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6th Year OL · Physics
Electromagnetic field · Electromagnetic induction

Generating the Electromotive Force

If you move a magnet towards a coil without touching it, a current appears in the coil. Nobody connected a cell or battery: the current seems to come from nowhere, or rather, from a changing magnetic flux. It is the same principle behind the generators at Ardnacrusha on the Shannon, which have fed the National Grid since 1929. This phenomenon, discovered by Faraday in 1831, is the basis of every electrical generator on the planet.

You can investigate it in the lab: connect a coil to a galvanometer and push a bar magnet into it. The needle deflects while the magnet moves and returns to zero when it stops. Reverse the motion and the needle deflects the other way. Move the magnet faster, or use more turns, and the deflection is bigger. The current in the coil is the induced current.

Record what you see in a table: the same magnet pushed in at a slow, medium and fast speed gives a larger galvanometer reading each time, and a coil with $100$ turns gives roughly twice the reading of a coil with $50$ turns. So the induced emf grows with the rate at which the flux changes and with the number of turns, which is what Faraday's law below states.

Magnet speed Galvanometer reading, $50$-turn coil (divisions) Galvanometer reading, $100$-turn coil (divisions)
Slow 2 4
Medium 5 10
Fast 8 16

This is primary data: a faster change of flux means a bigger induced emf, so a bigger current.

Magnetic flux

If you hold a hoop in front of a fan or in the wind on a Galway promenade, the air passing through it depends on the speed of the wind, the size of the hoop and whether you face it head-on or edge-on: how can we measure in a similar way how many magnetic field lines pass through a loop?

Vertical magnetic field lines passing through a flat loop; the angle α is measured between the field B and the normal vector to the loop. B normal α flat loop

Faraday's law: the induced emf

Graph of the magnetic flux Φ against time: it rises linearly, stays constant and falls linearly; the induced emf is non-zero only when the flux changes. t Φ Φ(t) Φ changing E ≠ 0 Φ constant E = 0 Φ changing E ≠ 0

Lenz's law: the direction of the induced current

If you drop a magnet down a vertical copper tube, it falls surprisingly slowly, as if floating in a viscous liquid, even though copper is not magnetically attracted to the magnet: what physical law explains why the induced current always slows the motion down?

Animation: the north pole of a magnet moves towards a loop, which lights up as an induced north face to repel it; then the magnet moves away and the loop lights up with the opposite polarity to hold it back. SN v N (opposes, repels) the loop "defends" its flux
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State, justifying your answer in each case, in which of these situations an electromotive force is induced in the coil: A magnet at rest inside a coil. A magnet entering…

State, justifying your answer in each case, in which of these situations an electromotive force is induced in the coil:

  • A magnet at rest inside a coil.

  • A magnet entering the coil.

  • A coil carrying a constant direct current, placed next to another coil.

  • The same pair of coils as in the previous case, at the moment the switch is opened and the current turns off.

Solve

Motional emf: a moving conductor

A Luas pulling out of a stop does not look like a magnet and a coil, yet a metal axle moving across the Earth's field can have an emf across it: how?

Animation of a conducting rod sliding at constant speed to the right along two rails in a magnetic field pointing into the page: the enclosed area grows, the flux is shaded and the induced current appears anticlockwise. × × × × × × v l = transverse side
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A laboratory trolley carries a rectangular loop and advances at $2\ \text{m/s}$ entering the region between the poles of an electromagnet, where the field is…

A laboratory trolley carries a rectangular loop and advances at $2\ \text{m/s}$ entering the region between the poles of an electromagnet, where the field is $0.4\ \text{T}$ and is perpendicular to the plane of the loop. The side of the loop transverse to the motion measures $15\ \text{cm}$. The situation is represented in the figure.

× × × × × × × × × × × × B = 0.4 T l = 15 cm v = 2 m/s emf = ?
  1. a) Calculate the induced emf while the loop is entering, using $\varepsilon = B \times l \times v$.
  2. b) Explain why the emf becomes zero when the loop is completely inside the field.
Solve
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A low-orbit satellite deploys a conducting cable $2\ \text{km}$ long and moves at $7.5\times10^{3}\ \text{m/s}$ in a direction perpendicular to the Earth's magnetic…

A low-orbit satellite deploys a conducting cable $2\ \text{km}$ long and moves at $7.5\times10^{3}\ \text{m/s}$ in a direction perpendicular to the Earth's magnetic field, which at that altitude is $3\times10^{-5}\ \text{T}$. The situation is represented in the figure.

· · · · · · B = 30 μT satellite v = 7.5·10³ m/s l = 2 km ΔV = ?
  1. a) Calculate the potential difference that appears across the ends of the cable, $\varepsilon = B \times l \times v$.
  2. b) Explain why an emf is only induced when the cable moves cutting the field lines, and not if it moves in the direction of the field.
Solve
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A square loop with $10\ \text{cm}$ sides and electrical resistance $R = 0.5\ \Omega$ moves at a constant speed of $1\ \text{m/s}$ and passes through a square region with…

A square loop with $10\ \text{cm}$ sides and electrical resistance $R = 0.5\ \Omega$ moves at a constant speed of $1\ \text{m/s}$ and passes through a square region with $20\ \text{cm}$ sides where there is a uniform magnetic field of $0.5\ \text{T}$, perpendicular to the plane of the loop and pointing inward (into the paper), as shown in the figure. The loop moves to the right.

Square loop with 10 cm sides moving at 1 m/s toward a square region with 20 cm sides with an inward magnetic field of 0,5 T × × × × × × × × × B = 0,5 T 20 cm 10 cm v = 1 m/s
  1. a) Calculate the emf while the loop is entering the region ($\varepsilon = B \times l \times v$) and the time the entry lasts.
  2. b) What is the emf while the loop is completely inside? Why?
  3. c) What emf is there while it exits, and what is the direction of the current compared with that during the entry? Apply Lenz's law.
  4. d) Draw in your notebook the graph of the current as a function of time over the whole crossing (from when the loop starts to enter until it finishes exiting), indicating the numerical values and the signs. Use the resistance of the loop, $R = 0.5\ \Omega$, and take the direction of the current during the entry as positive.

You can do it on paper and send a photo with your answer.

Solve
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To extract a rectangular loop at a constant speed from a field of $0.8\ \text{T}$, it must be pulled with a force, as shown in the figure. The loop has a resistance of…

To extract a rectangular loop at a constant speed from a field of $0.8\ \text{T}$, it must be pulled with a force, as shown in the figure. The loop has a resistance of $0.1\ \Omega$, its side transverse to the motion measures $20\ \text{cm}$ and it is extracted at $0.5\ \text{m/s}$.

× × × × × × × × × B = 0.8 T l = 20 cm R = 0.1 Ω F = ? v = 0.5 m/s
  1. a) Calculate the induced emf during extraction ($\varepsilon = B \times l \times v$).
  2. b) Calculate the induced current.
  3. c) Calculate the magnetic force on the transverse side, $F = I \times l \times B$, and explain why, according to Lenz's law, it opposes the extraction.
  4. d) Calculate the mechanical power of the external force, $P = F \times v$, and the electrical power dissipated in the loop, $P = \varepsilon \times I$; compare them and explain what physical principle this agreement illustrates.
Solve
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An aeroplane flies horizontally northwards at $900\ \text{km/h}$ with a wingspan of $35\ \text{m}$. In that area, the vertical component of the Earth's magnetic field is…

An aeroplane flies horizontally northwards at $900\ \text{km/h}$ with a wingspan of $35\ \text{m}$. In that area, the vertical component of the Earth's magnetic field is $4\times10^{-5}\ \text{T}$ and points towards the ground. Observe the diagram (top view):

× × × × B vertical towards the ground west east v = 900 km/h (north) 35 m ΔV = ?
  1. a) Express the speed in $\text{m/s}$.
  2. b) Calculate the potential difference that appears between the wingtips, $\varepsilon = B \times l \times v$.
  3. c) The positive charges in the metal experience the magnetic force from Lorentz's law: applying the right-hand rule to the vector product of $\vec{v}$ by $\vec{B}$, reason which wingtip, east or west, is at a higher potential.
Solve

Induced charge and eddy currents

If you make a solid copper plate swing between the poles of a switched-off electromagnet, it behaves like an ordinary pendulum, but when the electromagnet is switched on it stops dead almost instantly: how do braking currents arise inside a continuous block of metal with no wires or connections?

Comparison of eddy currents: in a solid metal piece they form large free circular loops; in a core of insulated laminations they are confined to small loops and are damped. B into the page (⊗), changing ⊗⊗ ⊗⊗ I solid piece: eddy currents flow freely insulated laminations: they are damped small currents, fewer losses
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A coil with $500$ turns of $40\ \text{cm}^2$ each and total resistance $20\ \Omega$ is in a magnetic field perpendicular to its plane that varies, in SI units, according…

A coil with $500$ turns of $40\ \text{cm}^2$ each and total resistance $20\ \Omega$ is in a magnetic field perpendicular to its plane that varies, in SI units, according to the law $B = 0.8 - 2 \times t$. The situation is represented in the figure.

× × × × × B = 0.8 − 2·t (T) into page and decreasing N = 500 turns i = ?
  1. a) Calculate the rate of change of the field, $\frac{\Delta B}{\Delta t}$.
  2. b) Calculate the induced emf in the coil and the current flowing through it.
  3. c) Calculate the charge passing through a cross-section of the circuit in $0.3\ \text{s}$.
  4. d) If the field points into the page and is decreasing, reason using Lenz's law whether the induced current is clockwise or anticlockwise as viewed from your side.
Solve
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In an industrial induction furnace, a coil around the crucible generates an alternating field: in half a cycle, the field changes from $-0.05\ \text{T}$ to…

In an industrial induction furnace, a coil around the crucible generates an alternating field: in half a cycle, the field changes from $-0.05\ \text{T}$ to $+0.05\ \text{T}$ in $1\times10^{-5}\ \text{s}$. A section of the metal in the crucible is equivalent to a loop of $100\ \text{cm}^2$ perpendicular to the field.

  1. a) Calculate the change in flux in half a cycle and the average induced emf in that equivalent loop.
  2. b) Explain why the furnace requires alternating current and would not work with a constant direct current.
  3. c) Explain why the metal in the crucible heats up due to the Joule effect, and explain what eddy currents are.
Solve
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To measure the field between the poles of a horseshoe magnet, a search coil of $50$ turns of $4\ \text{cm}^2$ each is used, with a total resistance of $10\ \Omega$,…

To measure the field between the poles of a horseshoe magnet, a search coil of $50$ turns of $4\ \text{cm}^2$ each is used, with a total resistance of $10\ \Omega$, connected to a charge meter. It is placed perpendicular to the field and abruptly removed in $0.2\ \text{s}$; the meter registers a charge of $4\times10^{-4}\ \text{C}$.

  1. a) Chain the relationships $q = I \times \Delta t$, $I = \frac{\varepsilon}{R}$ and $\varepsilon = N \times \frac{\Delta \Phi}{\Delta t}$ to deduce that $B = \frac{q \times R}{N \times S}$.
  2. b) Calculate the magnitude of the field between the poles of the magnet.
  3. c) Calculate the average emf and the average current during extraction.
Solve

Back to the start: The induced current does not come "from nowhere": when you move the magnet you change the magnetic flux in the coil, inducing an emf, and the electrical energy generated comes exactly from the mechanical work you do to overcome the magnetic repulsion predicted by Lenz's law.

Summary

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