---
title: "Electromagnetic Induction and Applications"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 29
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/29-electromagnetic-induction-and-applications
---

# Chapter 29 — Electromagnetic 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](#def-b1-induction-flux) drives an [electromotive force](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-sources). 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](#def-b1-induction-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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-induction/img-fb3c35ccfe21.jpg)

*An MRI scanner: a superconducting solenoid holding $1.5\,\mathrm{T}$ over a patient — a megajoule of [magnetic energy](#prop-b1-induction-inductor) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-sources)*

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

counted in the orientation of the circuit, i.e. acting as an ideal [voltage source](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-sources) 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](#thm-b1-induction-faraday)*: 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](#rem-b1-induction-twofaces)*: transformer, induction cooktop) or because the circuit moves or deforms in a steady field (*[Lorentz induction](#rem-b1-induction-twofaces)*: 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](#thm-b1-induction-faraday) 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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-induction/fig-3c625203a550.svg)

*Orientation and [Lenz’s law](#thm-b1-induction-faraday). 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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff) this is the [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) $u = L\,\dd i/\dd t$ of the inductor used since [Chapter 7](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#ch-b1-transient-regimes). 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](#prop-b1-induction-inductor) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff) on both, orientations chosen along the common flux): divide. The current relation is [Ampère’s law](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-ampere) 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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-induction/fig-ddf992c95756.svg)

*Left: a transformer — two coils on a closed iron core share the same flux; the [voltages](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) scale with the turns, the currents inversely. Right: two coupled coils in air and their [mutual inductance](#def-b1-induction-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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) $F = B\ell i = B^2\ell^2v/R$ *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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) $i\vect\ell\wedge\vect B$ on the current so oriented points against $\vect v$ ([Theorem 29.2](#thm-b1-induction-faraday), 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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $NBS\omega$ and the frequency of rotation. Into a resistance $R$ it delivers the mean power $(NBS\omega)^2/2R$, supplied by the [torque](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) 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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-dipoleinfield) 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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-induction/fig-968d3df38440.svg)

![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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-induction/fig-3b5b25027c02.svg)

*Left: the sliding bar as a generator — emf $B\ell v$, current through $R$, [Laplace force](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) braking the bar; the pusher’s power is the [resistor](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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](#ex-b1-induction-eddy)* 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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) $F = B\ell i$ and is the seat of the back-emf $e = -B\ell v$ ([receiver convention](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff): a [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) $B\ell v$). With a mass $m$, a suspension of stiffness $k$ and a mechanical damping $h$, fed by a [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) 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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-induction/fig-9d2d41e9b61d.svg)

*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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) stops — and the same coupling is what lets the microphone take energy from the sound ([Problem 29.1](#pb-b1-induction-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 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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace), force the pusher must apply, mechanical power, electrical power.

**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](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder) with a $1.0\,\Omega$ resistance; field inside and energy density, to check $B^2/2\mu_0$.

**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 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) — and the primary, a short, would burn.

**Exercise 29.5 ★★.**

[Lenz’s law](#thm-b1-induction-faraday): 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 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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) balances the weight, the magnet falls steadily and its lost [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) 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](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) into $10\,\Omega$; mean electrical power; compare with the mean mechanical power, and with the instantaneous [torque](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) at the moment the emf vanishes.

**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](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) vanishes — the [torque](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) 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](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder); distance travelled before stopping from $1.0\,\mathrm{m}/\mathrm{s}$; where did the [kinetic energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-ke) go?

**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](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-ke), $0.5\,\mathrm{J}$, 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](#def-b1-induction-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 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 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](#exo-b1-induction-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](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder). (c) Power balance: source power $=$ Joule $+$ mechanical; [efficiency](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-efficiency) as a function of $v$. (d) Numbers: $E = 2.0\,\mathrm{V}$, $m = 50\,\mathrm{g}$, the rest as in [Exercise 29.2](#exo-b1-induction-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](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#def-b1-units-dimensions-oom)).

**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](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage), high-current machine; current into $1\,\mathrm{m}\Omega$, and the braking [torque](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment). (d) Recover $e$ from [Faraday’s law](#thm-b1-induction-faraday) applied to the circuit brush–radius–rim–brush: which flux varies?

**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](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-moment) $= P/\omega = ei/\omega = 2500/314 = 8\,\mathrm{N}\,\mathrm{m}$. (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 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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.](https://one-course.com/images/onecourse/chapters/physics-3/b1-induction/img-25a69feba0de.jpg)

*A bottle dynamo: the wheel turns a magnet inside a coil, the flux through the coil changes, and the lamp lights — [Faraday’s law](#thm-b1-induction-faraday) 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 $R = 12\,\Omega$.

1. Flux through the coil and induced emf as functions of time; [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) and frequency at $v = 18\,\mathrm{km}/\mathrm{h}$ .
2. 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?
3. Mean mechanical power the cyclist supplies to the dynamo, and the corresponding force on the tyre (compare with rolling friction, about $2\,\mathrm{N}$ ).
4. Explain with [Lenz’s law](#thm-b1-induction-faraday) why the dynamo resists, and why a lamp that lights “for free” does not contradict energy conservation.
5. Now include $L$ : complex [impedance](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-impedance) of the circuit; show that the current [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) 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.
6. Without the inductance, what would the lamp receive at $54\,\mathrm{km}/\mathrm{h}$ downhill? With it?
7. Current and lamp power at $9\,\mathrm{km}/\mathrm{h}$ and at $36\,\mathrm{km}/\mathrm{h}$ ; comment on the self-regulation.
8. Real dynamos rotate a magnet inside a fixed coil. Does the analysis change? What practical problem does it remove?
9. 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](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-complex).

10. Force on the coil for a current $i$ ; back-emf for a speed $v$ (derive both, with signs).
11. Write the electrical and the mechanical equations.
12. Eliminate $i$ and show that $\underline v = \dfrac{B\ell\,\underline u}{(R + jL\omega)(jm\omega + h + k/j\omega) + B^2\ell^2}$ .
13. 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](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-secondorder) with and without the electrical damping.
14. The amplifier delivers $i$ of [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $1.0\,\mathrm{A}$ at $1\,\mathrm{kHz}$ : force, acceleration, displacement and velocity [amplitudes](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) of the cone (mass-controlled regime); back-emf [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) compared with $Ri$ .
15. Same current at the resonance frequency: velocity [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) (damping-controlled) and back-emf; comment.
16. Electrical power at $1\,\mathrm{kHz}$ ; mechanical power dissipated in $h$ (taken as the sound and suspension losses); [efficiency](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-efficiency) .
17. 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?
18. Why must the field be radial, and why is a tweeter small?
19. Sketch the modulus of the [impedance](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-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.**

20. The same device as a microphone: a sound moves the cone at $v = 1.0\,\mathrm{mm}/\mathrm{s}$ ; open-circuit [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-voltage) .
21. 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 $B^2\ell^2/(R + R_L)$ .
22. 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.
23. Tap the cone of a loudspeaker whose terminals are left open, then of one whose terminals are shorted: which one rings longer, and why?
24. Write the energy balance of the loudspeaker over one period at [steady state](https://one-course.com/books/physics/3/en/chapter/7-transient-regimes-first-and-second-order#def-b1-transient-regimes-firstorder) , naming every term.
25. Sum up: for each device, the law used and the two numbers worth remembering.

**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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $13.5\,\mathrm{V}$, 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](https://one-course.com/books/physics/3/en/chapter/18-non-inertial-frames-dynamics-on-earth#def-b1-non-inertial-frames-frames) 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](https://one-course.com/books/physics/3/en/chapter/27-electric-potential-conductors-and-capacitors#def-b1-potential-capacitors-capacitance) 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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#thm-b1-dc-circuits-kirchhoff): a [voltage](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-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](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) 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](https://one-course.com/books/physics/3/en/chapter/8-sinusoidal-steady-state-and-impedance#def-b1-sinusoidal-impedance-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](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-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](https://one-course.com/books/physics/3/en/chapter/6-dc-circuits-kirchhoffs-laws-and-theorems#def-b1-dc-circuits-resistor).

**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](https://one-course.com/books/physics/3/en/chapter/28-magnetostatics-field-forces-and-dipoles#thm-b1-magnetostatics-laplace) and back-emf — $3.1\,\mathrm{N}\,\mathrm{s}/\mathrm{m}$ of electrical damping, $0.08\%$ [efficiency](https://one-course.com/books/physics/3/en/chapter/24-heat-engines-and-machines#def-b1-heat-engines-efficiency). Microphone: the same coil backward — $5\,\mathrm{mV}$ per millimetre per second.