University Physics — Year 1 · Bachelor Year 1

29Electromagnetic Induction and Applications

Move a magnet toward a coil and a current flows in the coil; stop the magnet and the current stops. Faraday found this in 1831, after ten years of looking, and it is the fact on which the electrical world runs: every power station, every transformer, every motor and loudspeaker and microphone, every induction cooktop and wireless charger converts energy through the law that a changing magnetic flux drives an electromotive force. This chapter states the law with its signs, gives it the two faces it wears (a changing field in a fixed circuit, a circuit moving in a fixed field), defines self and mutual inductance, and then works through the electromechanical machines — the sliding bar, the alternator, the transformer, the loudspeaker — where the same law turns mechanical power into electrical power and back, never for free.

An MRI scanner: a superconducting solenoid holding 1.5\, T over a patient — a megajoule of magnetic energy in a field with nothing in it. Photograph: Jan Ainali, CC BY 3.0. — An MRI scanner: a superconducting solenoid holding $1.5\,\mathrm{T}$ over a patient — a megajoule of magnetic energy in a field with nothing in it. Photograph: Jan Ainali, CC BY 3.0.

29.1 Faraday’s law

Definition 29.1 (Magnetic flux through a circuit)

Orient a closed circuit $\Gamma$ (a sense of travel); its surface gets the normal $\vect n$ given by the right-hand rule. The magnetic flux through the circuit is $\Phi = \sum\vect B\cdot\vect n\,\dd S$ over any surface bounded by $\Gamma$ (weber, $1\,\mathrm{Wb} = 1\,\mathrm{T}\,\mathrm{m}^{2}$ ) — “any”, because the flux of $\vect B$ through a closed surface is zero. For a uniform field and a plane coil of $N$ turns and area $S$ , $\Phi = NBS\cos\alpha$ , $\alpha$ the angle between $\vect B$ and $\vect n$ .

Theorem 29.2 (Faraday’s law of induction)

Whenever the flux through a circuit varies, the circuit is the seat of an induced electromotive force

e = -\frac{\dd\Phi}{\dd t} ,

counted in the orientation of the circuit, i.e. acting as an ideal voltage source of emf $e$ in that sense; in a circuit of resistance $R$ it drives the current $i = e/R$ . The minus sign is Lenz’s law: the induced current flows so as to oppose the change of flux that produces it — its own field, and the forces it feels, fight the cause.

Proof. Admitted at this level. ∎

Remark 29.3 (Two faces of the same law)

The flux changes either because $\vect B$ varies in time through a fixed circuit (Neumann induction: transformer, induction cooktop) or because the circuit moves or deforms in a steady field (Lorentz induction: generator, loudspeaker). In the second case the emf can also be computed directly: the carriers of a conductor element $\dd\vect l$ moving at $\vect v$ feel the force $q\vect v\wedge \vect B$ , equivalent to an electric field $\vect v\wedge\vect B$ , so that $e = \oint(\vect v\wedge\vect B)\cdot\dd\vect l$ — and this always agrees with $-\dd\Phi/\dd t$ . Faraday’s law covers both; the Year 2 volume explains why the two mechanisms give one formula.

Orientation and Lenz’s law. Left: the circuit’s orientation fixes n and the sign of . Middle: a magnet approaching a ring increases the flux; the induced current creates a field opposing the magnet’s and the ring repels it. Right: a bar sliding on rails in a field pointing toward the reader; the flux of the circuit grows, the induced current flows so that the Laplace force on the bar opposes its motion. — Orientation and Lenz’s law. Left: the circuit’s orientation fixes $\vect n$ and the sign of $\Phi$ . Middle: a magnet approaching a ring increases the flux; the induced current creates a field opposing the magnet’s and the ring repels it. Right: a bar sliding on rails in a field pointing toward the reader; the flux of the circuit grows, the induced current flows so that the Laplace force on the bar opposes its motion.

Example 29.4 (Orders of magnitude)

A coil of $100$ turns and $10\,\mathrm{cm}^{2}$ in the Earth’s field, flipped in $0.1\,\mathrm{s}$ : $\Delta\Phi = 2 \times 100 \times 10^{-3} \times 5 \times 10^{-5} = 1 \times 10^{-5}\,\mathrm{Wb}$ , $e \sim 0.1\,\mathrm{mV}$ — measurable with a sensitive galvanometer, the way the Earth’s field was once mapped. The same coil in the $1\,\mathrm{T}$ gap of a magnet, pulled out in $0.1\,\mathrm{s}$ : $1\,\mathrm{V}$ . A transformer coil of $1000$ turns with $1\,\mathrm{T}$ at $50\,\mathrm{Hz}$ over $10\,\mathrm{cm}^{2}$ : $e_{\max} = 1000 \times 10^{-3} \times 314 = 314\,\mathrm{V}$ — the mains.

29.2 Self and mutual inductance

Definition 29.5 (Inductance)

The flux of a circuit’s own field through itself is proportional to its current: $\Phi_{\text{own}} = Li$ , with $L > 0$ its self-inductance (henry, $\mathrm{H}$ ). The flux that circuit 1 sends through circuit 2 is $\Phi_{1 \to 2} = Mi_1$ , and symmetrically $\Phi_{2 \to 1} = Mi_2$ , with $M$ their mutual inductance (sign fixed by the orientations). The total flux of circuit 1 is then $\Phi_1 = L_1i_1 + Mi_2$ .

Proposition 29.6 (The inductor; the solenoid)

A circuit of inductance $L$ carrying a varying current is the seat of the emf $e = -L\,\dd i/\dd t$ : in the receiver convention this is the voltage $u = L\,\dd i/\dd t$ of the inductor used since Chapter 7. It stores the energy $W = \tfrac12Li^2$ . A long solenoid of $n$ turns per metre, length $\ell$ , section $S$ , has $L = \mu_0n^2S\ell$ ; its stored energy is $\tfrac12\mu_0n^2I^2S\ell = \dfrac{B^2}{2\mu_0}\,S\ell$ : magnetic energy lives in the field at the density $B^2/2\mu_0$ , exactly as electric energy at $\varepsilon_0E^2/2$ .

Proof. Faraday with $\Phi = Li$ , $L$ constant. Energy: the power received by the inductor is $ui = Li\,\dd i/\dd t = \dd(\tfrac12Li^2)/\dd t$ . Solenoid: $B = \mu_0nI$ inside, flux $BS$ through each of the $n\ell$ turns, so $\Phi = \mu_0n^2S\ell I$ . The general density is admitted. ∎

Example 29.7 (An MRI magnet)

$1.5\,\mathrm{T}$ over about a cubic metre: $B^2/2\mu_0 = 2.25/2.5 \times 10^{-6} = 0.9\,\mathrm{MJ}/\mathrm{m}^{3}$ — the energy of a car battery, stored in empty space. With a coil current of $500\,\mathrm{A}$ , $L = 2W/I^2 = 7\,\mathrm{H}$ . If the superconductor suddenly turns resistive (a quench), that megajoule becomes heat in seconds and boils off hundreds of litres of liquid helium: the magnet room has a vent for it.

Proposition 29.8 (The ideal transformer)

Two coils of $N_1$ and $N_2$ turns wound on the same closed iron core, so that the same flux $\varphi$ threads every turn, obey

\frac{u_2}{u_1} = \frac{N_2}{N_1} , \qquad N_1i_1 + N_2i_2 = 0 \quad\text{(no load losses, no magnetizing current)},

so that $u_2i_2 = -u_1i_1$ : the power taken at the primary is delivered at the secondary; voltages step with the turns ratio, currents inversely. It works only with varying currents.

Proof. $u_1 = N_1\,\dd\varphi/\dd t$ and $u_2 = N_2\,\dd\varphi/\dd t$ (receiver convention on both, orientations chosen along the common flux): divide. The current relation is Ampère’s law on the core: its circulation $\mu_0\mu_r$ -times the total ampere-turns would have to create the flux; for an ideal core ( $\mu_r \to \infty$ ) that takes vanishing ampere-turns, hence $N_1i_1 + N_2i_2 = 0$ . Admitted in this form. ∎

Left: a transformer — two coils on a closed iron core share the same flux; the voltages scale with the turns, the currents inversely. Right: two coupled coils in air and their mutual inductance

M

; a coil’s own flux is

Li

29.3 Moving conductors: generators and motors

Proposition 29.9 (The sliding bar)

A bar of length $\ell$ slides at speed $v$ on two rails closed by a resistance $R$ , in a uniform field $B$ normal to the plane. It is the seat of the emf $e = B\ell v$ , drives $i = B\ell v/R$ , and feels the Laplace force $F = B\ell i = B^2\ell^2v/R$ opposing its motion. The mechanical power spent to push it, $Fv$ , equals the electrical power $ei = Ri^2$ dissipated: the bar is a generator. Fed instead by a source $E$ , it is a motor: the current $(E - B\ell v)/R$ pushes it until the back-emf $B\ell v$ balances $E$ , at the no-load speed $E/B\ell$ .

Proof. Flux $B\ell x$ , $e = -B\ell\dot x$ in the orientation for which $\vect n$ is along $\vect B$ ; the Laplace force $i\vect\ell\wedge\vect B$ on the current so oriented points against $\vect v$ (Theorem 29.2, Lenz). Multiply the electrical equation $e = Ri$ by $i$ and the mechanical one by $v$ : $Fv = B\ell iv = ei$ . The electromechanical power $ei$ and the Laplace power $-Fv$ are always opposite — the law of the conversion. ∎

Proposition 29.10 (The rotating loop: alternator)

A coil of $N$ turns and area $S$ rotating at $\omega$ about an axis perpendicular to a uniform $B$ has $\Phi = NBS\cos\omega t$ and

e = NBS\omega\sin\omega t :

a sinusoidal emf of amplitude $NBS\omega$ and the frequency of rotation. Into a resistance $R$ it delivers the mean power $(NBS\omega)^2/2R$ , supplied by the torque that keeps it turning against the Laplace couple $-\vect m\wedge\vect B$ of its own current.

Proof. Differentiate the flux; $\langle\sin^2\rangle = \tfrac12$ . The couple is Theorem 28.14 with $m = Ni S$ ; its mean power $\langle\Gamma\omega\rangle$ equals $\langle ei\rangle$ by the same computation as for the bar. ∎

Left: the sliding bar as a generator — emf B v, current through R, Laplace force braking the bar; the pusher’s power is the resistor’s heat. Middle and right: the rotating coil of an alternator and its sinusoidal emf. — Left: the sliding bar as a generator — emf $B\ell v$ , current through $R$ , Laplace force braking the bar; the pusher’s power is the resistor’s heat. Middle and right: the rotating coil of an alternator and its sinusoidal emf.

Example 29.11 (Eddy currents)

A bulk conductor moving in a non-uniform field, or sitting in a varying one, is threaded by changing flux along countless closed paths: eddy currents circulate in it, heating it and — by Lenz — braking it. A magnet dropped down a copper tube drifts down slowly; the brakes of trucks and roller coasters use it (no contact, no wear, no brake failure — but no braking at rest); the induction cooktop heats the pan by the same currents, at $25\,\mathrm{kHz}$ ; transformer cores are laminated precisely to cut them.

29.4 The loudspeaker: a coupled system

Proposition 29.12 (Electromechanical coupling of a coil in a radial field)

A coil of total wire length $\ell$ sits in a radial field $B$ so that every element of wire is perpendicular to $\vect B$ ; it moves along its axis at speed $v$ and carries $i$ . Then it feels the axial Laplace force $F = B\ell i$ and is the seat of the back-emf $e = -B\ell v$ (receiver convention: a voltage $B\ell v$ ). With a mass $m$ , a suspension of stiffness $k$ and a mechanical damping $h$ , fed by a voltage $u$ through the coil’s resistance $R$ and inductance $L$ :

u = Ri + L\frac{\dd i}{\dd t} + B\ell v , \qquad m\frac{\dd v}{\dd t} = -kx - hv + B\ell i .

Multiplying by $i$ and by $v$ and adding: the electrical power $ui$ equals the Joule loss $Ri^2$ plus the growth of the stored energies $\tfrac12Li^2 + \tfrac12mv^2 + \tfrac12kx^2$ plus the mechanical loss $hv^2$ (which includes the sound): the conversion term $B\ell iv$ cancels between the two equations. Run backward (motion imposed, $u$ read) the same device is a dynamic microphone.

Proof. Laplace force on each element $i\,\dd\vect l\wedge\vect B$ , all axial and of the same sign; emf from $(\vect v\wedge\vect B)\cdot\dd\vect l$ summed over the wire. The energy balance is the sum of the two equations multiplied as stated. ∎

A moving-coil loudspeaker in section: the voice coil sits in the annular gap of a magnet where the field is radial, so that a current gives an axial force whatever the position; the cone it drives is held by a stiff, damped suspension; the coil’s motion feeds a back-emf B v into the electrical circuit. — A moving-coil loudspeaker in section: the voice coil sits in the annular gap of a magnet where the field is radial, so that a current gives an axial force whatever the position; the cone it drives is held by a stiff, damped suspension; the coil’s motion feeds a back-emf $B\ell v$ into the electrical circuit.

Remark 29.13 (Why the coupling damps)

At low frequency ( $L\omega \ll R$ ), $i \approx (u - B\ell v)/R$ and the force is $B\ell u/R - (B^2\ell^2/R)\,v$ : the electrical circuit adds the damping $B^2\ell^2/R$ to the mechanical one. A loudspeaker connected to an amplifier (low output resistance) is heavily damped — its cone stops when the signal stops — and the same coupling is what lets the microphone take energy from the sound (Problem 29.1).

29.5 Exercises

Exercise 29.1 ★

A coil of $100$ turns and $10\,\mathrm{cm}^{2}$ in a uniform $0.20\,\mathrm{T}$ field at $30{}^{\circ}$ to its normal. Flux; mean emf if the field is switched off in $10\,\mathrm{ms}$ ; in $1\,\mathrm{ms}$ ; sense of the induced current relative to the field (Lenz).

Solution

Solution of Exercise 29.1.

$\Phi = NBS\cos30^\circ = 100 \times 0.2 \times 10^{-3} \times 0.866 = 1.7 \times 10^{-2}\,\mathrm{Wb}$ ; $\langle e\rangle = \Delta\Phi/\Delta t = 1.7\,\mathrm{V}$ in $10\,\mathrm{ms}$ , $17\,\mathrm{V}$ in $1\,\mathrm{ms}$ . The current flows so as to recreate the vanishing flux: its field is along the old $\vect B$ .

Exercise 29.2 ★

A bar $20\,\mathrm{cm}$ long slides at $2.0\,\mathrm{m}/\mathrm{s}$ on rails closed by $0.10\,\Omega$ , in $0.50\,\mathrm{T}$ . Emf, current, Laplace force, force the pusher must apply, mechanical power, electrical power.

Solution

Solution of Exercise 29.2.

$e = B\ell v = 0.5 \times 0.2 \times 2 = 0.20\,\mathrm{V}$ ; $i = 2.0\,\mathrm{A}$ ; $F = B\ell i = 0.20\,\mathrm{N}$ braking, so the pusher applies $0.20\,\mathrm{N}$ ; $Fv = 0.40\,\mathrm{W} = Ri^2$ .

Exercise 29.3 ★

Solenoid of $1000\,\mathrm{turns}/\mathrm{m}$ , length $20\,\mathrm{cm}$ , section $4.0\,\mathrm{cm}^{2}$ : inductance; energy at $5.0\,\mathrm{A}$ ; time constant with a $1.0\,\Omega$ resistance; field inside and energy density, to check $B^2/2\mu_0$ .

Solution

Solution of Exercise 29.3.

$L = \mu_0n^2S\ell = 4\pi \times 10^{-7} \times 10^6 \times 4 \times 10^{-4} \times 0.2 = 0.10\,\mathrm{mH}$ ; $W = \tfrac12LI^2 = 1.3\,\mathrm{mJ}$ ; $\tau = L/R = 0.1\,\mathrm{ms}$ . $B = \mu_0nI = 6.3\,\mathrm{mT}$ , $B^2/2\mu_0 = 15.7\,\mathrm{J}/\mathrm{m}^{3}$ over $S\ell = 8 \times 10^{-5}\,\mathrm{m}^{3}$ : $1.3\,\mathrm{mJ}$ .

Exercise 29.4 ★

A transformer from $230\,\mathrm{V}$ to $12\,\mathrm{V}$ : turns ratio; primary turns for a $50$ -turn secondary; primary current when the secondary delivers $2.0\,\mathrm{A}$ . Why does it not work on a car battery?

Solution

Solution of Exercise 29.4.

$N_2/N_1 = 12/230 = 0.052$ ; $N_1 = 50/0.052 = 960$ turns; $i_1 = 2 \times 0.052 = 0.10\,\mathrm{A}$ . A steady current makes a steady flux: no $\dd\varphi/\dd t$ , no secondary voltage — and the primary, a short, would burn.

Exercise 29.5 ★★

Lenz’s law: give the sense of the induced current and the direction of the force in each case — (a) a north pole approaches a ring along its axis; (b) a ring lies in a field that grows; (c) a ring is dropped onto a magnet; (d) a magnet falls through a long copper tube. For (d), explain with energy why it falls at a small constant speed.

Solution

Solution of Exercise 29.5.

(a) The flux (along the approach) grows: the current makes a field opposing it — its own north pole faces the magnet’s north — and the ring is pushed away. (b) Current whose field opposes the growth; the ring, a dipole anti-aligned with $\vect B$ , is pushed toward weaker field. (c) As it falls the flux grows: same current as (a), the ring is braked. (d) Each section of tube ahead of the magnet sees growing flux, behind it decreasing; both induced currents brake the magnet; at the speed where the Laplace force balances the weight, the magnet falls steadily and its lost potential energy becomes Joule heat in the tube.

Exercise 29.6 ★★

Alternator: $N = 200$ , $S = 20\,\mathrm{cm}^{2}$ , $B = 0.10\,\mathrm{T}$ , $50\,\mathrm{Hz}$ . Peak and rms emf; peak current and peak torque into $10\,\Omega$ ; mean electrical power; compare with the mean mechanical power, and with the instantaneous torque at the moment the emf vanishes.

Solution

Solution of Exercise 29.6.

$E_{\max} = NBS\omega = 200 \times 2 \times 10^{-3} \times 0.1 \times 314 = 12.6\,\mathrm{V}$ , rms $8.9\,\mathrm{V}$ ; $I_{\max} = 1.26\,\mathrm{A}$ , $\Gamma_{\max} = NBSI_{\max} = 0.050\,\mathrm{N}\,\mathrm{m}$ (when $\sin\omega t = 1$ ); $\langle P\rangle = E_{\max}^2/2R = 7.9\,\mathrm{W}$ $= \langle\Gamma\omega\rangle = \tfrac12 \times 0.05 \times 314$ ; when $e = 0$ the current is zero and the torque vanishes — the torque pulses at $100\,\mathrm{Hz}$ .

Exercise 29.7 ★★

Magnetic brake. A square loop of side $a = 10\,\mathrm{cm}$ and resistance $R = 1.0\,\mathrm{m}\Omega$ , mass $1.0\,\mathrm{kg}$ , leaves a region of uniform field $B = 1.0\,\mathrm{T}$ at speed $v$ (one side still in the field). Emf, current, braking force; equation of motion and time constant; distance travelled before stopping from $1.0\,\mathrm{m}/\mathrm{s}$ ; where did the kinetic energy go?

Solution

Solution of Exercise 29.7.

$e = Bav$ , $i = Bav/R$ , $F = B^2a^2v/R = 10\,v$ (N). $m\,\dd v/\dd t = -B^2a^2v/R$ : $v = v_0\eu^{-t/\tau}$ , $\tau = mR/B^2a^2 = 0.10\,\mathrm{s}$ ; distance $v_0\tau = 10\,\mathrm{cm}$ (if the field region is long enough). The kinetic energy, $0.5\,\mathrm{J}$ , is dissipated as $Ri^2$ in the loop.

Exercise 29.8 ★★

A small coil ( $N_2 = 50$ turns, section $2.0\,\mathrm{cm}^{2}$ ) sits inside a long solenoid ( $n_1 = 2000\,\mathrm{m}^{-1}$ ), axes aligned. Mutual inductance; emf in the coil when the solenoid current ramps at $100\,\mathrm{A}/\mathrm{s}$ , and when it is $1.0\,\mathrm{A}$ at $10\,\mathrm{kHz}$ . Does $M$ depend on which coil carries the current?

Solution

Solution of Exercise 29.8.

Flux of the solenoid’s field $\mu_0n_1i_1$ through the $N_2$ turns: $M = \mu_0n_1N_2S_2 = 4\pi \times 10^{-7} \times 2000 \times 50 \times 2 \times 10^{-4} = 25\,\text{µ}\mathrm{H}$ . Ramp: $e = M\,\dd i_1/\dd t = 2.5\,\mathrm{mV}$ ; at $10\,\mathrm{kHz}$ , $1\,\mathrm{A}$ : $e_{\max} = M\omega I = 25 \times 10^{-6} \times 6.3 \times 10^4 = 1.6\,\mathrm{V}$ . $M$ is symmetric: the same number gives the flux the small coil sends through the solenoid — far harder to compute directly.

Exercise 29.9 ★★

Energy in a field: $B = 1.5\,\mathrm{T}$ over $1.0\,\mathrm{m}^{3}$ , as in an MRI magnet. Energy; inductance for $I = 500\,\mathrm{A}$ ; height to which that energy would lift a car of $1\,\mathrm{t}$ ; volume of liquid helium it boils in a quench ( $2.6\,\mathrm{kJ}$ per litre). Energy of the Earth’s field ( $5 \times 10^{-5}\,\mathrm{T}$ ) in a $50\,\mathrm{m}^{3}$ room.

Solution

Solution of Exercise 29.9.

$W = B^2V/2\mu_0 = 2.25/(2.51 \times 10^{-6}) = 0.90\,\mathrm{MJ}$ ; $L = 2W/I^2 = 7.2\,\mathrm{H}$ ; $h = W/mg = 92\,\mathrm{m}$ ; helium: $9 \times 10^5/2600 = 350\,\mathrm{L}$ . Earth’s field: $(5 \times 10^{-5})^2/2\mu_0 = 1 \times 10^{-3}\,\mathrm{J}/\mathrm{m}^{3}$ , $50\,\mathrm{mJ}$ in the room.

Exercise 29.10 ★★★

Rails as a motor. The bar of Exercise 29.2 (mass $m$ , no friction) is connected to a source $E$ through the resistance $R$ . (a) Electrical and mechanical equations; show $m\,\dd v/\dd t = B\ell(E - B\ell v)/R$ . (b) Terminal speed and time constant. (c) Power balance: source power $=$ Joule $+$ mechanical; efficiency as a function of $v$ . (d) Numbers: $E = 2.0\,\mathrm{V}$ , $m = 50\,\mathrm{g}$ , the rest as in Exercise 29.2. (e) A rail gun accelerates a $10\,\mathrm{g}$ projectile with $1\,\mathrm{MA}$ over $5\,\mathrm{m}$ in $1\,\mathrm{T}$ : exit speed (order of magnitude).

Solution

Solution of Exercise 29.10.

(a) $E - B\ell v = Ri$ (the back-emf opposes the source), $m\,\dd v/\dd t = B\ell i$ : $m\,\dd v/\dd t = B\ell(E - B\ell v)/R$ . (b) $v_\infty = E/B\ell$ , $\tau = mR/B^2\ell^2$ . (c) Multiply the electrical equation by $i$ : $Ei = Ri^2 + B\ell vi$ , and $B\ell vi = Fv$ is the mechanical power; efficiency $Fv/Ei = B\ell v/E = v/v_\infty$ — $100\%$ only at no load, where nothing is delivered. (d) $v_\infty = 2/(0.5 \times 0.2) = 20\,\mathrm{m}/\mathrm{s}$ , $\tau = 0.05 \times 0.1/0.01 = 0.5\,\mathrm{s}$ . (e) $F = B\ell I = 1 \times 0.1 \times 10^6 = 1 \times 10^{5}\,\mathrm{N}$ (taking $\ell = 10\,\mathrm{cm}$ ), $a = 1 \times 10^{7}\,\mathrm{m}/\mathrm{s}^{2}$ , $v = \sqrt{2a \times 5} = 1 \times 10^{4}\,\mathrm{m}/\mathrm{s}$ — ten kilometres per second, the promise and the problem of rail guns (the rails erode).

Exercise 29.11 ★★★

Faraday’s disk. A copper disk of radius $R$ rotates at $\omega$ about its axis in a uniform $B$ along the axis; brushes touch the centre and the rim. (a) Emf between centre and rim by summing $(\vect v\wedge \vect B)\cdot\dd\vect l$ along a radius: $e = \tfrac12B\omega R^2$ . (b) Numbers: $R = 10\,\mathrm{cm}$ , $3000\,\mathrm{rpm}$ , $1.0\,\mathrm{T}$ . (c) Why is it a low-voltage, high-current machine; current into $1\,\mathrm{m}\Omega$ , and the braking torque. (d) Recover $e$ from Faraday’s law applied to the circuit brush–radius–rim–brush: which flux varies?

Solution

Solution of Exercise 29.11.

(a) At radius $r$ the metal moves at $v = \omega r$ ; $\vect v\wedge\vect B$ is radial, of magnitude $\omega rB$ : $e = \int_0^R\omega Br\,\dd r = \tfrac12B\omega R^2$ . (b) $\omega = 314$ : $e = 0.5 \times 1 \times 314 \times 0.01 = 1.6\,\mathrm{V}$ . (c) One “turn” only, so volts at most; into $1\,\mathrm{m}\Omega$ : $1.6\,\mathrm{kA}$ , torque $= P/\omega = ei/\omega = 2500/314 = 8\,\mathrm{N}\,\mathrm{m}$ . (d) The circuit brush–radius– rim–brush sweeps out area $\tfrac12R^2\omega\,\dd t$ per $\dd t$ : $\dd\Phi/\dd t = \tfrac12BR^2\omega$ — the flux through the moving, deforming circuit.

Exercise 29.12 ★★★

Induction cooktop. The coil under the glass carries $i_1 = I_1\cos\omega t$ with $I_1 = 30\,\mathrm{A}$ at $25\,\mathrm{kHz}$ ; the pan bottom acts as a single turn of resistance $R_p = 10\,\mathrm{m}\Omega$ coupled by $M = 1.0\,\text{µ}\mathrm{H}$ (inductance of the pan turn neglected). (a) Emf induced in the pan, peak value. (b) Current and mean power dissipated in the pan. (c) Why a high frequency, and why does the heat appear in the pan and not in the glass? (d) The pan’s current reacts on the coil: emf it induces there (peak), and the extra power the generator must supply — check it equals (b).

Solution

Solution of Exercise 29.12.

(a) $e_2 = -M\,\dd i_1/\dd t$ , peak $M\omega I_1 = 10^{-6} \times 1.57 \times 10^5 \times 30 = 4.7\,\mathrm{V}$ . (b) $I_2 = 4.7/0.01 = 470\,\mathrm{A}$ peak; $\langle P\rangle = R_pI_2^2/2 = 1.1\,\mathrm{kW}$ . (c) The emf is $\propto\omega$ : high frequency gives a useful emf with a modest $M$ ; the glass is an insulator, nothing flows in it — the heat is born inside the metal. (d) The pan current induces in the coil $M\omega I_2 = 10^{-6} \times 1.57 \times 10^5 \times 470 = 74\,\mathrm{V}$ peak, in phase with $i_1$ (the pan current lags $e_2$ by nothing, being resistive, and $e_2$ lags $i_1$ by $90{}^{\circ}$ … so the reflected emf is $-M\,\dd i_2/\dd t$ , in phase opposition to $i_1$ ’s driving voltage): the generator supplies $\tfrac12 \times 74 \times 30 = 1.1\,\mathrm{kW}$ more — exactly what the pan dissipates.

A bottle dynamo: the wheel turns a magnet inside a coil, the flux through the coil changes, and the lamp lights — Faraday’s law on a bicycle.

29.6 Problem: The bicycle dynamo and the loudspeaker

Problem 29.1

Weekend problem — two electromechanical converters taken apart: a dynamo that regulates itself, and a loudspeaker whose amplifier is also its brake

Part I — The bicycle dynamo. A coil of $N = 300$ turns and area $S = 3.0\,\mathrm{cm}^{2}$ turns in a uniform field $B = 0.30\,\mathrm{T}$ ; its axle carries a roller of radius $r = 1.0\,\mathrm{cm}$ pressed against the tyre, so that the coil turns at $\omega = v/r$ for a bicycle speed $v$ . Coil resistance $r_c = 4.0\,\Omega$ , inductance $L = 20\,\mathrm{mH}$ ; the lamp is a resistance $R = 12\,\Omega$ .

Flux through the coil and induced emf as functions of time; amplitude and frequency at $v = 18\,\mathrm{km}/\mathrm{h}$ .
Neglecting $L$ first: rms current and power in the lamp at $18\,\mathrm{km}/\mathrm{h}$ ; is a “ $6\,\mathrm{V}$ , $3\,\mathrm{W}$ ” lamp comfortable?
Mean mechanical power the cyclist supplies to the dynamo, and the corresponding force on the tyre (compare with rolling friction, about $2\,\mathrm{N}$ ).
Explain with Lenz’s law why the dynamo resists, and why a lamp that lights “for free” does not contradict energy conservation.
Now include $L$ : complex impedance of the circuit; show that the current amplitude is $I = NBS\omega/\sqrt{(R + r_c)^2 + L^2\omega^2}$ and that it saturates at high speed; its limit, and the speed at which it reaches $70\%$ of the limit.
Without the inductance, what would the lamp receive at $54\,\mathrm{km}/\mathrm{h}$ downhill? With it?
Current and lamp power at $9\,\mathrm{km}/\mathrm{h}$ and at $36\,\mathrm{km}/\mathrm{h}$ ; comment on the self-regulation.
Real dynamos rotate a magnet inside a fixed coil. Does the analysis change? What practical problem does it remove?
What does the lamp do when the bicycle stops at a red light, and how do modern lights fix it?

Part II — The loudspeaker. Voice coil: wire length $\ell = 5.0\,\mathrm{m}$ in a radial field $B = 1.0\,\mathrm{T}$ , resistance $R = 8.0\,\Omega$ , inductance $L = 0.50\,\mathrm{mH}$ ; moving mass (coil and cone) $m = 10\,\mathrm{g}$ ; suspension stiffness $k = 1000\,\mathrm{N}/\mathrm{m}$ , mechanical damping $h = 1.0\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ . Sinusoidal regime, complex amplitudes.

Force on the coil for a current $i$ ; back-emf for a speed $v$ (derive both, with signs).
Write the electrical and the mechanical equations.
Eliminate $i$ and show that $\underline v = \dfrac{B\ell\,\underline u}{(R + jL\omega)(jm\omega + h + k/j\omega) + B^2\ell^2}$ .
At frequencies where $L\omega \ll R$ , show that the coupling acts as an extra mechanical damping $B^2\ell^2/R$ ; value; resonance frequency of the cone and its quality factor with and without the electrical damping.
The amplifier delivers $i$ of amplitude $1.0\,\mathrm{A}$ at $1\,\mathrm{kHz}$ : force, acceleration, displacement and velocity amplitudes of the cone (mass-controlled regime); back-emf amplitude compared with $Ri$ .
Same current at the resonance frequency: velocity amplitude (damping-controlled) and back-emf; comment.
Electrical power at $1\,\mathrm{kHz}$ ; mechanical power dissipated in $h$ (taken as the sound and suspension losses); efficiency.
Why is the sound pressure proportional to the cone’s acceleration (admit it) the reason a mass-controlled cone gives a flat response above resonance?
Why must the field be radial, and why is a tweeter small?
Sketch the modulus of the impedance $\underline u/\underline i$ seen by the amplifier from $10\,\mathrm{Hz}$ to $10\,\mathrm{kHz}$ : value at low frequency, at resonance (use the damping-controlled velocity), and at $10\,\mathrm{kHz}$ .

Part III — Backward, and the books.

The same device as a microphone: a sound moves the cone at $v = 1.0\,\mathrm{mm}/\mathrm{s}$ ; open-circuit voltage.
Loaded by $R_L = 600\,\Omega$ : current, and the braking force on the cone; show that the microphone takes power from the sound through an effective damping $B^2\ell^2/(R + R_L)$ .
The coil is wound on an aluminium cylinder (the former), a closed turn of length $0.10\,\mathrm{m}$ and resistance $1.0\,\mathrm{m}\Omega$ : damping it adds; why formers are slit.
Tap the cone of a loudspeaker whose terminals are left open, then of one whose terminals are shorted: which one rings longer, and why?
Write the energy balance of the loudspeaker over one period at steady state, naming every term.
Sum up: for each device, the law used and the two numbers worth remembering.

Solution

Solution of Problem 29.1.

1. $\Phi = NBS\cos\omega t$ , $e = NBS\omega\sin\omega t$ ; $NBS = 300 \times 0.3 \times 3 \times 10^{-4} = 0.027\,\mathrm{Wb}$ ; $\omega = v/r = 5/0.01 = 500\,\mathrm{rad}/\mathrm{s}$ : amplitude $13.5\,\mathrm{V}$ , frequency $\omega/2\pi = 80\,\mathrm{Hz}$ .

2. $I_{\text{rms}} = 13.5/\sqrt2/16 = 0.60\,\mathrm{A}$ ; lamp $RI^2 = 4.3\,\mathrm{W}$ under $7.2\,\mathrm{V}$ : a $6\,\mathrm{V}$ , $3\,\mathrm{W}$ lamp is over-driven and will not last.

3. Mechanical $=$ electrical: $(R + r_c)I_{\text{rms}}^2 = 16 \times 0.36 = 5.7\,\mathrm{W}$ ; $F = P/v = 5.7/5 = 1.1\,\mathrm{N}$ — half the rolling friction; the cyclist feels it.

4. The induced current’s Laplace couple opposes the rotation (Lenz); the lamp’s energy comes from the cyclist’s legs, $5.7\,\mathrm{W}$ of it.

5. $\underline Z = R + r_c + jL\omega$ ; $I = NBS\omega/\sqrt{(R + r_c)^2 + L^2\omega^2}$ ; for $L\omega \gg R + r_c$ , $I \to NBS/L = 0.027/0.02 = 1.35\,\mathrm{A}$ . $70\%$ : $L\omega = 0.7(R + r_c)/\sqrt{1 - 0.49} = 15.7$ , $\omega = 784\,\mathrm{rad}/\mathrm{s}$ , $v = 7.8\,\mathrm{m}/\mathrm{s} = 28\,\mathrm{km}/\mathrm{h}$ .

6. Without $L$ : $e_{\max} = 3 \times 13.5 = 40.5\,\mathrm{V}$ , $I_{\text{rms}} = 1.8\,\mathrm{A}$ , $39\,\mathrm{W}$ in the lamp — it blows. With $L$ : $I = 40.5/\sqrt{256 + 900} = 1.2\,\mathrm{A}$ peak, $8.5\,\mathrm{W}$ : hot, but ten times less.

7. $9\,\mathrm{km}/\mathrm{h}$ : $e_{\max} = 6.75$ , $|Z| = \sqrt{256 + 25} = 16.8$ , $I = 0.40$ , $P = \tfrac12 \times 0.16 \times 12 = 1.0\,\mathrm{W}$ : dim. $36\,\mathrm{km}/\mathrm{h}$ : $e_{\max} = 27$ , $|Z| = 25.6$ , $I = 1.05$ , $P = 6.6\,\mathrm{W}$ . Four times the speed, $2.6$ times the current: the inductance regulates.

8. Only the relative motion matters: the flux through the fixed coil varies identically, same emf. No sliding contacts (brushes) are needed to bring the current out of a rotating coil.

9. No motion, no flux change, no light — hence the capacitor or small battery (“standlight”) charged while riding.

10. Each element of wire is perpendicular to the radial $\vect B$ : $F = B\ell i$ , axial. Moving at $v$ , each element sees $\vect v\wedge\vect B$ along the wire: $e = -B\ell v$ , opposing the current that produces the motion (receiver convention: a voltage $B\ell v$ ).

11. $u = Ri + L\,\dd i/\dd t + B\ell v$ ; $m\,\dd v/\dd t = -kx - hv + B\ell i$ .

12. $\underline i = (\underline u - B\ell\underline v)/(R + jL\omega)$ ; $\underline v\,(jm\omega + h + k/j\omega) = B\ell\,\underline i$ ; substitute and solve for $\underline v$ .

13. $i \approx (u - B\ell v)/R$ gives $m\dot v + (h + B^2\ell^2/R)v + kx = B\ell u/R$ : extra damping $25/8 = 3.1\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ . $f_0 = \tfrac1{2\pi}\sqrt{k/m} = 50\,\mathrm{Hz}$ ; $Q = \sqrt{km}/h_{\text{tot}}$ : $3.2$ open, $0.77$ with the amplifier.

14. $F = 5.0\,\mathrm{N}$ , $a = F/m = 500\,\mathrm{m}/\mathrm{s}^{2}$ , $x = a/\omega^2 = 13\,\text{µ}\mathrm{m}$ , $v = a/\omega = 0.080\,\mathrm{m}/\mathrm{s}$ ; $B\ell v = 0.40\,\mathrm{V}$ against $Ri = 8\,\mathrm{V}$ : $5\%$ .

15. With the current imposed, $v = F/h = 5\,\mathrm{m}/\mathrm{s}$ (an amplitude of $16\,\mathrm{mm}$ — a big excursion for $1\,\mathrm{A}$ ); $B\ell v = 25\,\mathrm{V}$ , three times $Ri$ : the amplifier must supply $33\,\mathrm{V}$ to push $1\,\mathrm{A}$ at resonance — the impedance peak. A voltage-source amplifier would instead let the electrical damping hold the cone.

16. $\tfrac12RI^2 = 4.0\,\mathrm{W}$ ; $\tfrac12hv^2 = \tfrac12 \times 0.0064 = 3.2\,\mathrm{mW}$ : efficiency $0.08\%$ — loudspeakers are heaters that whisper.

17. Above resonance $a = F/m = B\ell i/m$ is independent of frequency for a given current, so the pressure, $\propto a$ , is flat: the mass-controlled band is the working band.

18. Radial field: the force is axial and the same at every position of the coil in the gap. A tweeter must stay mass-controlled to $20\,\mathrm{kHz}$ with a tiny excursion: light coil and cone; and a small cone radiates high frequencies in all directions.

19. Low frequency: $v$ small, $|Z| \approx R = 8\,\Omega$ ; at $50\,\mathrm{Hz}$ : $u = Ri + B\ell v$ with $v = B\ell i/h$ : $Z = R + B^2\ell^2/h = 8 + 25 = 33\,\Omega$ ; at $10\,\mathrm{kHz}$ : $|R + jL\omega| = |8 + 31j| = 32\,\Omega$ . A peak at resonance, a flat $8\,\Omega$ floor, a rise with $L\omega$ .

20. $e = B\ell v = 5 \times 10^{-3} = 5\,\mathrm{mV}$ .

21. $i = 5 \times 10^{-3}/608 = 8.2\,\text{µ}\mathrm{A}$ ; $F = B\ell i = 41\,\text{µ}\mathrm{N}$ , opposing $v$ : $F = [B^2\ell^2/(R + R_L)]\,v$ , a damping of $0.041\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ ; the power $Fv = 4 \times 10^{-8}\,\mathrm{W}$ is taken from the sound and ends in the two resistors.

22. Former: $B^2\ell_f^2/R_f = 1 \times 0.01/10^{-3} = 10\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ , ten times $h$ : it would deaden the cone and waste power as heat; a slit breaks the closed turn.

23. Open: only $h$ damps, $Q = 3.2$ , it rings. Shorted: the back-emf drives a current through $R$ , damping $B^2\ell^2/R$ is added, $Q = 0.77$ : it stops at once. (Try it.)

24. Over a period, stored energies return to their values: $\langle ui\rangle = \langle Ri^2\rangle + \langle hv^2\rangle$ — electrical input $=$ Joule heat in the coil $+$ mechanical losses (sound radiated and suspension friction); the coupling term $B\ell iv$ appears in both equations with opposite signs and cancels.

25. Dynamo: Faraday on a rotating coil — $13.5\,\mathrm{V}$ at $18\,\mathrm{km}/\mathrm{h}$ , current capped at $1.35\,\mathrm{A}$ by its own inductance. Loudspeaker: Laplace force and back-emf — $3.1\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ of electrical damping, $0.08\%$ efficiency. Microphone: the same coil backward — $5\,\mathrm{mV}$ per millimetre per second.

Terms defined in this chapter

See all 393 terms in the glossary