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.
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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 measures how much magnetic field passes through a loop of area :
where is the angle between the field and the normal (perpendicular) to the loop. Flux is measured in webers (Wb). 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:
where is the number of turns in the coil and is the rate of change of flux. The faster the flux changes and the more turns the coil has, the larger the induced EMF. Note that a steady, unchanging flux induces nothing; only change matters.
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 (AC); 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:
A step-up transformer () raises voltage; a step-down transformer lowers it. Energy is conserved, so for an ideal transformer . 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.
2019 TASC2 marksAn Airbus 350 is flying north through the Earth's magnetic field, which has a flux density of 1.90 x 10^-5 T at an angle of declination of -70 degrees. The plane flies at 900 km/h with a wingspan of 64.75 m. Find the potential difference (motional EMF) generated across the wing.Show worked answer →
A conductor moving through a magnetic field has a motional EMF of EMF = B_perp L v, where B_perp is the component of the field perpendicular to both the wing and the velocity.
Convert the speed: 900 km/h = 900 / 3.6 = 250 m s-1.
For a horizontally moving wing, the vertical component of the Earth's field cuts the wing: B_vert = B sin70 = 1.90 x 10^-5 x sin70 = 1.90 x 10^-5 x 0.9397 = 1.785 x 10^-5 T.
EMF = B_vert L v = 1.785 x 10^-5 x 64.75 x 250 = 0.289 V.
A PD of about 0.29 V is induced across the wingtips. Markers want the vertical field component (using the dip/declination angle) and EMF = BLv.
2024 TASC5 marksA 30-turn coil 5.00 cm by 2.00 cm rotates at 50 Hz in a magnetic field of flux density 0.2 T. Show that the speed of the side AB is about 310 cm 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 cm side: r = 0.01 m. Its speed is v = 2 pi f r (circumference per revolution times rotation rate):
v = 2 pi x 50 x 0.01 = 3.14 m s-1 = 314 cm s-1, which is about 310 cm s-1 as required.
Maximum EMF of a single side occurs when it moves perpendicular to the field, EMF = B L v per turn. With L = 0.05 m (the length of side AB):
EMF per turn = 0.2 x 0.05 x 3.14 = 0.0314 V.
For 30 turns the side AB contributes EMF = 30 x 0.0314 = 0.942 V.
So each side AB produces a maximum EMF of about 0.94 V (the full coil adds the matching contribution from the opposite side). Markers want v = 2 pi f r and EMF = n B L v at the peak.