University Physics — Year 1 · Bachelor Year 1

28Magnetostatics: Field, Forces, and Dipoles

A compass needle swings when a wire nearby carries a current: that observation of Ørsted’s, in 1820, joined two sciences that had grown up apart, electricity and magnetism, and within months Ampère had the law of the force between currents and Biot and Savart the field of a wire. This chapter is the statics of the magnetic field: what currents create (field maps, the Biot–Savart law, and Ampère’s theorem, the magnetic twin of Gauss’s law), what the field does to currents (the Laplace force, which runs every motor and every meter), and the magnetic dipole, the current loop that explains the compass, the magnet and the Earth’s own field.

The oldest instrument of magnetism: a compass needle is a small magnetic dipole aligning with the Earth’s field.

Ørsted’s experiment: close the circuit and the needle swings — a current creates a magnetic field, circling the wire.

28.1 Sources and field maps

Definition 28.1 (Magnetostatic field)

Steady currents (charges in steady motion) create at every point a magnetic field $\vect B$ (tesla, $\mathrm{T}$ ), defined by the force $q\vect v\wedge\vect B$ it exerts on a moving test charge (Definition 17.1). Its field lines are tangent to $\vect B$ and oriented along it. Permanent magnets create fields of the same nature, from the microscopic currents of their atoms. Orders of magnitude: Earth $5 \times 10^{-5}\,\mathrm{T}$ ; a fridge magnet $0.01\,\mathrm{T}$ ; an MRI scanner $1.5\,\mathrm{T}$ to $3\,\mathrm{T}$ ; the largest steady laboratory fields $45\,\mathrm{T}$ .

Proposition 28.2 (Field maps)

A straight wire: circles centred on the wire, in planes perpendicular to it, oriented by the right-hand rule (thumb along the current, fingers along $\vect B$ ).
A circular loop: lines threading the loop along its axis and closing round outside; far away, the map of a small magnet.
A long solenoid: parallel, dense, uniform lines inside; almost no field outside; the map of a bar magnet, whose “north” face is the one the lines leave.
Magnetic field lines have no beginning and no end: they close on themselves or run to infinity. Equivalently the flux of $\vect B$ through any closed surface is zero — there are no magnetic charges.

Proof. Admitted at this level. ∎

Three field maps. A dot is a current toward the reader, a cross a current away. The lines of a wire are circles; those of a loop thread it and close outside; those of a long solenoid are straight and uniform inside and close far away, where the field is weak — in every case they have neither source nor sink.

Proposition 28.3 (Symmetries of $\vect B$ )

$\vect B$ behaves under mirror reflection oppositely to $\vect E$ (it is a pseudo-vector, built from a cross product): at a point of a plane of symmetry of the currents, $\vect B$ is perpendicular to the plane; at a point of a plane of antisymmetry (the mirror image reverses the currents), $\vect B$ lies in the plane. Invariances transfer as for $\vect E$ .

Proof. The field of a current element is $\propto I\,\dd\vect l\wedge\vect e_r$ (below); a mirror reverses one factor of a cross product relative to the image of a true vector, so the image of $\vect B$ is the opposite of the mirror image of the arrow. On a symmetry plane $\vect B$ must equal the opposite of its own mirror image, hence be normal to the plane; on an antisymmetry plane, the opposite of the opposite: in the plane. ∎

Example 28.4 (Using the symmetries)

For a straight wire, any plane containing the wire is a symmetry plane, so at every point $\vect B$ is perpendicular to the plane through the wire and the point: orthoradial, $\vect B = B(r)\,\vect e_\theta$ by invariance along and around the wire. For a loop, the plane of the loop is a symmetry plane: on it $\vect B$ is normal to the loop; the axis lies in every antisymmetry plane containing it: on the axis $\vect B$ is along the axis.

28.2 The Biot–Savart law

Theorem 28.5 (Biot–Savart)

A circuit carrying the current $I$ creates at $M$ the field

\vect B(M) = \frac{\mu_0}{4\pi}\sum_{\text{circuit}} \frac{I\,\dd\vect l\wedge\vect e_{PM}}{PM^2} ,

sum over the elements $\dd\vect l$ of the circuit at points $P$ , oriented along the current, with $\vect e_{PM}$ the unit vector from $P$ to $M$ , and $\mu_0 = 4\pi \times 10^{-7}\,\mathrm{T}\,\mathrm{m}/\mathrm{A}$ the permeability of vacuum. Each element contributes a field perpendicular to itself and to the line $PM$ , falling as $1/PM^2$ .

Proof. Admitted at this level. ∎

Proposition 28.6 (Wire, loop, solenoid)

Infinite straight wire, at distance $r$ : $B = \dfrac{\mu_0I}{2\pi r}$ .
Circular loop of radius $R$ , on its axis at distance $z$ from the centre: $B_z = \dfrac{\mu_0IR^2}{2(R^2 + z^2)^{3/2}} = \dfrac{\mu_0I}{2R} \sin^3\alpha$ , with $\alpha$ the half-angle under which the loop is seen from $M$ ; at the centre $B = \mu_0I/2R$ .
Solenoid of $n$ turns per metre, on its axis: $B = \tfrac12\mu_0nI (\cos\alpha_1 - \cos\alpha_2)$ , with $\alpha_1, \alpha_2$ the angles under which the two ends are seen; for a long solenoid, $B = \mu_0nI$ inside, half that at an end.

Proof. (1) $M$ at distance $r$ from the wire, $P$ at abscissa $z$ along it: $\dd\vect l\wedge\vect e_{PM}$ has magnitude $\dd z\,\sin\beta = \dd z\cdot r/PM$ with $PM = \sqrt{r^2 + z^2}$ , and all contributions are orthoradial: $B = \dfrac{\mu_0I}{4\pi}\displaystyle\int_{-\infty}^{\infty}\dfrac{r\,\dd z}{(r^2 + z^2)^{3/2}} = \dfrac{\mu_0I}{4\pi}\,\dfrac{2}{r}$ . (2) Each element of the loop is at the same distance $\sqrt{R^2 + z^2}$ from $M$ , perpendicular to $\vect e_{PM}$ ; its field has magnitude $\mu_0I\,\dd l/4\pi(R^2 + z^2)$ , and only the axial component, the fraction $R/\sqrt{R^2 + z^2}$ of it, survives the sum round the loop: $B_z = \mu_0I\cdot2\pi R\cdot R/4\pi(R^2 + z^2)^{3/2}$ . (3) A slice $\dd z'$ of solenoid is a loop of current $nI\,\dd z'$ ; with $z' - z = R/\tan\alpha$ , $\dd z' = -R\,\dd\alpha/\sin^2\alpha$ , and $B = \dfrac{\mu_0nI}{2}\displaystyle\int\sin^3\alpha\,\dfrac{\dd\alpha}{\sin^2\alpha}$ over the solenoid $= \tfrac12\mu_0nI(\cos\alpha_1 - \cos\alpha_2)$ ; infinitely long, $\alpha_1 \to 0$ , $\alpha_2 \to \pi$ . ∎

The three Biot–Savart computations of the chapter: the element l of a wire and its contribution at M, perpendicular to the plane of the figure; the loop seen from M under the half-angle ; the solenoid, whose ends are seen under _1 and _2. — The three Biot–Savart computations of the chapter: the element $\dd\vect l$ of a wire and its contribution at $M$ , perpendicular to the plane of the figure; the loop seen from $M$ under the half-angle $\alpha$ ; the solenoid, whose ends are seen under $\alpha_1$ and $\alpha_2$ .

Example 28.7 (Orders of magnitude)

A $10\,\mathrm{A}$ wire at $1\,\mathrm{cm}$ : $B = 2 \times 10^{-7} \times 10/0.01 = 2 \times 10^{-4}\,\mathrm{T}$ , four times the Earth’s field — the compass does swing. A coil of $100$ turns, radius $5\,\mathrm{cm}$ , $2\,\mathrm{A}$ : $\mu_0NI/2R = 2.5\,\mathrm{mT}$ at its centre. A solenoid of $1000$ turns per metre at $5\,\mathrm{A}$ : $6.3\,\mathrm{mT}$ ; with an iron core of relative permeability $1000$ , several tesla would follow if iron did not saturate near $2\,\mathrm{T}$ — the ceiling of every electromagnet, which is why the strongest fields come from superconducting coils without iron.

28.3 Ampère’s law

Theorem 28.8 (Ampère)

The circulation of $\vect B$ along any closed oriented curve $\Gamma$ equals $\mu_0$ times the total current crossing any surface bounded by $\Gamma$ , counted positively when it crosses in the direction given by the right-hand rule from the orientation of $\Gamma$ :

\oint_\Gamma \vect B\cdot\dd\vect l = \mu_0 I_{\text{enclosed}} .

Proof. For a circle of radius $r$ centred on a straight wire, $\vect B$ is tangent with magnitude $\mu_0I/2\pi r$ everywhere, so the circulation is $2\pi r \times \mu_0I/2\pi r = \mu_0I$ , independent of $r$ ; for a curve not enclosing the wire the contributions cancel. The general case is admitted (Year 2 volume). ∎

Method 28.9 (Using Ampère’s law)

As for Gauss: (1) symmetries give the direction of $\vect B$ and the variable it depends on; (2) choose an Amperian loop through $M$ on which $\vect B$ is either tangent with constant magnitude or perpendicular; (3) count the current through it; (4) solve. It works for infinite wires and cylinders, infinite solenoids and toroids, and infinite current sheets.

Proposition 28.10 (Cylinder, solenoid, toroid)

A cylindrical wire of radius $a$ carrying $I$ uniformly: $B = \dfrac{\mu_0Ir}{2\pi a^2}$ inside, $\dfrac{\mu_0I}{2\pi r}$ outside — the maximum is at the surface.
An infinite solenoid of $n$ turns per metre: $\vect B = \mu_0nI\, \vect e_z$ uniform inside, $\vect 0$ outside, whatever the shape of its cross-section.
A toroid of $N$ turns: $B = \dfrac{\mu_0NI}{2\pi r}$ inside the windings, zero outside — a solenoid closed on itself, with no stray field.

Proof. (1) Circle of radius $r$ : $2\pi rB = \mu_0I(r/a)^2$ or $\mu_0I$ . (2) $\vect B$ is axial by symmetry and, by Ampère on a rectangle with both long sides outside, uniform outside, hence zero (it vanishes at infinity); a rectangle with one long side inside, of length $\ell$ , gives $B\ell = \mu_0n\ell I$ . (3) Circle of radius $r$ inside the torus: $2\pi rB = \mu_0NI$ ; outside, the enclosed current is zero. ∎

Left: the field of a thick wire, linear inside and in 1/r outside, largest at the surface. Right: the Amperian rectangle that gives the field of an infinite solenoid — only the side inside contributes to the circulation. — Left: the field of a thick wire, linear inside and in $1/r$ outside, largest at the surface. Right: the Amperian rectangle that gives the field of an infinite solenoid — only the side inside contributes to the circulation.

28.4 The Laplace force

Theorem 28.11 (Laplace force)

A conductor element $\dd\vect l$ carrying $I$ in a field $\vect B$ receives the force $\dd\vect F = I\,\dd\vect l\wedge\vect B$ ; a straight segment $\vect L$ in a uniform field receives $\vect F = I\vect L\wedge\vect B$ , of magnitude $ILB\sin\theta$ , perpendicular to both the wire and the field.

Proof. The charge carriers in $\dd\vect l$ (number $n\,S\,\dd l$ , charge $q$ , drift velocity $\vect v$ ) each feel $q\vect v\wedge\vect B$ ; their sum is $nSq\,\dd l\,\vect v\wedge\vect B = I\,\dd\vect l\wedge\vect B$ since $I = nqvS$ and $\vect v$ is along $\dd\vect l$ . The carriers transmit the force to the lattice of the wire through their collisions. ∎

Proposition 28.12 (Force between parallel wires)

Two long parallel wires a distance $d$ apart carrying $I_1$ and $I_2$ attract each other if the currents are parallel, repel if antiparallel, with the force per unit length

\frac{F}{\ell} = \frac{\mu_0I_1I_2}{2\pi d} .

Two wires $1\,\mathrm{m}$ apart carrying $1\,\mathrm{A}$ feel $2 \times 10^{-7}$ N per metre — the definition of the ampere from 1948 to 2019, and the origin of the value $\mu_0 = 4\pi \times 10^{-7}$ .

Proof. Wire 1 creates $\mu_0I_1/2\pi d$ at wire 2, perpendicular to it; the Laplace force on a length $\ell$ of wire 2 is $I_2\ell \times \mu_0I_1/2\pi d$ , directed toward wire 1 when the currents are parallel (check with the right-hand rule). ∎

The Laplace force: between parallel wires (left, middle) and on a sliding bar carrying a current across a field (right) — the Laplace rails, whose energy balance is the subject of . — The Laplace force: between parallel wires (left, middle) and on a sliding bar carrying a current across a field (right) — the Laplace rails, whose energy balance is the subject of Chapter 29.

28.5 The magnetic dipole

Definition 28.13 (Magnetic moment)

A plane loop of area $S$ carrying $I$ , oriented by the right-hand rule (normal $\vect n$ along the thumb when the fingers follow $I$ ), has the magnetic moment $\vect m = IS\,\vect n$ ( $\mathrm{A}\,\mathrm{m}^{2}$ ); for $N$ turns, $NIS\,\vect n$ . Seen from distances $r \gg \sqrt S$ the loop is a magnetic dipole: its field has exactly the form of the electric dipole’s with $\vect p/\varepsilon_0 \to \mu_0\vect m$ :

B_r = \frac{\mu_0}{4\pi}\,\frac{2m\cos\theta}{r^3}, \qquad B_\theta = \frac{\mu_0}{4\pi}\,\frac{m\sin\theta}{r^3} .

A bar magnet, an atom, the Earth are magnetic dipoles: the Earth’s moment is $8 \times 10^{22}\,\mathrm{A}\,\mathrm{m}^{2}$ , an electron’s $9.3 \times 10^{-24}\,\mathrm{A}\,\mathrm{m}^{2}$ (the Bohr magneton).

Proof. On the axis the loop field $\mu_0IR^2/2z^3$ for $z \gg R$ equals $\mu_0 \cdot 2m/4\pi z^3$ with $m = I\pi R^2$ , matching $B_r$ at $\theta = 0$ ; the full angular form is admitted. ∎

Theorem 28.14 (Dipole in a field)

A magnetic moment $\vect m$ in a uniform field $\vect B$ feels no net force, a torque $\vect\Gamma = \vect m\wedge\vect B$ that aligns it with the field, and has the potential energy $E_p = -\vect m\cdot\vect B$ . In a non-uniform field it is pulled toward the strong-field region when aligned (and pushed away when anti-aligned).

Proof. For a rectangular loop $a \times b$ with $\vect n$ at angle $\theta$ to $\vect B$ : the Laplace forces on the two sides of length $b$ parallel to the axis of rotation are $\pm IbB$ , opposite, a distance $a\sin\theta$ apart: a couple of moment $IabB\sin\theta = mB\sin\theta$ , tending to reduce $\theta$ ; the forces on the other two sides are opposite and collinear. Energy: $\Gamma = -\dd E_p/\dd\theta$ with $\Gamma = -mB\sin\theta$ gives $E_p = -mB\cos\theta$ . Any plane loop is a sum of small rectangles; the non-uniform case is admitted. ∎

Left: a current loop in a uniform field, seen along its axis of rotation — the Laplace forces on the two sides form a couple that aligns the moment with the field. Right: the field lines of a bar magnet are those of a magnetic dipole, identical in shape to the electric dipole’s, leaving the north face and returning to the south.

Example 28.15 (The compass and the Earth)

A compass needle is a small magnet, moment $m \sim 0.1\,\mathrm{A}\,\mathrm{m}^{2}$ ; in the horizontal component of the Earth’s field, $B_h \approx 2 \times 10^{-5}\,\mathrm{T}$ , the torque $mB_h\sin\theta$ turns it north, and it oscillates about north with period $2\pi\sqrt{J/mB_h} \sim 1\,\mathrm{s}$ (Exercise 28.9). The Earth’s own dipole, $8 \times 10^{22}\,\mathrm{A}\,\mathrm{m}^{2}$ , is equivalent to a current of a few billion amperes circling in the liquid iron core; the field it makes at the surface is the weakest of the chapter and the first one anybody used.

28.6 Exercises

Exercise 28.1 ★

Field at $1.0\,\mathrm{cm}$ from a straight wire carrying $10\,\mathrm{A}$ ; current needed to reach $1\,\mathrm{T}$ at that distance; distance at which a $100\,\mathrm{A}$ wire’s field equals the Earth’s $5 \times 10^{-5}\,\mathrm{T}$ .

Solution

Solution of Exercise 28.1.

$B = \mu_0I/2\pi r = 2 \times 10^{-7} \times 10/0.01 = 2 \times 10^{-4}\,\mathrm{T}$ . For $1\,\mathrm{T}$ : $I = 2\pi rB/\mu_0 = 5 \times 10^{4}\,\mathrm{A}$ . Earth’s value at $r = 2 \times 10^{-7} \times 100/5 \times 10^{-5} = 0.4\,\mathrm{m}$ .

Exercise 28.2 ★

Field at the centre of a coil of $100$ turns, radius $5.0\,\mathrm{cm}$ , carrying $2.0\,\mathrm{A}$ ; on its axis at $5.0\,\mathrm{cm}$ and at $50\,\mathrm{cm}$ from the centre.

Solution

Solution of Exercise 28.2.

Centre: $\mu_0NI/2R = 4\pi \times 10^{-7} \times 200/0.1 = 2.5\,\mathrm{mT}$ . At $z = R$ : $\times(R^2/2R^2)^{3/2} = 2^{-3/2}$ : $0.89\,\mathrm{mT}$ . At $z = 10R$ : $\times(1/101)^{3/2} = 10^{-3}$ : $2.5\,\text{µ}\mathrm{T}$ , already the $1/z^3$ dipole law.

Exercise 28.3 ★

A solenoid of $1000$ turns per metre carries $5.0\,\mathrm{A}$ : field inside, far from the ends; at the very end. Number of turns of $0.5\,\mathrm{mm}$ wire one can wind per metre in a single layer, and the field then at $5\,\mathrm{A}$ .

Solution

Solution of Exercise 28.3.

$B = \mu_0nI = 4\pi \times 10^{-7} \times 1000 \times 5 = 6.3\,\mathrm{mT}$ ; at the end, half: $3.1\,\mathrm{mT}$ . Wire of $0.5\,\mathrm{mm}$ : $2000$ turns per metre, so $12.6\,\mathrm{mT}$ at $5\,\mathrm{A}$ .

Exercise 28.4 ★

Force on a $1.0\,\mathrm{m}$ wire carrying $10\,\mathrm{A}$ perpendicular to a $0.10\,\mathrm{T}$ field; on the same wire at $30{}^{\circ}$ to the field. A motor has $100$ such wires on a rotor of radius $5\,\mathrm{cm}$ : torque (order of magnitude).

Solution

Solution of Exercise 28.4.

$F = ILB = 10 \times 1 \times 0.1 = 1.0\,\mathrm{N}$ ; at $30{}^{\circ}$ , $\sin30^\circ$ : $0.5\,\mathrm{N}$ . Motor: $100 \times 1\,\mathrm{N} \times 0.05\,\mathrm{m} = 5\,\mathrm{N}\,\mathrm{m}$ — a small industrial motor.

Exercise 28.5 ★★

Field of a finite straight segment at distance $d$ from its line, its ends seen under angles $\alpha_1$ and $\alpha_2$ (measured from the perpendicular): $B = \dfrac{\mu_0I}{4\pi d}(\sin\alpha_2 - \sin\alpha_1)$ . Check the infinite-wire limit. Field at the centre of a square loop of side $a$ .

Solution

Solution of Exercise 28.5.

With $z = d\tan\alpha$ : $B = \dfrac{\mu_0I}{4\pi}\displaystyle\int\dfrac{d\,\dd z}{(d^2 + z^2)^{3/2}} = \dfrac{\mu_0I}{4\pi d}\displaystyle\int_{\alpha_1}^{\alpha_2}\cos\alpha\,\dd\alpha = \dfrac{\mu_0I}{4\pi d} (\sin\alpha_2 - \sin\alpha_1)$ ; infinite wire, $\alpha_{1,2} = \mp\pi/2$ : $\mu_0I/2\pi d$ . Square: each side at $d = a/2$ seen under $\pm45^\circ$ gives $\mu_0I\sqrt2/2\pi a$ ; four sides: $B = 2\sqrt2\,\mu_0I/\pi a = 0.90\,\mu_0I/a$ (a circle of the same “radius” $a/2$ would give $\mu_0I/a$ ).

Exercise 28.6 ★★

Helmholtz coils. Two identical coils ( $N$ turns, radius $R$ ) on a common axis, a distance $R$ apart, carry the same current in the same sense. Field at the midpoint; show that the first and second derivatives of $B$ along the axis vanish there. Numbers: $N = 100$ , $R = 15\,\mathrm{cm}$ , $I = 1.0\,\mathrm{A}$ .

Solution

Solution of Exercise 28.6.

$B(z) = f(z - R/2) + f(z + R/2)$ with $f(u) = \mu_0NIR^2/2(R^2 + u^2)^{3/2}$ . Midpoint: $2f(R/2) = \mu_0NIR^2/(5R^2/4)^{3/2} = (4/5)^{3/2}\mu_0NI/R$ . $f'$ is odd, so $B'(0) = f'(-R/2) + f'(R/2) = 0$ ; $f''(u) \propto (4u^2 - R^2)(R^2 + u^2)^{-7/2}$ vanishes at $u = R/2$ , so $B''(0) = 2f''(R/2) = 0$ : the field is uniform to third order. Numbers: $0.716 \times 4\pi \times 10^{-7} \times 100/0.15 = 0.60\,\mathrm{mT}$ .

Exercise 28.7 ★★

A copper wire of radius $2.0\,\mathrm{mm}$ carries $100\,\mathrm{A}$ . Field at the surface, at $1.0\,\mathrm{mm}$ from the axis, at $1.0\,\mathrm{cm}$ . Sketch $B(r)$ . The wire is now the inner conductor of a coaxial cable whose sheath carries the return current: field outside the sheath.

Solution

Solution of Exercise 28.7.

Surface: $\mu_0I/2\pi a = 2 \times 10^{-7} \times 100/0.002 = 10\,\mathrm{mT}$ ; at $1\,\mathrm{mm}$ , inside, $\times r/a$ : $5\,\mathrm{mT}$ ; at $1\,\mathrm{cm}$ : $2\,\mathrm{mT}$ . $B$ rises linearly to the surface, then falls as $1/r$ . Coaxial: outside the sheath the enclosed current is zero, $B = 0$ .

Exercise 28.8 ★★

Force per metre between two wires $1\,\mathrm{m}$ apart at $1\,\mathrm{A}$ ; between two bus bars $10\,\mathrm{cm}$ apart carrying $10\,\mathrm{kA}$ ; under a short circuit of $100\,\mathrm{kA}$ . What must hold the bars?

Solution

Solution of Exercise 28.8.

$2 \times 10^{-7}$ N/m. Bus bars: $2 \times 10^{-7} \times 10^8/0.1 = 200\,\mathrm{N}/\mathrm{m}$ ; at $100\,\mathrm{kA}$ : $2 \times 10^{4}\,\mathrm{N}/\mathrm{m}$ — two tonnes per metre. The insulating supports of switchgear are sized for the fault current, not the working one.

Exercise 28.9 ★★

A compass needle, moment $m = 0.10\,\mathrm{A}\,\mathrm{m}^{2}$ , moment of inertia $J = 1.0 \times 10^{-7}\,\mathrm{kg}\,\mathrm{m}^{2}$ , in a horizontal field $B_h = 2.0 \times 10^{-5}\,\mathrm{T}$ . Torque at $90{}^{\circ}$ ; work to turn it from north to south; period of small oscillations. How does one measure $B_h$ with it?

Solution

Solution of Exercise 28.9.

$\Gamma = mB_h = 2 \times 10^{-6}\,\mathrm{N}\,\mathrm{m}$ ; $W = 2mB_h = 4 \times 10^{-6}\,\mathrm{J}$ ; $J\ddot\theta = -mB_h\theta$ : $T = 2\pi\sqrt{J/mB_h} = 2\pi\sqrt{10^{-7}/2 \times 10^{-6}} = 1.4\,\mathrm{s}$ . Time the oscillations: $B_h = 4\pi^2J/mT^2$ once $m$ and $J$ are known (Gauss measured $m$ separately from the deflection it gives a second needle).

Exercise 28.10 ★★★

The Earth as a dipole. (a) Show that on its axis, far away, a loop’s field is $B = \mu_0m/2\pi z^3$ with $m = I\pi R^2$ ; compare with the electric dipole. (b) The field at the magnetic pole is about $6 \times 10^{-5}\,\mathrm{T}$ : deduce the Earth’s dipole moment. (c) Current in a loop of radius $3000\,\mathrm{km}$ (the core) that would produce it. (d) Field at the equator on the dipole model; the angle of dip of the field at latitude $45{}^{\circ}$ (use $B_r$ and $B_\theta$ ).

Solution

Solution of Exercise 28.10.

(a) $\mu_0IR^2/2z^3 = \mu_0(I\pi R^2)/2\pi z^3 = \mu_0m/2\pi z^3 = (\mu_0/4\pi)\,2m/z^3$ : the electric $2p/4\pi\varepsilon_0z^3$ with $1/\varepsilon_0 \to \mu_0$ . (b) $m = 2\pi R_E^3B/ \mu_0 = 2\pi \times 2.58 \times 10^{20} \times 6 \times 10^{-5}/1.26 \times 10^{-6} = 7.7 \times 10^{22}\,\mathrm{A}\,\mathrm{m}^{2}$ . (c) $I = m/\pi R^2 = 7.7 \times 10^{22}/(\pi \times 9 \times 10^{12}) = 2.7 \times 10^{9}\,\mathrm{A}$ . (d) Equator ( $\theta = 90^\circ$ ): $B_\theta = \mu_0m/4\pi R_E^3 = 3 \times 10^{-5}\,\mathrm{T}$ , half the polar value. At $\theta = 45^\circ$ : $\tan(\text{dip}) = B_r/B_\theta = 2\cot\theta = 2$ , dip $= 63{}^{\circ}$ — the field plunges steeply into the ground at mid-latitudes.

Exercise 28.11 ★★★

Finite solenoid. Length $L = 20\,\mathrm{cm}$ , radius $R = 2.0\,\mathrm{cm}$ , $n = 2000\,\mathrm{m}^{-1}$ , $I = 3.0\,\mathrm{A}$ . Field at the centre and at an end from the course formula; fraction of the infinite-solenoid value; at what distance outside along the axis has the field fallen to $1\%$ of the central value? Sketch $B(z)$ along the axis.

Solution

Solution of Exercise 28.11.

$\mu_0nI = 4\pi \times 10^{-7} \times 2000 \times 3 = 7.5\,\mathrm{mT}$ . Centre: $\tan\alpha_1 = R/(L/2) = 0.2$ , $\cos\alpha_1 = -\cos\alpha_2 = 0.981$ : $B = 0.981\,\mu_0nI = 7.4\,\mathrm{mT}$ ( $98\%$ ). End: $\cos\alpha_1 = 0$ , $\tan\alpha_2 = R/L$ , $\cos\alpha_2 = -0.995$ : $B = 0.497\,\mu_0nI = 3.7\,\mathrm{mT}$ ( $50\%$ ). Outside at $x$ beyond the end, both ends on the same side: $B = \tfrac12\mu_0nI(\cos\alpha_1 - \cos\alpha_2) \approx \tfrac12\mu_0nI\,\tfrac{R^2}{2}[1/x^2 - 1/(x + L)^2]$ ; $1\%$ of the centre value at $x \approx 10\,\mathrm{cm}$ , half a length out. $B(z)$ : a flat top over most of the length, falling to half at the ends and to a few percent a length away.

Exercise 28.12 ★★★

Magnetic pressure. In a long solenoid the field is $B$ inside and $0$ outside; the windings carry the surface current $K = nI$ per metre of length. (a) Show that the field at the windings, the average of inside and outside, is $B/2$ , and that the Laplace force on them is an outward pressure $p = B^2/2\mu_0$ . (b) Numbers for $1\,\mathrm{T}$ , $10\,\mathrm{T}$ , $45\,\mathrm{T}$ ; compare with atmospheric pressure and with the tensile strength of steel ( $1\,\mathrm{GPa}$ ). (c) What limits the strongest steady magnets?

Solution

Solution of Exercise 28.12.

(a) The windings’ own field jumps from $0$ to $B$ across them; the field acting on them is the mean, $B/2$ (the sheet cannot push on itself). Force on an element of winding of length $\dd l$ and width $\dd z$ carrying $K\,\dd z$ : $K\,\dd z\,\dd l\,B/2$ , radial outward (check with $I\,\dd\vect l \wedge\vect B$ ): pressure $p = KB/2 = B^2/2\mu_0$ since $K = nI = B/\mu_0$ . (b) $1\,\mathrm{T}$ : $4 \times 10^{5}\,\mathrm{Pa}$ = $4\,\mathrm{bar}$ ; $10\,\mathrm{T}$ : $400\,\mathrm{bar}$ ; $45\,\mathrm{T}$ : $8 \times 10^{8}\,\mathrm{Pa}$ = $8000\,\mathrm{bar}$ , close to the $1\,\mathrm{GPa}$ of steel. (c) The windings must hold a pressure growing as $B^2$ : above $40\,\mathrm{T}$ or so no material and no cooling scheme copes — the limit of steady magnets is mechanical (and, for superconductors, their critical field).

A high-voltage transmission line: three conductors per circuit whose currents sum to zero, so that their fields cancel at a distance. Photograph: Stefan Andrej Shambora, CC BY 2.0.

28.7 Problem: The cable, the line and the meter

Problem 28.1

Weekend problem — Ampère’s law on a coaxial cable, Laplace forces on a power line, and the design of a moving-coil galvanometer

$\mu_0 = 4\pi \times 10^{-7}$ SI; copper resistivity $\rho = 1.7 \times 10^{-8}\,\Omega\,\mathrm{m}$ .

Part I — The coaxial cable. A cable carries $I = 1.0\,\mathrm{A}$ along a solid copper core of radius $a = 0.50\,\mathrm{mm}$ and back along a sheath of inner radius $b = 2.5\,\mathrm{mm}$ and outer radius $c = 3.0\,\mathrm{mm}$ ; currents are uniform over each conductor’s section.

Symmetries and invariances: direction of $\vect B$ and the variable it depends on.
$B(r)$ in the core, between the conductors, and outside the cable, by Ampère’s law.
Values at $r = a$ and $r = b$ .
The core is replaced by a thin copper tube of the same radius $a$ carrying the same current: what changes in each region?
$B(r)$ inside the sheath ( $b < r < c$ ); check the values at $b$ and $c$ ; sketch $B(r)$ from $0$ to $4\,\mathrm{mm}$ .
Laplace force on the sheath: direction (attraction or repulsion from the core?) and magnitude per unit area, taking the field at the sheath as the average of its values at $b$ and $c$ . Same for a pulsed current of $10\,\mathrm{kA}$ .
Why is the coaxial geometry used for signals? Compare with a pair of parallel wires $d = 2\,\mathrm{mm}$ apart: field at a distance $D = 10\,\mathrm{cm}$ from the pair (superpose two wires).

Part II — The power line. Two horizontal conductors $d = 1.0\,\mathrm{m}$ apart, $h = 10\,\mathrm{m}$ above the ground, carry $I = 500\,\mathrm{A}$ in opposite directions.

Field of one conductor at the point of the ground below the midpoint; show that the two fields combine into a vertical field and give its value; compare with the Earth’s and with the $100\,\text{µ}\mathrm{T}$ exposure guideline.
Show that at distances $D \gg d$ the field of the pair falls as $\mu_0Id/2\pi D^2$ : why faster than a single wire?
Field at $1.0\,\mathrm{m}$ below one conductor, where a technician might work.
Force per metre between the conductors; attraction or repulsion? Same under a $20\,\mathrm{kA}$ short-circuit current.
Force on a $300\,\mathrm{m}$ span, normal and in short circuit; compare with the span’s weight ( $1.0\,\mathrm{kg}/\mathrm{m}$ ); what do the lines do during a short circuit?
A three-phase line carries three currents of the same amplitude shifted by $120{}^{\circ}$ : show that their sum vanishes at every instant, and what this implies for the far field.
A compass at the point of question 8, where the Earth’s horizontal field is $2.0 \times 10^{-5}\,\mathrm{T}$ : does the line’s field deflect it? (Consider its direction.)

Part III — The moving-coil galvanometer. A rectangular coil, $N = 200$ turns, sides $a = 2.0\,\mathrm{cm}$ (parallel to the axis of rotation) and $b = 3.0\,\mathrm{cm}$ , turns about its axis in the gap between cylindrical pole pieces where the field, $B = 0.20\,\mathrm{T}$ , is radial: always in the plane of the coil, perpendicular to its sides $a$ . A spiral spring exerts the restoring torque $-C\theta$ with $C = 1.0 \times 10^{-5}\,\mathrm{N}\,\mathrm{m}/\mathrm{rad}$ . Moment of inertia of the coil and needle: $J = 6.7 \times 10^{-8}\,\mathrm{kg}\,\mathrm{m}^{2}$ .

Laplace forces on the four sides; which ones produce a torque about the axis?
Torque for a current $I$ ; why does the radial field make it independent of $\theta$ ?
Deflection $\theta$ at equilibrium; sensitivity in radians per milliampere; current for a full-scale deflection of $1.2\,\mathrm{rad}$ .
The coil is wound with copper wire of diameter $0.10\,\mathrm{mm}$ : resistance of the coil; voltage across it and power dissipated at full scale.
Equation of motion of the coil; natural period of the oscillations about the equilibrium.
The motion of the coil in the field induces a braking torque $-\alpha\dot\theta$ (Chapter 29); value of $\alpha$ for critical damping, and why one wants it.
Series resistor to make a $10\,\mathrm{V}$ full-scale voltmeter; shunt resistor to make a $1.0\,\mathrm{A}$ full-scale ammeter.
Magnetic moment of the coil at full scale; energy $-\vect m\cdot \vect B$ it would have, aligned, in a uniform $0.20\,\mathrm{T}$ field; compare with the energy stored in the spring at full scale.
If the field were uniform instead of radial, show that the scale would no longer be linear: write the equilibrium condition with $\theta$ measured from the position where the coil’s plane contains $\vect B$ .
A $50\,\mathrm{Hz}$ current of $0.5\,\mathrm{mA}$ amplitude is fed to the meter: what does the needle show, and why? (Compare $50\,\mathrm{Hz}$ with the natural frequency.)
Sum up the three scenes in three numbers, and say what single law produced each.

Solution

Solution of Problem 28.1.

1. Every plane containing the axis is a plane of symmetry of the currents: $\vect B$ is perpendicular to it, orthoradial; invariance along and around the axis: $\vect B = B(r)\,\vect e_\theta$ .

2. Circle of radius $r$ : $2\pi rB = \mu_0I_{\text{enc}}$ . Core: $I_{\text{enc}} = Ir^2/a^2$ , $B = \mu_0Ir/2\pi a^2$ ; between: $\mu_0I/2\pi r$ ; outside ( $r > c$ ): $I - I = 0$ , $B = 0$ .

3. $B(a) = 2 \times 10^{-7}/5 \times 10^{-4} = 4.0 \times 10^{-4}\,\mathrm{T}$ ; $B(b) = 2 \times 10^{-7}/2.5 \times 10^{-3} = 8.0 \times 10^{-5}\,\mathrm{T}$ .

4. Inside the tube no current is enclosed: $B = 0$ there instead of the linear rise; everywhere else nothing changes — the field outside a cylindrical current depends only on the total current.

5. $I_{\text{enc}} = I - I(r^2 - b^2)/(c^2 - b^2) = I(c^2 - r^2)/(c^2 - b^2)$ : $B = \dfrac{\mu_0I}{2\pi r}\,\dfrac{c^2 - r^2}{c^2 - b^2}$ , equal to $\mu_0I/2\pi b$ at $b$ and to $0$ at $c$ : continuous. Sketch: linear to $4 \times 10^{-4}\,\mathrm{T}$ at $0.5\,\mathrm{mm}$ , $1/r$ down to $8 \times 10^{-5}\,\mathrm{T}$ at $2.5\,\mathrm{mm}$ , falling to zero at $3\,\mathrm{mm}$ , zero beyond.

6. The sheath’s current is antiparallel to the core’s: repulsion, an outward pressure. Surface current $K = I/2\pi b = 64\,\mathrm{A}/\mathrm{m}$ ; mean field $\tfrac12(8 \times 10^{-5} + 0) = 4 \times 10^{-5}\,\mathrm{T}$ : $p = KB = 2.5 \times 10^{-3}\,\mathrm{Pa}$ . At $10\,\mathrm{kA}$ : $\times10^8$ , $2.5 \times 10^{5}\,\mathrm{Pa}$ = $2.5\,\mathrm{bar}$ — pulsed-power cables are built to take it.

7. No field outside: the cable neither radiates nor picks up magnetic interference from its neighbours. A pair of wires: the two fields nearly cancel, leaving $\approx \mu_0Id/2\pi D^2 = 2 \times 10^{-7} \times 2 \times 10^{-3}/10^{-2} = 4 \times 10^{-8}\,\mathrm{T}$ at $10\,\mathrm{cm}$ — small, but the coax gives zero.

8. $D = \sqrt{h^2 + d^2/4} = 10.0\,\mathrm{m}$ : each gives $\mu_0I/2\pi D = 1.0 \times 10^{-5}\,\mathrm{T}$ , perpendicular to the line from the conductor. The currents being opposite, the components parallel to the line joining the conductors cancel and the vertical ones add: $B = 2 \times 10^{-5} \times (d/2)/D = 1.0 \times 10^{-6}\,\mathrm{T}$ , vertical. Fifty times below the Earth’s field, a hundred times below the guideline.

9. For $D \gg d$ , $B \approx \dfrac{\mu_0I}{2\pi}\Bigl(\dfrac1{D_1} - \dfrac1{D_2}\Bigr) \sim \dfrac{\mu_0I}{2\pi}\dfrac{d}{D^2}$ : the two $1/D$ fields cancel to first order and only their difference, of relative size $d/D$ , survives — a “two-wire dipole”.

10. Own conductor at $1\,\mathrm{m}$ : $1.0 \times 10^{-4}\,\mathrm{T}$ horizontal; the other at $1.41\,\mathrm{m}$ : $7.1 \times 10^{-5}\,\mathrm{T}$ at $45{}^{\circ}$ , opposing: resultant $(5 \times 10^{-5}, 5 \times 10^{-5})$ , $7 \times 10^{-5}\,\mathrm{T}$ — $70\,\text{µ}\mathrm{T}$ , below the public guideline and far below the occupational one.

11. $F/\ell = \mu_0I^2/2\pi d = 2 \times 10^{-7} \times 2.5 \times 10^5 = 0.05\,\mathrm{N}/\mathrm{m}$ , repulsive; at $20\,\mathrm{kA}$ : $80\,\mathrm{N}/\mathrm{m}$ .

12. $15\,\mathrm{N}$ normally; $24\,\mathrm{kN}$ in short circuit, eight times the span’s $2.9\,\mathrm{kN}$ weight: the conductors are flung apart, swing and may clash when they fall back — hence spacers and breakers that open within a few cycles.

13. $I\cos\omega t + I\cos(\omega t - 2\pi/3) + I\cos(\omega t + 2\pi/3) = I\cos\omega t\,(1 + 2\cos2\pi/3) = 0$ : the three phasors sum to zero. The $1/D$ fields cancel as for the coax; the far field falls as $1/D^2$ or faster.

14. The line’s field there is vertical; a compass responds to the horizontal component only: no deflection (the dip changes by $10^{-6}/(2 \times 10^{-5})$ , three degrees, which the compass cannot show).

15. Sides $a$ (parallel to the axis): $\vect B$ radial is perpendicular to them, force $NIaB$ on each, perpendicular to the coil’s plane, opposite on the two sides: a couple. Sides $b$ : $\dd\vect l$ is radial, parallel to $\vect B$ : no force.

16. $\Gamma = 2 \times NIaB \times b/2 = NIabB = NISB = 0.024\,I$ (N m, $I$ in A). The radial field is always in the coil’s plane and perpendicular to the sides $a$ : the forces are always normal to the plane and their lever arm $b/2$ never changes.

17. $C\theta = NISB$ : $\theta = 2400\,I$ rad, i.e. $2.4\,\mathrm{rad}/\mathrm{mA}$ ; full scale $1.2\,\mathrm{rad}$ at $I = 0.50\,\mathrm{mA}$ .

18. Length $200 \times 2(0.02 + 0.03) = 20\,\mathrm{m}$ , section $\pi(5 \times 10^{-5})^2 = 7.9 \times 10^{-9}\,\mathrm{m}^{2}$ : $R = 1.7 \times 10^{-8} \times 20/7.9 \times 10^{-9} = 43\,\Omega$ ; $U = 43 \times 5 \times 10^{-4} = 22\,\mathrm{mV}$ ; $P = UI = 11\,\text{µ}\mathrm{W}$ .

19. $J\ddot\theta = -C\theta + NISB\,I$ : $\omega_0 = \sqrt{C/J} = \sqrt{10^{-5}/6.7 \times 10^{-8}} = 12\,\mathrm{rad}/\mathrm{s}$ , $T = 0.51\,\mathrm{s}$ .

20. $J\ddot\theta + \alpha\dot\theta + C\theta = NISB\,I$ : critical for $\alpha = 2\sqrt{JC} = 2\sqrt{6.7 \times 10^{-13}} = 1.6 \times 10^{-6}\,\mathrm{N}\,\mathrm{m}\,\mathrm{s}$ ; the needle then reaches its reading fastest without overshoot.

21. Voltmeter: $R_s = 10/5 \times 10^{-4} - 43 \approx 20\,\mathrm{k}\Omega$ . Ammeter: the shunt carries $0.9995\,\mathrm{A}$ under $22\,\mathrm{mV}$ : $R_{\text{sh}} = 22\,\mathrm{m}\Omega$ .

22. $m = NIS = 200 \times 5 \times 10^{-4} \times 6 \times 10^{-4} = 6 \times 10^{-5}\,\mathrm{A}\,\mathrm{m}^{2}$ ; $-mB = -1.2 \times 10^{-5}\,\mathrm{J}$ ; spring $\tfrac12C\theta^2 = \tfrac12 \times 10^{-5} \times 1.44 = 7.2 \times 10^{-6}\,\mathrm{J}$ : the same order — the spring stores what the magnetic torque worked while the coil turned.

23. In a uniform field the torque is $mB\cos\theta$ (maximal when the plane contains $\vect B$ , zero when the normal is along it): $C\theta = NISB\,I\cos\theta$ — linear only for small $\theta$ , compressed at the top of the scale.

24. $50\,\mathrm{Hz}$ against $\omega_0/2\pi = 2\,\mathrm{Hz}$ : the coil cannot follow; it sits at the mean of the current, zero, with a quiver of amplitude $\approx \theta_{\text{static}}(\omega_0/\omega)^2 = 1.2 \times (12/314)^2 \approx 2 \times 10^{-3}\,\mathrm{rad}$ , invisible. A moving-coil meter reads the mean — for alternating current it needs a rectifier.

25. $4 \times 10^{-4}\,\mathrm{T}$ at the core of a $1\,\mathrm{A}$ coax and nothing outside (Ampère’s law); $1\,\text{µ}\mathrm{T}$ under a $500\,\mathrm{A}$ line and $0.05\,\mathrm{N}/\mathrm{m}$ between its conductors (Biot–Savart plus Laplace); $2.4\,\mathrm{rad}/\mathrm{mA}$ for the meter (Laplace couple against a spring).

Terms defined in this chapter

See all 393 terms in the glossary