse. A coil is rotated in a magnetic field, so the flux through it changes continuously, inducing an alternating EMF. With slip rings the output is alternating current; with a split-ring commutator it can be made into pulsed DC.

A transformer changes the voltage of an AC supply. An alternating current in the primary coil produces a changing flux in an iron core, which induces an EMF in the secondary coil. For an ideal transformer:

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}

A step-up transformer (Ns>NpN_s > N_p) raises voltage; a step-down transformer lowers it. Energy is conserved, so for an ideal transformer VpIp=VsIsV_p I_p = V_s I_s. Transformers only work with AC because they need a continuously changing flux.

When tackling induction questions, identify what is causing the flux to change, apply Faraday's law for the size of the EMF, and use Lenz's law to argue the direction. For transformers, remember the turns ratio sets the voltage ratio and that real transformers lose some energy to heat in the core and windings.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20192 marksAn Airbus A350 is flying north through the Earth's magnetic field, which has a flux density of 1.90×105 T1.90 \times 10^{-5}\ \text{T} at a dip angle of 7070^\circ. The plane flies at 900 km h1900\ \text{km h}^{-1} with a wingspan of 64.75 m64.75\ \text{m}. Find the motional EMF generated across the wing.
Show worked answer →

A conductor moving through a field has a motional EMF of ε=BLv\varepsilon = B_\perp L v, where BB_\perp is the component of the field perpendicular to both the wing and the velocity.

Convert the speed: 900 km h1=9003.6=250 m s1900\ \text{km h}^{-1} = \dfrac{900}{3.6} = 250\ \text{m s}^{-1}.

For a horizontally moving wing, the vertical component of the Earth's field cuts the wing: Bvert=Bsin70=(1.90×105)(0.9397)=1.785×105 TB_\text{vert} = B\sin 70^\circ = (1.90\times10^{-5})(0.9397) = 1.785 \times 10^{-5}\ \text{T}.

ε=BvertLv=(1.785×105)(64.75)(250)=0.289 V.\varepsilon = B_\text{vert} L v = (1.785\times10^{-5})(64.75)(250) = 0.289\ \text{V}.

A potential difference of about 0.29 V0.29\ \text{V} is induced across the wingtips. Markers want the vertical field component (using the dip angle) and ε=BLv\varepsilon = BLv.

TCE 20245 marksA 3030-turn coil 5.00 cm5.00\ \text{cm} by 2.00 cm2.00\ \text{cm} rotates at 50 Hz50\ \text{Hz} in a magnetic field of flux density 0.2 T0.2\ \text{T}. Show that the speed of the side AB is about 3.1 m s13.1\ \text{m s}^{-1}, then calculate the maximum EMF generated by the side AB.
Show worked answer →

The side AB moves in a circle of radius equal to half the 2.00 cm2.00\ \text{cm} side: r=0.01 mr = 0.01\ \text{m}. Its speed is v=2πfrv = 2\pi f r:

v=2π(50)(0.01)=3.14 m s1,v = 2\pi (50)(0.01) = 3.14\ \text{m s}^{-1},

which is about 3.1 m s13.1\ \text{m s}^{-1} as required.

The maximum EMF of a single side occurs when it moves perpendicular to the field, ε=BLv\varepsilon = BLv per turn, with L=0.05 mL = 0.05\ \text{m}:

εturn=(0.2)(0.05)(3.14)=0.0314 V.\varepsilon_\text{turn} = (0.2)(0.05)(3.14) = 0.0314\ \text{V}.

For 3030 turns the side AB contributes ε=(30)(0.0314)=0.94 V\varepsilon = (30)(0.0314) = 0.94\ \text{V}.

So each side AB produces a maximum EMF of about 0.94 V0.94\ \text{V}. Markers want v=2πfrv = 2\pi f r and ε=NBLv\varepsilon = NBLv at the peak.

ExamExplained