6.8 Maxwell's equations, conservation laws and heat

Gauss's law and Ampère's law are the divergence theorem and Stokes' theorem applied to the electric and magnetic fields, Maxwell's four equations are their local forms plus Faraday's law and one correction forced by d squared equals zero, and the same two theorems turn any conservation law into a continuity equation, giving incompressible flow and the heat equation.

Why does light travel at the speed it does? In 1865 Maxwell answered by writing down a system of equations about two vector fields and showing that they have wave solutions travelling at a speed that depends on two measured constants of electricity and magnetism, and that this speed is the speed of light. Those equations are statements about flux and circulation, and the two theorems of 6.6 translate them between a statement about a whole region and a statement about each point. The same translation produces the continuity equation for any conserved quantity, the condition for an incompressible fluid, and the heat equation. This chapter is the one place in Unit 6 where the integrals are answers to questions about the real world.

Before you start

6.6: the divergence theorem, and the flux of $r/\lvert r\rvert^3$ out of every surface enclosing the origin, which is $4\pi$ . 6.6: Stokes’ theorem. 6.5: flux. 6.7: $d^2 = 0$ , that is $\operatorname{div}\operatorname{curl} = 0$ .

Quick checkWhat is the outward flux of

G = r/\lvert r\rvert^3

through a closed surface that encloses the origin?

Quick checkStokes' theorem turns

\oint_CB\cdot d\ell

into which integral?

Quick checkFor a

C^2

field

B

, what is

\operatorname{div}(\operatorname{curl}B)

The picture: the field that circles a wire

The field of the picture is $F = (-y, x, 0)/(x^2 + y^2)$ . Up to a constant factor, it is the magnetic field of a long straight wire carrying current along the $z$ -axis: it circles the wire, and its size falls like $1/\text{distance}$ . Put the paddle wheel anywhere except on the axis and it does not turn, so the curl is $0$ there. Yet the circulation of the field around any loop that goes once round the wire is the same non-zero number. That number is not a mystery any more: by Stokes’ theorem it is the flux of the curl through a disc, and the curl is zero everywhere except on the wire itself, where all of it is concentrated. The wire carries the current, and the current is the curl. That sentence is Ampère’s law.

The idea, in words

A physical law of the form ‘the total amount of something in a region changes only because of what flows through its boundary’ is an integral statement: it is about a whole region $E$ and its skin $\partial E$ . The divergence theorem converts the skin into the inside, so the statement becomes ‘the integral over $E$ of (rate of change $+$ divergence of the flux) is $0$ ’. If that is true for every region, however small, the integrand must be zero at every point, and we have a differential equation. Likewise a law about the circulation of a field round a loop $C$ and the flux of something through a surface bounded by $C$ is converted by Stokes’ theorem into a statement about the curl at each point. So there is a recipe: integral law $\Rightarrow$ (theorem) $\Rightarrow$ integral of a local quantity over an arbitrary region is zero $\Rightarrow$ local quantity is zero. The last step needs a small lemma, which we prove first.

Lemma B standard; used in Apostol 12.20

A continuous function with zero integral over every region is zero

Let $h$ be continuous on an open set $U\subset\mathbb R^3$ with $\iiint_Bh\,dV = 0$ for every ball $B\subset U$ . Then $h\equiv0$ on $U$ . The same holds with surfaces and double integrals.

In one lineIf $h$ is continuous on an open set and $\iiint_Eh\,dV = 0$ for every ball $E$ inside it, then $h = 0$ everywhere.

Why it is trueIt is the step that turns every 'for all regions' physical law into a differential equation.

Proofstandard

The idea: If h were non-zero at a point, continuity would make it keep the same sign on a small ball, and the integral over that ball could not be zero.

Assume

h(p)>0

Suppose

h(p)\neq0

at some point

p

, say

h(p) > 0

(otherwise replace

h

-h

There is

\delta>0

with

h(q) > h(p)/2

for every

q

in the ball

B

of radius

\delta

about

p

Continuity of

h

p

with tolerance

h(p)/2

\iiint_Bh\,dV\ge\frac{h(p)}2\operatorname{Vol}(B) > 0

, contradicting the hypothesis.

\blacksquare

The integral of a function that is larger than a positive constant is larger than that constant times the volume.

Where this came from· Maxwell, 1865

Michael Faraday’s experiments of the 1830s showed that changing magnetic effects produce electric ones. James Clerk Maxwell built a mathematical theory on them. In his 1865 paper A Dynamical Theory of the Electromagnetic Field he gave a system of equations for the electric and magnetic fields. He had added a term, the displacement current, which he first introduced in his earlier work on mechanical models of the field in 1861 and 1862; in the 1865 paper he used it to derive that the fields obey a wave equation, with a speed close to the measured speed of light, and concluded that light is an electromagnetic wave. Maxwell’s own presentation had twenty equations in components, collected in eight groups. The compact form of four vector equations that textbooks print was found later by Oliver Heaviside, around 1884, in the vector notation that he and Willard Gibbs developed. The language of curl, divergence and the two theorems of 6.6 is what makes four equations out of many.

Gauss’s law: charge and the divergence theorem

Definition B

Electric field

The electric field $E$ is the force per unit charge: a charge $q$ at position $r$ feels the force $qE(r)$ . For a point charge $q$ at the origin, Coulomb’s law gives $E(r) = \dfrac{q}{4\pi\varepsilon_0}\dfrac{r}{\lvert r\rvert^3}$ , with $\varepsilon_0\approx8.854\times10^{-12}$ F/m. Fields of several charges add (superposition).

In one lineThe vector field $E$ such that a small charge $q$ at a point feels the force $qE$ ; for a point charge $q$ at the origin, $E = \dfrac{q}{4\pi\varepsilon_0}\dfrac{r}{\lvert r\rvert^3}$ (Coulomb's law).

Why it existsIt turns 'force between charges' into a field that exists whether or not there is a test charge, and fields are what the integral theorems act on.

Definition B

Charge density

The charge density $\rho$ (charge per unit volume) is defined by $Q_E = \iiint_E\rho\,dV$ for the charge in the region $E$ . (In this chapter $\rho$ means charge density, not the spherical radius of 6.4.)

In one lineThe function $\rho(r)$ such that the charge in a region $E$ is $Q = \iiint_E\rho\,dV$ .

Why it existsA cloud of charge is a function; the total charge is its integral.

Definition B

Gauss's law

Gauss’s law: for every closed surface $S = \partial E$ with outward normal $n$ , $\displaystyle\iint_SE\cdot n\,dS = \frac1{\varepsilon_0}\iiint_E\rho\,dV$ . In differential form, $\nabla\cdot E = \dfrac\rho{\varepsilon_0}$ .

In one lineThe flux of $E$ out of a closed surface $S$ is the enclosed charge divided by $\varepsilon_0$ : $\iint_SE\cdot n\,dS = Q_{enc}/\varepsilon_0$ . Its local form is $\nabla\cdot E = \rho/\varepsilon_0$ .

Why it existsIt says charge is the source of the electric field, and that the field lines of a charge neither start nor end anywhere else.

Theorem B Griffiths 2.2; the computation of 6.6

Gauss's law follows from Coulomb's law

Let $E = \sum_i\dfrac{q_i}{4\pi\varepsilon_0}\dfrac{r - p_i}{\lvert r - p_i\rvert^3}$ .

In one lineFor charges at points $p_i$ with values $q_i$ and a closed surface $S$ not passing through any of them, $\iint_SE\cdot n\,dS = \frac1{\varepsilon_0}\sum_{p_i\text{ inside}}q_i$ .

Why it is trueIt shows the integral law is a consequence of the inverse square law, and it is the reason that the force law has the exponent two: only then is the flux through every surface independent of its shape.

ProofMATH 376 lecture of 4 March and 6.6

The idea: By linearity treat one charge. If the charge is inside, the flux through any enclosing surface equals the flux through a small sphere around it, which is 4 pi times the coefficient. If it is outside, the divergence theorem applies to the field and gives zero.

\iint_SE\cdot n\,dS = \sum_i\iint_SE_i\cdot n\,dS

Flux is linear in the field, so it is enough to take one charge

q

p

p

is outside

S

G

C^1

on the solid inside, so the divergence theorem gives flux

\iiint0\,dV = 0

From 6.6:

G = (r - p)/\lvert r - p\rvert^3

has

\operatorname{div}G = 0

for

r\neq p

p

is inside

S

, the flux of

G

through

S

equals the flux through the small sphere,

\frac1{\varepsilon^2}\cdot4\pi\varepsilon^2 = 4\pi

From 6.6, remove a small ball of radius

\varepsilon

about

p

and apply the divergence theorem to the region between. On the small sphere

G\cdot n = 1/\varepsilon^2

The flux of

E

\frac1{4\pi\varepsilon_0}\cdot4\pi\sum_{inside}q_i = \frac1{\varepsilon_0}\sum_{inside}q_i

\blacksquare

Multiply by the coefficient

\frac q{4\pi\varepsilon_0}

and add over the charges inside.With a charge density

\rho

the sum is the integral

\iiint_E\rho\,dV

Corollary B Apostol 12.20

The differential form of Gauss's law

If $\iint_{\partial B}E\cdot n\,dS = \frac1{\varepsilon_0}\iiint_B\rho\,dV$ for every ball $B$ , then $\nabla\cdot E = \rho/\varepsilon_0$ .

In one lineIf Gauss's law holds for every closed surface, and $E$ is $C^1$ and $\rho$ continuous, then $\nabla\cdot E = \rho/\varepsilon_0$ at every point.

Why it is trueIt is the recipe of the previous section in action: the divergence theorem, then the lemma.

ProofApostol 12.20

The idea: Replace the surface integral with a volume integral by the divergence theorem, move everything to one side, and apply the lemma.

\iint_{\partial B}E\cdot n\,dS = \iiint_B\nabla\cdot E\,dV

Divergence theorem for

E

on the ball (valid since

E

C^1

\iiint_B\big(\nabla\cdot E - \rho/\varepsilon_0\big)\,dV = 0

for every ball

B

Gauss's law in integral form.

\nabla\cdot E - \rho/\varepsilon_0 = 0

everywhere.

\blacksquare

The integrand is continuous, so the lemma of the previous section applies.

Using Gauss’s law: choose a surface where $E\cdot n$ is constant

Gauss’s law is most useful when symmetry makes $E\cdot n$ constant on a surface, so the flux is $E\cdot n$ times the area, which is the shortcut of 6.5.

Worked example Griffiths 2.2; the shortcut of 6.5 and 6.6

A solid sphere of radius

a

carries total charge

Q

spread uniformly. Find

E

inside and outside.

Solution.

Symmetry. The charge distribution looks the same from every direction, so $E$ points radially and depends only on the distance $r$ : $E = E(r)\hat r$ . Rotating the sphere changes nothing, so the field cannot prefer a direction other than the radial one.
Gaussian surface. A sphere of radius $r$ concentric with the charge. On it $E\cdot n = E(r)$ is constant, so the flux is $E(r)\cdot4\pi r^2$ .
Outside ( $r>a$ ). The enclosed charge is $Q$ , so $E(r)\,4\pi r^2 = Q/\varepsilon_0$ , which gives $E(r) = \dfrac{Q}{4\pi\varepsilon_0r^2}$ . Same as the field of a point charge at the centre.
Inside ( $r<a$ ). The enclosed charge is the fraction $\dfrac{r^3}{a^3}$ of the volume, $Q\dfrac{r^3}{a^3}$ . So $E(r)\,4\pi r^2 = \dfrac{Qr^3}{\varepsilon_0a^3}$ and $E(r) = \dfrac{Qr}{4\pi\varepsilon_0a^3}$ . Inside, the field grows linearly with the distance from the centre.
Check with the differential form. The charge density is $\rho = \dfrac{3Q}{4\pi a^3}$ . Inside, $E = \dfrac\rho{3\varepsilon_0}(x, y, z)$ has divergence $\dfrac\rho{3\varepsilon_0}\cdot3 = \dfrac\rho{\varepsilon_0}$ . Outside, $E\propto r/\lvert r\rvert^3$ has divergence $0$ . Both agree with $\nabla\cdot E = \rho/\varepsilon_0$ .

Answer. $E = \dfrac{Q}{4\pi\varepsilon_0r^2}$ for $r\ge a$ and $E = \dfrac{Qr}{4\pi\varepsilon_0a^3}$ for $r\le a$ , the two agreeing at $r = a$ . For $Q = 2$ nC, $a = 0.1$ m and $r = 0.05$ m, with $\frac1{4\pi\varepsilon_0} = 8.988\times10^9$ N m $^2$ /C $^2$ : $E = 8.988\times10^9\cdot2\times10^{-9}\cdot0.05/0.001\approx899$ V/m.

Worked example Griffiths 2.2

An infinite straight line carries charge

\lambda

per unit length. Find

E

at distance

s

. Then do the same for an infinite plane with charge

\sigma

per unit area.

Solution.

Symmetry for the line. By symmetry under rotation about the line and translation along it, $E$ points away from the line and depends only on the distance $s$ .
Gaussian surface. A coaxial cylinder of radius $s$ and length $L$ , closed with two discs. It has to be a closed surface for the law to apply.
Flux. On the two end discs $E\cdot n = 0$ (the field is perpendicular to the axis). On the side wall $E\cdot n = E(s)$ is constant, with area $2\pi sL$ . So the flux is $E(s)\cdot2\pi sL$ .
Gauss. The enclosed charge is $\lambda L$ , so $E(s)\,2\pi sL = \lambda L/\varepsilon_0$ and $E(s) = \dfrac\lambda{2\pi\varepsilon_0s}$ .
The plane. By symmetry $E$ points perpendicular to the plane. Take a pillbox: two discs of area $A$ parallel to the plane on either side, with side walls perpendicular to the plane. The flux is $2EA$ and the enclosed charge is $\sigma A$ , so $2EA = \sigma A/\varepsilon_0$ and $E = \dfrac\sigma{2\varepsilon_0}$ , independent of the distance.

Answer. $E = \dfrac\lambda{2\pi\varepsilon_0s}$ for the line, falling like $1/s$ , and $E = \dfrac\sigma{2\varepsilon_0}$ for the plane, constant. For $\lambda = 1$ nC/m at $s = 1$ m: $E = 2\cdot8.988\times10^9\cdot10^{-9}/1\approx18.0$ V/m.

Your answerA long straight wire carries a uniform charge of

1

nC per metre (

10^{-9}

C/m). What is the electric field, in V/m, at a distance of

1

m? Use

\frac1{4\pi\varepsilon_0} = 8.988\times10^9

N m

^2

^2

Quick checkWhy can you not use Gauss's law with a Gaussian cube to find the field of a point charge at its centre?

Quick checkInside a uniformly charged solid sphere, the electric field

Magnetism: no sources, and Ampère’s law

Definition B

Magnetic field

The magnetic field $B$ is the field for which a charge $q$ with velocity $v$ at that point feels the extra force $qv\times B$ (note the cross product of 6.5). Experiments show no isolated magnetic charges, which gives Gauss’s law for magnetism: $\iint_{\partial E}B\cdot n\,dS = 0$ for every closed surface, and so $\nabla\cdot B = 0$ by the same argument as above. In the language of 6.6, $B$ is a solenoidal field.

In one lineThe vector field $B$ such that a charge $q$ moving with velocity $v$ feels the force $qv\times B$ ; its flux through every closed surface is $0$ .

Why it existsIt describes the force on moving charges, and the fact that it has no sources is the statement that there are no magnetic charges.

Definition B

Current density

The current density $J$ is the charge flow per unit area per unit time, so that the current crossing the oriented surface $S$ is $I = \iint_SJ\cdot n\,dS$ . In a wire of cross-section $A$ carrying a uniform current $I$ along the wire, $\lvert J\rvert = I/A$ .

In one lineThe vector field $J$ such that the current (charge per unit time) through a surface $S$ is $\iint_SJ\cdot n\,dS$ .

Why it existsA current in a wire is a flux of charge, so the current through a surface is the flux of a vector field. It is the source of the magnetic field.

Definition B

Ampere's law

Ampère’s law (the version for steady currents): for a closed loop $C$ bounding the oriented surface $S$ (induced orientation), $\displaystyle\oint_CB\cdot d\ell = \mu_0\iint_SJ\cdot n\,dS = \mu_0I_{enc}$ . By Stokes’ theorem the left side is $\iint_S(\nabla\times B)\cdot n\,dS$ , so for every surface $\nabla\times B = \mu_0J$ . Here $\mu_0\approx4\pi\times10^{-7}$ N/A $^2$ .

In one lineFor the magnetic field of steady currents, the circulation of $B$ round a closed loop $C$ is $\mu_0$ times the current through any surface bounded by $C$ : $\oint_CB\cdot d\ell = \mu_0\iint_SJ\cdot n\,dS$ , locally $\nabla\times B = \mu_0J$ .

Why it existsIt says current is the source of the circulating magnetic field, exactly as Stokes' theorem equates circulation to the flux of a curl.

Worked example MATH 376 worksheet 14 problem 6

A wire has a cross-section the disc of radius

2

and current density

J = (0, 0, 20r)

, where

r

is the distance from the axis. Find the circulation of

B

round the edge of the cross-section, with

\nabla\times B = \mu_0J

Solution.

Surface. $S$ is the cross-sectional disc of radius $2$ with $n = (0, 0, 1)$ ; its edge $C$ has the induced orientation.
Stokes, then Ampère. $\oint_CB\cdot d\ell = \iint_S(\nabla\times B)\cdot n\,dS = \mu_0\iint_SJ\cdot n\,dS$ .
Dot. $J\cdot n = 20r$ .
Integrate. $\int_0^{2\pi}\!\int_0^2\mu_0\,20r\cdot r\,dr\,d\theta = 2\pi\mu_0\cdot20\cdot\frac{8}{3} = \dfrac{320\pi\mu_0}3$ . The extra $r$ is the area element $r\,dr\,d\theta$ .

Answer. $\dfrac{320\pi\mu_0}3$ . The circulation equals $\mu_0$ times the total current. The total current is $\iint J\cdot n\,dS = 2\pi\int_0^220r^2\,dr = \dfrac{320\pi}3$ in the units of $J$ .

Worked example Griffiths 5.3

A long straight wire of radius

a

carries a steady current

I

spread uniformly over its cross-section. Find

B

inside and outside. Then find

B

inside a long solenoid with

n

turns per metre carrying current

I

Solution.

Symmetry for the wire. $B$ circles the axis and depends only on the distance $s$ : $B = B(s)\hat\varphi$ . By symmetry about the axis and along it, $B$ can depend only on $s$ . The direction is the right-hand rule for the current (Biot and Savart’s law gives it). A radial part would put a net magnetic flux through a coaxial cylinder, which $\nabla\cdot B = 0$ forbids.
Amperian loop. A circle of radius $s$ about the axis, with $B\cdot d\ell = B(s)\,ds$ along it. The circulation is $B(s)\cdot2\pi s$ .
Outside ( $s>a$ ). The loop encloses the whole current: $B(s)\,2\pi s = \mu_0I$ , so $B = \dfrac{\mu_0I}{2\pi s}$ .
Inside ( $s<a$ ). The current through the loop’s disc is $I\dfrac{s^2}{a^2}$ , so $B(s)\,2\pi s = \mu_0I\dfrac{s^2}{a^2}$ and $B = \dfrac{\mu_0Is}{2\pi a^2}$ .
Outside is the field of the picture. $B = \dfrac{\mu_0I}{2\pi}\dfrac{(-y, x)}{x^2 + y^2}$ : curl $0$ away from the wire, circulation $\mu_0I$ round any loop enclosing it.
Solenoid. Take a rectangular loop with one long side of length $\ell$ inside the solenoid, the opposite long side far outside where $B\approx0$ , and short sides perpendicular to the axis. The circulation is $B\ell$ and the enclosed current is $nI\ell$ , so $B = \mu_0nI$ .

Answer. $B = \dfrac{\mu_0I}{2\pi s}$ outside, $\dfrac{\mu_0Is}{2\pi a^2}$ inside, and $B = \mu_0nI$ in a long solenoid. For $I = 100$ A and $s = 5$ cm outside the wire: $B = \dfrac{4\pi\times10^{-7}\cdot100}{2\pi\cdot0.05} = 4\times10^{-4}$ T. For $n = 1000$ per metre and $I = 2$ A: $B = 2.5\times10^{-3}$ T.

Maxwell’s correction: $d^2 = 0$ forces a new term

Ampère’s law $\nabla\times B = \mu_0J$ is correct for steady currents. For changing ones it contradicts the conservation of charge. First the conservation law.

Definition B

Continuity equation

For a conserved quantity with density $u$ (amount per unit volume) and flux density $j$ (amount crossing unit area per unit time), the continuity equation is $\dfrac{\partial u}{\partial t} + \nabla\cdot j = 0$ . For electric charge, $u = \rho$ and $j = J$ . For mass of a fluid with velocity $v$ , $u$ is the mass density and $j = u\,v$ .

In one lineIf a conserved quantity has density $u$ and flux density $j$ , then $\partial_tu + \nabla\cdot j = 0$ . For charge: $\partial_t\rho + \nabla\cdot J = 0$ .

Why it existsIt is the local form of 'what is inside changes only by what crosses the boundary', and it is the equation behind every conservation law in continuum physics.

Theorem B Apostol 12.20; Griffiths 5.3

The continuity equation

Let $u$ be the density and $j$ the flux density of a quantity that is neither created nor destroyed, with $u$ and $j$ of class $C^1$ .

In one lineIf the amount of a quantity in every fixed region $E$ changes only through the flux across $\partial E$ , then $\partial_tu + \nabla\cdot j = 0$ .

Why it is trueIt is a conservation law written for every point at once, and it is the model for the derivation of the heat equation and the equation of fluid flow.

ProofGriffiths 5.3

The idea: The amount in a fixed ball changes by minus the flux out of its skin. The divergence theorem turns the flux into a volume integral, and the lemma turns the equality for all balls into the equation at each point.

\dfrac d{dt}\iiint_Bu\,dV = -\iint_{\partial B}j\cdot n\,dS

The amount of the quantity in the fixed ball

B

\iiint_Bu\,dV

\iiint_B\partial_tu\,dV = -\iiint_B\nabla\cdot j\,dV

Differentiate under the integral sign on the left (the ball does not move,

u

C^1

), and use the divergence theorem on the right.

\iiint_B\big(\partial_tu + \nabla\cdot j\big)\,dV = 0

for every ball

B

Move everything to one side.

\partial_tu + \nabla\cdot j = 0

everywhere.

\blacksquare

The integrand is continuous, so the lemma applies.

Proof workshopGive the reason for each line: the continuity equationGriffiths 5.3

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\dfrac d{dt}\displaystyle\iiint_Bu\,dV = -\iint_{\partial B}j\cdot n\,dS$ .
$\displaystyle\iiint_B\partial_tu\,dV = -\iiint_B\nabla\cdot j\,dV$ .
$\displaystyle\iiint_B\big(\partial_tu + \nabla\cdot j\big)\,dV = 0$ for every ball.
$\partial_tu + \nabla\cdot j = 0$ at every point.

Now the problem. Take the divergence of Ampère’s law $\nabla\times B = \mu_0J$ . The left side is $\nabla\cdot(\nabla\times B) = 0$ by $d^2 = 0$ (6.6, 6.7). So $\nabla\cdot J = 0$ . But charge conservation says $\nabla\cdot J = -\partial_t\rho$ . For a charging capacitor, charge piles up on the plates, $\partial_t\rho\neq0$ , and the two statements are inconsistent. Maxwell’s repair uses Gauss’s law, $\rho = \varepsilon_0\nabla\cdot E$ .

Definition B

Displacement current

Maxwell’s correction. Replace $J$ by $J + \varepsilon_0\partial_tE$ in Ampère’s law:

\nabla\times B = \mu_0J + \mu_0\varepsilon_0\frac{\partial E}{\partial t}.

The extra term is the displacement current. Check the divergence: $0 = \nabla\cdot(\nabla\times B) = \mu_0\nabla\cdot J + \mu_0\varepsilon_0\partial_t(\nabla\cdot E) = \mu_0\nabla\cdot J + \mu_0\partial_t\rho = \mu_0(\nabla\cdot J + \partial_t\rho)$ , which is $0$ by the continuity equation. The law is now consistent.

In one lineThe term $\varepsilon_0\,\partial E/\partial t$ that Maxwell added to the current density in Ampère's law, so that $\nabla\times B = \mu_0\big(J + \varepsilon_0\partial_tE\big)$ .

Why it existsIt makes Ampère's law consistent with charge conservation, and it makes a changing electric field a source of magnetic field, which is what lets light propagate.

Proof workshopRebuild the argument that forces Maxwell's displacement currentMaxwell 1865; Griffiths 7.3

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Worked example Griffiths 7.3 (the standard example of the displacement current)

A circular parallel-plate capacitor with plates of radius

R = 5

cm is charged by a steady current

I = 2

A in the wires. Find the magnetic field at distance

r

from the axis between the plates, at

r = R

Solution.

No current between the plates. There $J = 0$ , so Ampère’s law with Maxwell’s term reads $\oint B\cdot d\ell = \mu_0\varepsilon_0\dfrac d{dt}\iint E\cdot n\,dS$ .
The electric field. The charge $Q(t)$ on a plate gives $E = \dfrac{Q}{\varepsilon_0\pi R^2}$ between the plates, so $\dfrac{dE}{dt} = \dfrac{I}{\varepsilon_0\pi R^2}$ (with $I = dQ/dt$ ), numerically $\dfrac{2}{8.854\times10^{-12}\cdot7.854\times10^{-3}}\approx2.9\times10^{13}$ V/(m s).
Amperian loop of radius $r<R$ . The flux of $E$ through the disc is $E\pi r^2$ , so $\mu_0\varepsilon_0\dfrac d{dt}(E\pi r^2) = \mu_0\dfrac{Ir^2}{R^2}$ . The circulation is $B\,2\pi r$ .
Solve. $B(r) = \dfrac{\mu_0Ir}{2\pi R^2}$ for $r\le R$ .
At the edge. $B(R) = \dfrac{\mu_0I}{2\pi R} = \dfrac{4\pi\times10^{-7}\cdot2}{2\pi\cdot0.05} = 8\times10^{-6}$ T.

Answer. $B(R) = 8\ \mu$ T. This is exactly the field the current $I$ in the wire produces at distance $R$ : the displacement current carries on from where the real current stops, and the circulation does not jump.

Your answerIn the charging capacitor of the example (plates of radius

5

cm, current

2

A) what is the magnetic field, in microteslas, at the rim of the plates, between them?

Faraday’s law and the four equations

Definition B

Faraday's law

Faraday’s law of induction: for a fixed loop $C$ bounding the surface $S$ with the induced orientation, $\displaystyle\oint_CE\cdot d\ell = -\frac d{dt}\iint_SB\cdot n\,dS$ . By Stokes’ theorem the left side is $\iint_S(\nabla\times E)\cdot n\,dS$ , and for a fixed surface the right side is $-\iint_S\partial_tB\cdot n\,dS$ ; since this holds for every surface, $\nabla\times E = -\dfrac{\partial B}{\partial t}$ . The circulation is the electromotive force round the loop.

In one lineThe circulation of $E$ round a loop $C$ equals minus the rate of change of the magnetic flux through a surface bounded by $C$ : $\oint_CE\cdot d\ell = -\dfrac d{dt}\iint_SB\cdot n\,dS$ , locally $\nabla\times E = -\partial_tB$ .

Why it existsIt says a changing magnetic field drives an electric field that circulates, which is how generators and transformers work.

Definition B

Maxwell's equations

Maxwell’s equations in vacuum with charges and currents:

Law	Integral form	Differential form	Theorem that converts
Gauss	$\displaystyle\iint_{\partial E}E\cdot n\,dS = \frac{Q_{enc}}{\varepsilon_0}$	$\nabla\cdot E = \dfrac\rho{\varepsilon_0}$	divergence theorem
Gauss for magnetism	$\displaystyle\iint_{\partial E}B\cdot n\,dS = 0$	$\nabla\cdot B = 0$	divergence theorem
Faraday	$\displaystyle\oint_{\partial S}E\cdot d\ell = -\frac d{dt}\iint_SB\cdot n\,dS$	$\nabla\times E = -\dfrac{\partial B}{\partial t}$	Stokes’ theorem
Ampère-Maxwell	$\displaystyle\oint_{\partial S}B\cdot d\ell = \mu_0I_{enc} + \mu_0\varepsilon_0\frac d{dt}\iint_SE\cdot n\,dS$	$\nabla\times B = \mu_0J + \mu_0\varepsilon_0\dfrac{\partial E}{\partial t}$	Stokes’ theorem

The first two rows are about closed surfaces $\partial E$ with outward normal, the last two about loops $\partial S$ with the induced orientation.

In one lineThe four laws: $\nabla\cdot E = \rho/\varepsilon_0$ , $\nabla\cdot B = 0$ , $\nabla\times E = -\partial_tB$ , $\nabla\times B = \mu_0J + \mu_0\varepsilon_0\partial_tE$ , together with the force law on a charge, $q(E + v\times B)$ .

Why it existsFour equations contain all of classical electricity, magnetism and optics. Each has an integral form, which is a statement about a region, and a differential form, which is a statement about a point.

Worked example Griffiths 7.2

A circular loop of wire of radius

0.1

m lies in a uniform magnetic field perpendicular to its plane,

B(t) = 0.5\sin(2\pi\cdot50\,t)

tesla. Find the largest electromotive force round the loop.

Solution.

Flux. The disc has area $\pi\cdot0.1^2 = 0.0314$ m $^2$ , so the flux is $\Phi(t) = 0.0314\cdot0.5\sin(100\pi t)$ weber.
Faraday. The EMF is $\oint E\cdot d\ell = -\dfrac{d\Phi}{dt}$ , of size $0.0314\cdot0.5\cdot100\pi\cos(100\pi t)$ .
Maximum. $0.5\cdot\pi\cdot0.01\cdot100\pi = 0.5\pi^2\approx4.93$ V.

Answer. About $4.9$ V. That is why a coil of $N$ turns in a $50$ Hz field makes $N$ times the voltage: the flux through the surface bounded by the whole coil is $N$ times the flux through one loop.

Your answerA circular loop of radius

0.1

m sits perpendicular to the field

B(t) = 0.5\sin(100\pi t)

tesla. What is the peak EMF round the loop, in volts?

Quick checkWhich theorem converts the integral form of Faraday's law into

\nabla\times E = -\partial_tB

Quick checkWhat does the equation

\nabla\cdot B = 0

say, in words?

Light

Theorem B Maxwell 1865; Griffiths 9.2

Electromagnetic waves

In vacuum, Maxwell’s equations give $\nabla^2E = \mu_0\varepsilon_0\dfrac{\partial^2E}{\partial t^2}$ , the wave equation of 5.9 in three space variables.

In one lineIn empty space ( $\rho = 0$ , $J = 0$ ) each Cartesian component of $E$ satisfies the wave equation $\partial_t^2E = c^2\nabla^2E$ with $c = 1/\sqrt{\mu_0\varepsilon_0}\approx2.998\times10^8$ m/s.

Why it is trueTwo measured constants of electricity and magnetism produce a speed, and the speed is that of light. Maxwell's conclusion was that light is an electromagnetic wave.

ProofGriffiths 9.2

The idea: Take the curl of Faraday's law, swap the order of the derivatives, and use Ampère-Maxwell. The identity curl curl E = grad div E minus Laplacian E, together with div E = 0 in empty space, finishes it.

\nabla\times(\nabla\times E) = -\nabla\times\partial_tB = -\partial_t(\nabla\times B)

Faraday's law, then curl of both sides.

= -\mu_0\varepsilon_0\dfrac{\partial^2E}{\partial t^2}

Ampère-Maxwell in empty space:

\nabla\times B = \mu_0\varepsilon_0\partial_tE

\nabla\times(\nabla\times E) = \nabla(\nabla\cdot E) - \nabla^2E

The identity

\nabla\times(\nabla\times E) = \nabla(\nabla\cdot E) - \nabla^2E

, componentwise.Check of the first component, with

E = (P, Q, R)

: curl curl has first component

\partial_y(Q_x - P_y) - \partial_z(P_z - R_x) = Q_{xy} + R_{xz} - P_{yy} - P_{zz}

; and grad div minus Laplacian has

\partial_x(P_x + Q_y + R_z) - (P_{xx} + P_{yy} + P_{zz}) = Q_{xy} + R_{xz} - P_{yy} - P_{zz}

, the same.

\nabla\times(\nabla\times E) = -\nabla^2E

Gauss's law in empty space:

\nabla\cdot E = 0

\nabla^2E = \mu_0\varepsilon_0\partial_t^2E

, a wave equation with speed squared

c^2 = 1/(\mu_0\varepsilon_0)

\blacksquare

Equate the two expressions for

\nabla\times(\nabla\times E)

. The same steps with the roles of

E

and

B

exchanged give the same equation for

B

Your answerWith

\mu_0 = 4\pi\times10^{-7}

T m/A and

\varepsilon_0 = 8.854\times10^{-12}

F/m, what is

c = 1/\sqrt{\mu_0\varepsilon_0}

in units of

10^8

m/s?

Conservation laws in fluids and heat

Incompressible flow

Definition B

Incompressible flow

For a fluid of mass density $\rho$ and velocity $v$ , the mass flux is $\rho v$ and the continuity equation is $\partial_t\rho + \nabla\cdot(\rho v) = 0$ . If the density is a constant $\rho_0$ , this is $\nabla\cdot v = 0$ : the flow is incompressible. By the divergence theorem the volume flux through every closed surface in the fluid is $0$ : what goes in comes out.

In one lineA flow of constant density $\rho_0$ with velocity field $v$ ; mass conservation then gives $\nabla\cdot v = 0$ , so $v$ is solenoidal and the flux through every closed surface is zero.

Why it existsWater in a pipe is nearly incompressible, and $\nabla\cdot v = 0$ is the condition that makes the speed rise where a pipe narrows.

Worked example new (Apostol 12.20 applications; 6.6 definitions)

Water flows through a pipe of diameter

10

cm at

1.5

m/s and then through a narrowing of diameter

5

cm. What is the speed in the narrow part? Is

v = (x, -y, 0)

a possible incompressible flow, and is it a potential flow?

Solution.

Closed surface. Take the closed surface made of the wide cross-section, the narrow cross-section and the pipe wall between them. The flux through the wall is $0$ (the fluid does not cross the wall). Incompressibility makes the total flux $0$ , so the flux in through the wide end equals the flux out through the narrow end: $A_1v_1 = A_2v_2$ .
Speed. The area is proportional to the diameter squared, so $v_2 = v_1\dfrac{A_1}{A_2} = 1.5\cdot\dfrac{10^2}{5^2} = 6$ m/s.
Divergence of $(x, -y, 0)$ . $\partial_xx + \partial_y(-y) = 1 - 1 = 0$ , so it is incompressible.
Curl. $Q_x - P_y = 0 - 0 = 0$ with $P = x$ , $Q = -y$ , $R = 0$ , and the other components vanish, so it is irrotational, a potential flow with $v = \nabla\dfrac{x^2 - y^2}2$ .
The potential is harmonic. $\nabla^2\dfrac{x^2 - y^2}2 = 1 - 1 = 0$ . For an incompressible potential flow, $0 = \nabla\cdot v = \nabla\cdot\nabla\varphi = \nabla^2\varphi$ . Laplace’s equation.

Answer. $6$ m/s. The field $(x, -y, 0)$ is a possible incompressible irrotational flow (it describes the flow approaching a flat wall along the $y$ axis: it comes in along $y$ and leaves along $x$ ), and its potential is harmonic.

Heat

Definition B

Heat flux

The heat flux $q$ is the energy crossing a surface per unit area per unit time. Fourier’s law says it is proportional to the temperature gradient and opposite to it: $q = -\kappa\nabla T$ with $\kappa>0$ the thermal conductivity (watts per metre per kelvin). The heat flowing out of a closed surface $S$ per unit time is $\iint_Sq\cdot n\,dS = \iiint_E\nabla\cdot q\,dV$ by the divergence theorem.

In one lineThe vector field $q$ such that the heat crossing a surface per unit time is $\iint q\cdot n\,dS$ ; Fourier's law says $q = -\kappa\nabla T$ for a material of thermal conductivity $\kappa$ .

Why it existsHeat flows from hot to cold, down the gradient of the temperature. The minus sign is that statement.

The heat equation of 5.9 is $\dfrac{\partial T}{\partial t} = \alpha\nabla^2T$ . In 5.9 it was stated and its solutions studied. Here we derive it: for the temperature $T(r, t)$ of a homogeneous material of density $\rho$ , specific heat $c$ and conductivity $\kappa$ , without sources, the constant is $\alpha = \dfrac{\kappa}{\rho c}$ , the thermal diffusivity.

Theorem B Ostrogradsky's problem (6.6); Apostol 12.20; Fourier 1822

The heat equation from conservation of energy

Let $T$ be $C^2$ in space and $C^1$ in time, and the material have constant $\rho$ , $c$ , $\kappa$ and no sources.

In one lineEnergy conservation (continuity) plus Fourier's law gives $\rho c\,\partial_tT = \kappa\nabla^2T$ .

Why it is trueIt is the continuity equation of this chapter with the energy density $\rho cT$ and the flux $q = -\kappa\nabla T$ .

ProofFourier's law and 6.8

The idea: Energy density is rho c T and the flux is q. Apply the continuity equation.

The continuity equation with

u = \rho cT

and

j = q

\partial_t(\rho cT) + \nabla\cdot q = 0

The energy per unit volume of a body at temperature

T

\rho cT

(up to a constant).

\rho c\,\partial_tT - \nabla\cdot(\kappa\nabla T) = 0

Fourier's law.

\rho c\,\partial_tT = \kappa\nabla^2T

\blacksquare

The conductivity is constant, and

\nabla\cdot\nabla T = \nabla^2T

(6.6).

Worked example MATH 376 worksheet 14 problem 2; the check is new

The temperature in a solid is

T = x^2 + y^2 + z^2

and the heat flow is

q = -\nabla T

. Compute

\nabla\cdot q

and interpret it. Then check that

T(x, t) = e^{-\alpha t}\sin x

solves the one dimensional heat equation

T_t = \alpha T_{xx}

with diffusivity

\alpha

Solution.

Heat flow. $\nabla T = (2x, 2y, 2z)$ , so $q = (-2x, -2y, -2z)$ .
Divergence. $\nabla\cdot q = -2 - 2 - 2 = -6$ . $\partial_x(-2x) = -2$ and similarly for $y$ and $z$ .
Interpret. Negative divergence: the heat flow gathers; energy accumulates at every point, so the temperature there would rise at the rate $\rho c\,\partial_tT = -\nabla\cdot q = 6$ per unit volume (with $\kappa = 1$ ). Indeed $\kappa\nabla^2T = 6$ . The worksheet calls it a sink for the heat flow: heat flows in toward each point.
The one dimensional solution. $T_t = -\alpha e^{-\alpha t}\sin x$ and $T_{xx} = -e^{-\alpha t}\sin x$ , so $T_t = \alpha T_{xx}$ . A sine profile decays exponentially while keeping its shape: the heat equation smooths out hot and cold spots, the bigger the faster.

Answer. $\nabla\cdot q = -6$ , and $T = e^{-\alpha t}\sin x$ is a solution.

Worked example new (Fourier's law and the divergence theorem)

Heat flows through a brick wall,

5

m by

2.5

m, thickness

0.2

m, conductivity

\kappa = 0.7

W/(m K), with

20\,^\circ

C inside and

0\,^\circ

C outside. How much power does the wall lose? Then a pipe carrying water at

80\,^\circ

C, radius

2

cm, is wrapped in insulation of conductivity

0.04

W/(m K) out to radius

4

cm; the outside of the insulation is

20\,^\circ

C. How much heat does each metre of pipe lose?

Solution.

Steady wall. In steady state $\nabla\cdot q = 0$ , and the temperature depends only on the coordinate $x$ across the wall, so $q = -\kappa T'(x)$ is constant and $T$ is linear. $\nabla\cdot q = q' = 0$ .
Flux. $q = \kappa\dfrac{\Delta T}{L} = 0.7\cdot\dfrac{20}{0.2} = 70$ W/m $^2$ , and the power through $12.5$ m $^2$ is $70\cdot12.5 = 875$ W.
Steady pipe. By symmetry $q$ is radial, $q = q(s)\hat s$ with $q(s) = -\kappa T'(s)$ . The heat out of a cylinder of radius $s$ and length $L$ is $P = q(s)\cdot2\pi sL$ , and by the divergence theorem in the region between two such cylinders ( $\nabla\cdot q = 0$ ) it is the same for every $s$ . The same argument as Gauss’s law for a line charge.
Solve. $-\kappa T'(s) = \dfrac{P}{2\pi sL}$ , so $T(s) = T_1 - \dfrac{P}{2\pi\kappa L}\ln\dfrac s{s_1}$ .
Apply the temperatures. $T(s_1) = 80$ , $T(s_2) = 20$ with $s_2/s_1 = 2$ : $60 = \dfrac{P}{2\pi\kappa L}\ln2$ , so $\dfrac PL = \dfrac{2\pi\kappa\cdot60}{\ln2} = \dfrac{2\pi\cdot0.04\cdot60}{0.693}\approx21.8$ W/m.

Answer. $875$ W through the wall and about $21.8$ W per metre of pipe. Doubling the insulation radius from $4$ to $8$ cm would give $\dfrac{2\pi\cdot0.04\cdot60}{\ln4}\approx10.9$ W/m, half as much, because the logarithm doubles.

Your answerHow many watts per metre does a pipe lose if the water-side surface is at

80\,^\circ

C at radius

2

cm, and the insulation (

\kappa = 0.04

W/(m K)) ends at radius

4

cm at

20\,^\circ

Quick checkIn the heat equation

\partial_tT = \alpha\nabla^2T

, what happens at a point where

T

has a strict local maximum in space?

Python labGauss's law on a cube, for charges inside, outside and in pairs

import numpy as np
# Units with q / epsilon_0 = 1: a point charge q has E = q (r - p) / (4 pi |r - p|^3), and Gauss's law says flux = q for charges inside.
n = 300
t = (np.arange(n) + 0.5) * 2 / n - 1; A, B = np.meshgrid(t, t, indexing="ij"); dA = (2 / n) ** 2

def E(x, y, z, p):                                  # field at (x, y, z) of a unit charge at p
    dx, dy, dz = x - p[0], y - p[1], z - p[2]
    return (dx, dy, dz) / (4 * np.pi * (dx**2 + dy**2 + dz**2) ** 1.5)

def cube_flux(charges):                             # charges = [(q, position), ...]
    total = 0.0
    for axis in range(3):
        for sign in (+1, -1):
            X = [A, B]; X.insert(axis, sign * np.ones_like(A))          # the points of one face
            for q, p in charges:
                total += q * sign * E(*X, p)[axis].sum() * dA           # E . n with n = sign * e_axis
    return round(total, 4) + 0.0

print("charge at the centre            :", cube_flux([(1, (0, 0, 0))]))
print("charge off-centre, inside       :", cube_flux([(1, (0.7, -0.4, 0.5))]))
print("charge outside, at (2, 0, 0)    :", cube_flux([(1, (2, 0, 0))]))
print("+1 and -1 both inside (a dipole):", cube_flux([(1, (0.3, 0, 0)), (-1, (-0.3, 0, 0))]))
print("+1 inside, -1 outside           :", cube_flux([(1, (0, 0, 0)), (-1, (3, 0, 0))]))
print("+2 and +3 inside                :", cube_flux([(2, (0.2, 0.2, 0)), (3, (-0.5, 0, 0.4))]))

Try this. Run it. The program adds up E . n over a 300 by 300 grid on each of the six faces of the cube [-1,1]^3, in units where a unit charge has flux 1. A charge at the centre gives 1.0, a charge placed anywhere else inside also gives 1.0, a charge outside gives 0.0, a +1 and a -1 charge both inside give 0.0, a +1 inside and a -1 outside give 1.0, and charges 2 and 3 inside give 5.0. The flux counts only the charge inside, whatever its position and whatever is outside. Move the outside charge to (1.05, 0, 0), close to the surface, and see that the answer is still near 0 (the numerical error grows when the charge is very near a face). Move the inside charge to (0.95, 0, 0) and look at the accuracy. Replace the exponent 1.5 by 1.4 in the denominator of E to make a field that is not an inverse square law, and see Gauss's law fail.

Where it is used

Used inside maths· Inside maths: Poisson's equation, harmonic functions and the Dirichlet problem

Static fields have $\nabla\times E = 0$ (Faraday with $\partial_tB = 0$ ). On a simply connected region this makes $E = -\nabla V$ for a potential $V$ (6.2), and Gauss’s law becomes Poisson’s equation $\nabla^2V = -\rho/\varepsilon_0$ , and Laplace’s equation $\nabla^2V = 0$ in empty space. The harmonic functions of Laplace’s equation have the mean value property (the value at the centre of a ball is the average over its surface; a proof uses the divergence theorem on the flux of $\nabla V$ ), no interior maximum, and are determined by their values on the boundary: the Dirichlet problem, which is where harmonic analysis and complex analysis meet. The fundamental solution is $V = \dfrac{q}{4\pi\varepsilon_0\lvert r\rvert}$ , the potential of the point charge of this chapter, whose flux $4\pi$ was computed in 6.6. Every step of this theory is Gauss’s flux integral with a different region.

Used in the real world· Used in the real world: wall losses, pipes and coaxial cable

Gauss’s law with cylindrical symmetry is not only about charges. A coaxial cable has an inner conductor of radius $a$ and an outer shield of radius $b$ with an insulator of relative permittivity $\varepsilon_r$ between them. Put charge $\lambda$ per metre on the inner conductor. Gauss’s law on a coaxial cylinder gives $E(s) = \dfrac{\lambda}{2\pi\varepsilon_0\varepsilon_rs}$ in the insulator, so the voltage between the conductors is $V = \int_a^bE\,ds = \dfrac{\lambda}{2\pi\varepsilon_0\varepsilon_r}\ln\dfrac ba$ , and the capacitance per metre is $\dfrac\lambda V = \dfrac{2\pi\varepsilon_0\varepsilon_r}{\ln(b/a)}$ . For $a = 0.45$ mm, $b = 1.47$ mm and polyethylene with $\varepsilon_r = 2.25$ this is $\dfrac{2\pi\cdot8.854\times10^{-12}\cdot2.25}{\ln3.27}\approx106$ pF per metre, the number a cable designer needs to match the cable to its signal source. The same calculation with heat in place of charge gave the insulation loss of a hot water pipe: $21.8$ W per metre for $2$ cm of insulation at conductivity $0.04$ W/(m K), and $875$ W for a $12.5$ m $^2$ brick wall with $20$ K across it. Heat flow, electrostatics and the flow of an incompressible fluid share one mathematical structure: a flux of a vector field with zero divergence, and the same three theorems.

Your answerA coaxial cable has inner radius

0.45

mm, outer radius

1.47

mm and a polyethylene insulator with relative permittivity

2.25

. What is its capacitance per metre in picofarads, using

\varepsilon_0 = 8.854

pF/m?

Traps

Watch out· The theorem needs the right kind of surface

Gauss’s law and the divergence theorem need a closed surface and the outward normal; Ampère’s and Faraday’s laws need a loop bounding an oriented surface with the induced orientation (right-hand rule). If the symmetry is not there, the law is still true but does not determine the field by itself. ‘Choose the surface where $E\cdot n$ is constant’ is a technique, not a law.

Watch out· A singular source inside breaks the hypothesis, not the law

The point charge field $r/\lvert r\rvert^3$ is not $C^1$ at the origin, so the divergence theorem does not apply to a region containing the origin as stated; Gauss’s law for the point charge is obtained by excising a small ball (done in 6.6), and for a distribution of charge it holds because $\rho$ is a source term in $\nabla\cdot E = \rho/\varepsilon_0$ . The same warning as for the angle field in Stokes’ theorem: check where the field is defined before you use a theorem.

Watch out· Same letter, different things

In this chapter $\rho$ is a charge or mass density, in 6.4 and 6.6 $\rho$ was the spherical radius, and $\varphi$ is a potential, an angle, or the azimuthal direction depending on context. $\mu_0$ and $\varepsilon_0$ are constants. Write what each symbol means before you compute. And in 376 notation call the components of a field $P, Q, R$ , not $F_x, F_y, F_z$ , which look like derivatives.

Watch out· Sign conventions: the minus signs carry the physics

The minus in Fourier’s law $q = -\kappa\nabla T$ says heat flows down the gradient. The minus in Faraday’s law $\nabla\times E = -\partial_tB$ says the induced field opposes the change (Lenz’s law). Dropping the first gives a heat equation run backwards, in which heat flows towards hot spots and the fastest wiggles grow fastest. Dropping the second changes the proof of the wave equation: with $\nabla\times E = +\partial_tB$ the same steps give $\nabla^2E = -\mu_0\varepsilon_0\partial_t^2E$ , which has no travelling waves at all. When a sign looks arbitrary, ask which direction the physical quantity moves.

Problems

checknew (Gauss's law)

A closed surface encloses charges $+3$ nC and $-1$ nC, and there is a $+5$ nC charge outside it. What is the flux of $E$ through the surface, in N m $^2$ /C? Use $\varepsilon_0 = 8.854\times10^{-12}$ .

Solution

Only the enclosed charge counts: $Q_{enc} = 3 - 1 = 2$ nC, and the flux is $Q_{enc}/\varepsilon_0 = 2\times10^{-9}/8.854\times10^{-12}\approx226$ N m $^2$ /C.

checknew (the continuity equation)

Check that $\rho = e^{-t}\sin x$ with $J = (-e^{-t}\cos x,\ 0,\ 0)$ satisfies the continuity equation $\partial_t\rho + \nabla\cdot J = 0$ , and that $J = (e^{-t}\cos x,\ 0,\ 0)$ does not.

Solution

$\partial_t\rho = -e^{-t}\sin x$ . For $J = (-e^{-t}\cos x, 0, 0)$ , $\nabla\cdot J = \partial_x(-e^{-t}\cos x) = e^{-t}\sin x$ , and the sum is $0$ : the equation holds. For $J = (e^{-t}\cos x, 0, 0)$ , $\nabla\cdot J = -e^{-t}\sin x$ and the sum is $-2e^{-t}\sin x\neq0$ : the equation fails. In the first case the charge in $[0, \pi]$ , $\int_0^\pi e^{-t}\sin x\,dx = 2e^{-t}$ , decreases at the rate $2e^{-t}$ , and the current $J_x$ at $x = \pi$ is $e^{-t}$ out and at $x = 0$ is $-e^{-t}$ , that is $e^{-t}$ out of the left end as well, a total outflow of $2e^{-t}$ .

checkMATH 376 worksheet 14 problem 2

For $T = x^2 + y^2 + z^2$ , find $\nabla\cdot(-\nabla T)$ and say whether the point $(1, 2, 3)$ is a source or a sink of heat flow.

Solution

$-\nabla T = (-2x, -2y, -2z)$ , with divergence $-6$ everywhere. Negative: a sink (the flow gathers), at every point including $(1, 2, 3)$ .

check

A uniform wire carries $10$ A in a circular cross-section of radius $2$ mm. Find $B$ at $1$ mm from the axis (inside) and at $1$ cm (outside), with $\mu_0 = 4\pi\times10^{-7}$ .

Solution

Inside: $B = \dfrac{\mu_0Is}{2\pi a^2} = \dfrac{4\pi\times10^{-7}\cdot10\cdot10^{-3}}{2\pi\cdot4\times10^{-6}} = 5\times10^{-4}$ T. Outside: $B = \dfrac{\mu_0I}{2\pi s} = \dfrac{4\pi\times10^{-7}\cdot10}{2\pi\cdot0.01} = 2\times10^{-4}$ T.

usenew (Gauss's law with a shell)

A thin spherical shell of radius $a$ carries total charge $Q$ spread uniformly over its surface. Find $E$ inside and outside, and the jump in $E$ across the shell. Express the jump with the surface charge density $\sigma = Q/(4\pi a^2)$ .

Solution

Gaussian spheres of radius $r$ . Inside ( $r<a$ ) the enclosed charge is $0$ , so $E = 0$ . Outside the enclosed charge is $Q$ , so $E = \dfrac{Q}{4\pi\varepsilon_0r^2}$ . The jump at $r = a$ is $\dfrac{Q}{4\pi\varepsilon_0a^2} = \dfrac\sigma{\varepsilon_0}$ , the field immediately outside a charged surface.

usenew (Faraday)

A square loop of side $0.2$ m is pulled out of a region with uniform field $B = 0.8$ T perpendicular to the loop, so that its overlap with the field region shrinks at $0.5$ m/s (one side of length $0.2$ m moves out). Find the EMF.

Solution

The flux is $B\cdot0.2\cdot w(t)$ where $w$ is the width inside the field, with $w' = -0.5$ . $\Phi' = 0.8\cdot0.2\cdot(-0.5) = -0.08$ Wb/s, so the EMF has size $0.08$ V.

usenew (incompressible flow)

Which of these velocity fields are incompressible: $v_1 = (x, y, -2z)$ , $v_2 = (y, -x, 0)$ , $v_3 = (x^2, -2xy, 0)$ ? Which are irrotational?

Solution

$\nabla\cdot v_1 = 1 + 1 - 2 = 0$ ; $\nabla\cdot v_2 = 0 + 0 + 0 = 0$ ; $\nabla\cdot v_3 = 2x - 2x + 0 = 0$ . All three are incompressible. Curls: $v_1$ : $(0, 0, 0)$ , irrotational; $v_2$ : $(0, 0, -1 - 1) = (0, 0, -2)$ , not irrotational (a rotation); $v_3$ : third component $\partial_x(-2xy) - \partial_y(x^2) = -2y\neq0$ , not irrotational.

usenew (heat in a cylinder)

A hot wire of radius $1$ mm carries heat out at $50$ W per metre. It is surrounded by a thick cylindrical shell of material with $\kappa = 0.5$ W/(m K), from radius $1$ mm to $10$ mm. Find the temperature drop across the material.

Solution

$\dfrac PL = \dfrac{2\pi\kappa\,\Delta T}{\ln(s_2/s_1)}$ , so $\Delta T = \dfrac{P}{L}\cdot\dfrac{\ln10}{2\pi\kappa} = 50\cdot\dfrac{2.3026}{3.1416} = 36.6$ K.

provenew (a lemma and the divergence theorem)

Use the divergence theorem and the lemma to prove that a $C^2$ function $V$ with $\nabla^2V = 0$ on a region has the property that the flux of $\nabla V$ through every sphere inside it is zero, and conversely that if the flux of $\nabla V$ through every small sphere is zero then $\nabla^2V = 0$ .

Solution

Forward: for a ball $B$ , $\iint_{\partial B}\nabla V\cdot n\,dS = \iiint_B\nabla\cdot\nabla V\,dV = \iiint_B\nabla^2V\,dV = 0$ . Converse: if the flux is zero for every ball, then $\iiint_B\nabla^2V\,dV = 0$ for every ball, and the lemma ( $\nabla^2V$ is continuous) gives $\nabla^2V = 0$ .

provenew (Poisson's equation)

Assume $\nabla\times E = 0$ on $\mathbb R^3$ and $E$ is $C^1$ . Show that there is a function $V$ with $E = -\nabla V$ , and that Gauss’s law $\nabla\cdot E = \rho/\varepsilon_0$ becomes $\nabla^2V = -\rho/\varepsilon_0$ .

Solution

On $\mathbb R^3$ , which is simply connected (indeed star-shaped), a field with curl zero is a gradient (the Poincaré lemma of 6.7, or the theorem of 6.2): $E = \nabla W$ for some $W$ ; put $V = -W$ . Then $\nabla\cdot E = -\nabla\cdot\nabla V = -\nabla^2V$ , and Gauss’s law says this equals $\rho/\varepsilon_0$ .

provenew (magnetic flux through a closed surface)

Show that if $B = \nabla\times A$ for a $C^2$ field $A$ (a vector potential), then $\nabla\cdot B = 0$ and the flux of $B$ through every closed surface is $0$ . Is the field $G = r/\lvert r\rvert^3$ on $\mathbb R^3\setminus\{0\}$ of this form?

Solution

$\nabla\cdot B = \nabla\cdot(\nabla\times A) = 0$ by $d^2 = 0$ . By the divergence theorem the flux through the boundary of any solid in the domain is $0$ . $G$ has flux $4\pi\neq0$ through the unit sphere, although $\nabla\cdot G = 0$ away from the origin. So $G$ is not a curl on $\mathbb R^3\setminus\{0\}$ : that domain is not star-shaped, and the Poincaré lemma does not apply. This is why a magnetic monopole would require a different law.

stretchnew (the wave equation, plane waves)

Check that $E = (0,\ \cos(kx - \omega t),\ 0)$ satisfies the wave equation $\nabla^2E = \mu_0\varepsilon_0\partial_t^2E$ exactly when $\omega/k = c$ . Find $B$ from Faraday’s law, and verify $\nabla\cdot E = 0$ and $\nabla\cdot B = 0$ .

Solution

$\nabla^2E = (0, -k^2\cos(kx - \omega t), 0)$ and $\partial_t^2E = (0, -\omega^2\cos(kx - \omega t), 0)$ , so $k^2 = \mu_0\varepsilon_0\omega^2$ , i.e. $\omega/k = 1/\sqrt{\mu_0\varepsilon_0} = c$ . For $B$ : $\nabla\times E = (R_y - Q_z,\ P_z - R_x,\ Q_x - P_y) = (0, 0, -k\sin(kx - \omega t))$ since $Q = \cos(kx - \omega t)$ gives $Q_x = -k\sin(kx - \omega t)$ . Faraday: $\partial_tB = -\nabla\times E = (0, 0, k\sin(kx - \omega t))$ , so $B = (0, 0, \frac k\omega\cos(kx - \omega t))$ (up to a constant). $\nabla\cdot E = \partial_y\cos(kx - \omega t) = 0$ and $\nabla\cdot B = \partial_z(\dots) = 0$ . $E$ and $B$ are perpendicular to each other and to the direction $x$ of travel, and $\lvert B\rvert = \lvert E\rvert/c$ .

stretchnew (the heat equation in one dimension, conservation of total heat)

A rod of length $L$ has temperature $T(x, t)$ with $T_t = \alpha T_{xx}$ , and both ends are insulated, $T_x(0, t) = T_x(L, t) = 0$ . Show that the total heat $\int_0^LT\,dx$ is constant in time.

Solution

$\dfrac d{dt}\int_0^LT\,dx = \int_0^LT_t\,dx = \alpha\int_0^LT_{xx}\,dx = \alpha\big[T_x\big]_0^L = 0$ . This is the one dimensional divergence theorem: the flux $-\kappa T_x$ is zero at both ends, so no heat enters or leaves.

Summary

Every conservation law has two forms: an integral law about a region, and a differential equation at a point. Divergence theorem, then the lemma (a continuous function with zero integral over every ball is zero) turn a flux law into a differential equation; Stokes’ theorem does the same for a circulation law.
Gauss’s law $\iint_SE\cdot n\,dS = Q_{enc}/\varepsilon_0$ , locally $\nabla\cdot E = \rho/\varepsilon_0$ . It follows from the inverse square law because the flux of $r/\lvert r\rvert^3$ is $4\pi$ through every surface enclosing the source. With symmetry it determines the field: sphere $\dfrac{Qr}{4\pi\varepsilon_0a^3}$ inside and $\dfrac Q{4\pi\varepsilon_0r^2}$ outside, line $\dfrac\lambda{2\pi\varepsilon_0s}$ , plane $\dfrac\sigma{2\varepsilon_0}$ .
Ampère’s law $\oint B\cdot d\ell = \mu_0I_{enc}$ , locally $\nabla\times B = \mu_0J$ ; a straight wire gives $B = \dfrac{\mu_0I}{2\pi s}$ outside, a circulating field with curl zero away from the wire. $\nabla\cdot B = 0$ : no magnetic charges. Faraday: $\nabla\times E = -\partial_tB$ .
Maxwell’s correction. $\nabla\cdot(\nabla\times B) = 0$ forces $\nabla\cdot J = 0$ , which contradicts charge conservation $\partial_t\rho + \nabla\cdot J = 0$ for a charging capacitor; adding $\mu_0\varepsilon_0\partial_tE$ repairs it using Gauss’s law. The four equations then give waves $\nabla^2E = \mu_0\varepsilon_0\partial_t^2E$ at the speed of light $c = 1/\sqrt{\mu_0\varepsilon_0}$ .
The continuity equation $\partial_tu + \nabla\cdot j = 0$ is the divergence theorem on a fixed region. Incompressible flow is $\nabla\cdot v = 0$ (flux conserved: narrower pipe, faster flow), and for a potential flow the potential is harmonic.
Heat: Fourier’s law $q = -\kappa\nabla T$ plus continuity gives the heat equation $\rho c\,\partial_tT = \kappa\nabla^2T$ . Steady heat through a wall is linear, through a pipe it is logarithmic: the same calculation as Gauss’s law for a line charge.
From 376: the circulation of $B$ round a wire with $J = (0, 0, 20r)$ is $\dfrac{320\pi\mu_0}3$ and the heat flow of $T = x^2 + y^2 + z^2$ has divergence $-6$ .

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

6.8 Maxwell's equations, conservation laws and heat

The picture: the field that circles a wire

The idea, in words

A continuous function with zero integral over every region is zero

Gauss’s law: charge and the divergence theorem

Electric field

Charge density

Gauss's law

Gauss's law follows from Coulomb's law

The differential form of Gauss's law

Using Gauss’s law: choose a surface where $E\cdot n$ is constant

Magnetism: no sources, and Ampère’s law

Magnetic field

Current density

Ampere's law

Maxwell’s correction: $d^2 = 0$ forces a new term

Continuity equation

The continuity equation

Displacement current

Faraday’s law and the four equations

Faraday's law

Maxwell's equations

Light

Electromagnetic waves

Conservation laws in fluids and heat

Incompressible flow

Incompressible flow

Heat

Heat flux

The heat equation from conservation of energy

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere

6.8 Maxwell's equations, conservation laws and heat

The picture: the field that circles a wire

The idea, in words

A continuous function with zero integral over every region is zero

Gauss’s law: charge and the divergence theorem

Electric field

Charge density

Gauss's law

Gauss's law follows from Coulomb's law

The differential form of Gauss's law

Using Gauss’s law: choose a surface where E⋅nE\cdot nE⋅n is constant

Magnetism: no sources, and Ampère’s law

Magnetic field

Current density

Ampere's law

Maxwell’s correction: d2=0d^2 = 0d2=0 forces a new term

Continuity equation

The continuity equation

Displacement current

Faraday’s law and the four equations

Faraday's law

Maxwell's equations

Light

Electromagnetic waves

Conservation laws in fluids and heat

Incompressible flow

Incompressible flow

Heat

Heat flux

The heat equation from conservation of energy

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere

Using Gauss’s law: choose a surface where $E\cdot n$ is constant

Maxwell’s correction: $d^2 = 0$ forces a new term