Physics · Book 4 · Bachelor Year 2

University Physics — Year 2

University Physics — Year 2 · Bachelor Year 2

10Charges, Currents and Conduction

Flip the switch and the lamp lights at once; yet the electrons in the copper wire drift toward it at a fraction of a millimetre per second, and would take hours to arrive. A lightning bolt carries thirty thousand amperes through a channel the width of a finger; the nerve that moves that finger carries picoamperes. The Year 1 volume treated currents as numbers in wires, I=U/RI = U/R; this chapter describes them as a field — a current density at every point of matter — writes the law that charge is never created or destroyed, and derives Ohm’s law from the jostling of electrons in a metal: the microscopic picture behind the resistor, the fuse and the Hall sensor.

Copper, the conductor of the electrical world: in the strands of this cable some 1029 electrons per cubic metre drift at a fraction of a millimetre per second.
Copper, the conductor of the electrical world: in the strands of this cable some 102910^{29} electrons per cubic metre drift at a fraction of a millimetre per second.

10.1 Charge and current densities

Definition 10.1 (Charge density; current density)

At the mesoscopic scale, matter carries a volume charge density ρ(M,t)\rho(M, t) (C/m3\mathrm{C}/\mathrm{m}^{3}): the charge ρ  ⁣dτ\rho\,\dd\tau in  ⁣dτ\dd\tau. A surface may carry a surface density σ\sigma (C/m2\mathrm{C}/\mathrm{m}^{2}), a wire a linear density λ\lambda (C/m\mathrm{C}/\mathrm{m}). If the carriers of species ii (charge qiq_i, number density nin_i) move at the mean velocity v→i\vect v_i, the current density is

j→=∑iniqi v→i(A/m2),\vect j = \sum_in_iq_i\,\vect v_i \qquad (\mathrm{A}/\mathrm{m}^{2}) ,

and the intensity through an oriented surface SS is the flux I=∬Sj→⋅n→  ⁣dSI = \iint_S\vect j\cdot\vect n\,\dd S: the charge crossing SS per unit time. For a single species, j→=ρmv→\vect j = \rho_m\vect v with ρm=nq\rho_m = nq the density of mobile charge.

Proof. In  ⁣dt\dd t the carriers of species ii that cross  ⁣dS\dd S are those in the oblique cylinder of base  ⁣dS\dd S and generator v→i ⁣dt\vect v_i\dd t: number niv→i⋅n→  ⁣dS  ⁣dtn_i\vect v_i\cdot\vect n\,\dd S\,\dd t, charge qiq_i times that. ∎

Example 10.2 (How slowly electrons drift)

Copper has one conduction electron per atom: n=ρCuNA/M=8900×6.02×1023/0.0635=8.5×1028 m−3n = \rho_{\text{Cu}}N_A/M = 8900 \times 6.02 \times 10^{23}/0.0635 = 8.5 \times 10^{28}\,\mathrm{m}^{-3}. A current of 10 A10\,\mathrm{A} in a 1.5 mm21.5\,\mathrm{mm}^{2} wire is j=6.7×106 A/m2j = 6.7 \times 10^{6}\,\mathrm{A}/\mathrm{m}^{2}, and the drift velocity is v=j/ne=6.7×106/(8.5×1028×1.6×10−19)=0.5 mm/sv = j/ne = 6.7 \times 10^6/(8.5 \times 10^{28} \times 1.6 \times 10^{-19}) = 0.5\,\mathrm{mm}/\mathrm{s} — two hours per metre. The lamp lights at once because the field that pushes the electrons is set up along the whole wire within nanoseconds (it travels at nearly the speed of light, as the cable of Chapter 8 showed): all the electrons start together, like water in a full hose.

Left: a tube of current — in a stationary regime the same intensity crosses every section. Right: an electron in a metal zigzags between collisions at about 106 m/s; the field adds a slow drift, opposite to E, of a fraction of a millimetre per second.
Left: a tube of current — in a stationary regime the same intensity crosses every section. Right: an electron in a metal zigzags between collisions at about 10610^6 m/s; the field adds a slow drift, opposite to E→\vect E, of a fraction of a millimetre per second.

Example 10.3 (Orders of magnitude)

Lightning, 30 kA30\,\mathrm{kA} in a channel of 1 cm1\,\mathrm{cm} radius: j≈1×108 A/m2j \approx 1 \times 10^{8}\,\mathrm{A}/\mathrm{m}^{2}. A household wire: 10610^6–10710^7. An electron beam in a cathode-ray tube, 1 mA1\,\mathrm{mA} over 1 mm21\,\mathrm{mm}^{2}: 10310^3. A nerve fibre, 1 nA1\,\mathrm{nA} through 10 µm210\,\text{µ}\mathrm{m}^{2}: 10210^2. The beam of a particle accelerator, 1 A1\,\mathrm{A} in a 0.1 mm20.1\,\mathrm{mm}^{2} spot: 10710^7 — in vacuum, with no collisions at all.

10.2 Conservation of charge

Theorem 10.4 (Local conservation of charge)

At every point,

∂ρ∂t+div⁡j→=0;\frac{\partial\rho}{\partial t} + \operatorname{div}\vect j = 0 ;

integrated over a fixed volume VV bounded by the closed surface Σ\Sigma,  ⁣dQV/ ⁣dt=−Iout\dd Q_V/\dd t = -I_{\text{out}}: the charge inside changes only by what crosses the boundary. In a stationary regime (∂tρ=0\partial_t\rho = 0), div⁡j→=0\operatorname{div}\vect j = 0: the flux of j→\vect j is conserved along a tube of current, the same intensity crosses every section of a wire, and the currents entering a node equal those leaving (Kirchhoff’s node law).

Proof. Word for word the mass balance of Theorem 2.8 with ρ\rho the charge density and j→\vect j its current: the net outflow of charge through the six faces of a fixed box, per unit volume, is div⁡j→\operatorname{div}\vect j, and no charge is created. The integral form sums the boxes (the interior faces cancel). For a tube between two sections S1S_1, S2S_2 with no flux through its side, I1=I2I_1 = I_2; a node is a small volume with several wires: ∑Iin=∑Iout\sum I_{\text{in}} = \sum I_{\text{out}}. ∎

Remark 10.5 (Where the current stops: the capacitor)

A wire feeding a capacitor plate is a tube of current that ends: the flux of j→\vect j into the plate is I= ⁣dQ/ ⁣dtI = \dd Q/\dd t, and charge accumulates — div⁡j→≠0\operatorname{div}\vect j \ne 0 there, the regime is not stationary. Yet the same current II flows in the wire on the other side: something must carry the balance across the gap. Chapter 11 names it — the displacement current — and makes the conservation of charge the keystone of Maxwell’s equations.

10.3 The Drude model and Ohm’s law

Proposition 10.6 (Drude model; local Ohm’s law)

Model the conduction electrons of a metal as free particles (mass mm, charge −e-e, density nn) that feel the field E→\vect E and, on average, lose their drift momentum by collisions with the lattice every τ\tau seconds — a friction force −mv→/τ-m\vect v/\tau on the mean velocity:

m ⁣dv→ ⁣dt=−eE→−mτv→.m\frac{\dd\vect v}{\dd t} = -e\vect E - \frac m\tau\vect v .

In a steady field the drift settles, in a few τ\tau, at v→=−eτE→/m\vect v = -e\tau \vect E/m, and the current density is

j→=γ E→,γ=ne2τm:\vect j = \gamma\,\vect E , \qquad \gamma = \frac{ne^2\tau}{m} :

the local Ohm’s law, with γ\gamma the conductivity (S/m\mathrm{S}/\mathrm{m}; its inverse ρe=1/γ\rho_e = 1/\gamma is the resistivity, Ω m\Omega\,\mathrm{m}). The ratio μ=eτ/m\mu = e\tau/m of the drift speed to the field is the mobility (m2/(V s)\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})). For a conductor of length LL and uniform section SS carrying a uniform j→\vect j, integrating E=j/γE = j/\gamma along the length gives U=EL=(L/γS)IU = EL = (L/\gamma S)I: the resistance R=L/γS=ρeL/SR = L/\gamma S = \rho_eL/S of the Year 1 volume.

Proof. With v→=−eτE→/m\vect v = -e\tau\vect E/m, j→=−nev→=(ne2τ/m)E→\vect j = -ne\vect v = (ne^2\tau/m)\vect E. The transient v→(t)=v→∞(1−e−t/τ)\vect v(t) = \vect v_\infty(1 - \eu^{-t/\tau}) lasts τ∼1×10−14 s\tau \sim 1 \times 10^{-14}\,\mathrm{s}, instantaneous on every circuit time scale. The resistance follows from ∫E→⋅ ⁣dl→\int\vect E\cdot\dd\vect l between the ends. ∎

Example 10.7 (Copper, from the measured conductivity)

Copper: γ=6.0×107 S/m\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}, n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}: τ=mγ/ne2=9.1×10−31×6×107/(8.5×1028×2.56×10−38)=2.5×10−14 s\tau = m\gamma/ne^2 = 9.1 \times 10^{-31} \times 6 \times 10^7/(8.5 \times 10^{28} \times 2.56 \times 10^{-38}) = 2.5 \times 10^{-14}\,\mathrm{s} — twenty-five femtoseconds between collisions. The electrons’ random speed in a metal is about 1.6×106 m/s1.6 \times 10^{6}\,\mathrm{m}/\mathrm{s} (a quantum effect, not the thermal 3kBT/m≈1×105 m/s\sqrt{3k_BT/m} \approx 1 \times 10^{5}\,\mathrm{m}/\mathrm{s}), so the mean free path is ℓ=vτ≈40 nm\ell = v\tau \approx 40\,\mathrm{nm}, a hundred atomic spacings: the electrons do not collide with the atoms themselves (a perfect crystal is transparent to them) but with its defects and thermal vibrations — which is why γ\gamma falls when a metal is heated or alloyed. Mobility eτ/m=4.4×10−3 m2/(V s)e\tau/m = 4.4 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s}): 4 mm/s4\,\mathrm{mm}/\mathrm{s} per volt per metre.

Proposition 10.8 (Ohm’s law at high frequency)

For a sinusoidal field E→‾ eiωt\underline{\vect E}\,\eu^{\iu\omega t} the Drude equation gives j→‾=γ‾(ω)E→‾\underline{\vect j} = \underline\gamma(\omega)\underline{\vect E} with

γ‾(ω)=γ01+iωτ,γ0=ne2τm:\underline\gamma(\omega) = \frac{\gamma_0}{1 + \iu\omega\tau} , \qquad \gamma_0 = \frac{ne^2\tau}{m} :

the conductivity is real and equal to its DC value as long as ωτ≪1\omega\tau \ll 1, i.e. up to about 1×1013 Hz1 \times 10^{13}\,\mathrm{Hz} for copper — through the whole radio and microwave range; in the infrared and visible (ωτ≫1\omega\tau \gg 1) it becomes imaginary, γ‾≈ne2/imω\underline\gamma \approx ne^2/\iu m\omega: the electrons oscillate freely, out of phase with the field, and the metal behaves as the plasma of Chapter 14.

Proof. iωmv→‾=−eE→‾−mv→‾/τ\iu\omega m\underline{\vect v} = -e\underline{\vect E} - m\underline{\vect v}/\tau, whence v→‾=−eτE→‾/m(1+iωτ)\underline{\vect v} = -e\tau\underline{\vect E}/m(1 + \iu\omega\tau) and j→‾=−nev→‾\underline{\vect j} = -ne\underline{\vect v}. ∎

Left: the Drude conductivity against — real and constant at low frequency, imaginary beyond 1/. Right: the Hall effect in a conducting strip — the carriers are pushed sideways by the field, charge the edges, and the transverse Hall field balances the magnetic force. Left: the Drude conductivity against — real and constant at low frequency, imaginary beyond 1/. Right: the Hall effect in a conducting strip — the carriers are pushed sideways by the field, charge the edges, and the transverse Hall field balances the magnetic force.
Left: the Drude conductivity against ωτ\omega\tau — real and constant at low frequency, imaginary beyond 1/τ1/\tau. Right: the Hall effect in a conducting strip — the carriers are pushed sideways by the field, charge the edges, and the transverse Hall field balances the magnetic force.

Remark 10.9 (Conductors, semiconductors, electrolytes)

The same law j→=∑niqiμiE→\vect j = \sum n_iq_i\mu_i\vect E holds for every conductor; what changes is nn and μ\mu. Metals: n∼1029n \sim 10^{29}, γ∼107\gamma \sim 10^7 S/m. Semiconductors: n∼1016n \sim 10^{16}–102310^{23}, set by doping and rising steeply with temperature, with two kinds of carriers (electrons and holes) — whence thermistors and diodes. Electrolytes: ions of both signs, μ∼10−8\mu \sim 10^{-8}–10−710^{-7} m2^2/(V s); sea water γ≈5 S/m\gamma \approx 5\,\mathrm{S}/\mathrm{m}, tap water 10−210^{-2}, pure water 10−510^{-5}. Insulators (glass, polymers): γ≲10−12\gamma \lesssim 10^{-12} S/m. The Hall effect of the Year 1 volume, in local form E→H=−j→∧B→/nq\vect E_H = -\vect j\wedge\vect B/nq, measures nn and the sign of qq: in aluminium or zinc the carriers turn out to be positive — holes — and no classical model explains that.

10.4 Energy: the local Joule law

Proposition 10.10 (Local Joule law)

The electric field delivers to the carriers, per unit volume and time, the power j→⋅E→\vect j\cdot\vect E; in an ohmic conductor this is

p=j→⋅E→=γE2=j2γ  (W/m3),p = \vect j\cdot\vect E = \gamma E^2 = \frac{j^2}\gamma \ \ (\mathrm{W}/\mathrm{m}^{3}) ,

all of it turned into heat by the collisions (the drift kinetic energy is negligible and constant). Integrated over a wire, ∫j2/γ  ⁣dτ=(L/γS)I2=RI2\int j^2/\gamma\,\dd\tau = (L/\gamma S)I^2 = RI^2.

Proof. The force qE→q\vect E on each carrier works at the rate qE→⋅v→q\vect E\cdot\vect v; summed over the n  ⁣dτn\,\dd\tau carriers: nqv→⋅E→  ⁣dτ=j→⋅E→  ⁣dτnq\vect v\cdot\vect E\,\dd\tau = \vect j\cdot \vect E\,\dd\tau (the magnetic force does no work). In the Drude steady state this power equals that dissipated by the friction term, i.e. transferred to the lattice. ∎

Example 10.11 (A wire heating)

The 1.5 mm21.5\,\mathrm{mm}^{2} wire at 10 A10\,\mathrm{A}: p=j2/γ=(6.7×106)2/6×107=7.4×105 W/m3p = j^2/\gamma = (6.7 \times 10^6)^2/6 \times 10^7 = 7.4 \times 10^{5}\,\mathrm{W}/\mathrm{m}^{3}, i.e. 1.1 W1.1\,\mathrm{W} per metre — easily evacuated by the air around it. Left to itself (no cooling), copper (c=385 J/(kg K)c = 385\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K}), ρ=8900 kg/m3\rho = 8900\,\mathrm{kg}/\mathrm{m}^{3}) would warm at p/ρc=0.2 K/sp/\rho c = 0.2\,\mathrm{K}/\mathrm{s}. At 100 A100\,\mathrm{A} in the same wire the rate is a hundred times larger, 20 K/s20\,\mathrm{K}/\mathrm{s}: it melts in a minute — the principle of the fuse, whose thin wire is designed to melt first.

Proposition 10.12 (Relaxation of charge in a conductor)

In an ohmic conductor a volume charge density decays as ρ(t)=ρ0e−t/τr\rho(t) = \rho_0\eu^{-t/\tau_r} with

τr=ε0γ:\tau_r = \frac{\varepsilon_0}{\gamma} :

1.5×10−19 s1.5 \times 10^{-19}\,\mathrm{s} for copper, 2×10−12 s2 \times 10^{-12}\,\mathrm{s} for sea water, 10 s10\,\mathrm{s} for glass. Up to frequencies of order 1/τr1/\tau_r a conductor carries no volume charge: any excess charge runs to its surface. The word "conductor" thus depends on the time scale: sea water is a conductor for radio waves and an insulator for visible light.

Proof. Charge conservation with j→=γE→\vect j = \gamma\vect E and Gauss’s law in local form, div⁡E→=ρ/ε0\operatorname{div}\vect E = \rho/\varepsilon_0 (Chapter 11): ∂tρ=−γdiv⁡E→=−(γ/ε0)ρ\partial_t\rho = -\gamma\operatorname{div}\vect E = -(\gamma/\varepsilon_0)\rho. ∎

Method 10.13 (From the local to the integral law)

To find the resistance of a conductor of arbitrary shape: (1) use the symmetry to write j→\vect j (stationary: its flux is conserved, so in a tube j×j \times section =I= I); (2) E→=j→/γ\vect E = \vect j/\gamma; (3) integrate E→\vect E along a line from one electrode to the other: U=∫E→⋅ ⁣dl→U = \int\vect E\cdot \dd\vect l, and R=U/IR = U/I; (4) check with the Joule power ∫j2/γ  ⁣dτ=RI2\int j^2/\gamma\, \dd\tau = RI^2. The same steps give the leakage of a cable, the resistance of the ground around a lightning rod, or of a conical contact.

10.5 Exercises

Exercise 10.1 ★

A 2.5 mm22.5\,\mathrm{mm}^{2} copper wire carries 16 A16\,\mathrm{A}. Current density, drift speed (n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}), time for an electron to travel the 20 m20\,\mathrm{m} to a socket; electric field in the wire (γ=6.0×107 S/m\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}), voltage drop over 20 m20\,\mathrm{m} and power lost.

Solution

Solution of Exercise 10.1.

j=16/2.5×10−6=6.4×106 A/m2j = 16/2.5 \times 10^{-6} = 6.4 \times 10^{6}\,\mathrm{A}/\mathrm{m}^{2}; v=j/ne=0.47 mm/sv = j/ne = 0.47\,\mathrm{mm}/\mathrm{s}; 20 m20\,\mathrm{m} in 4.3×1044.3 \times 10^4 s, twelve hours. E=j/γ=0.11 V/mE = j/\gamma = 0.11\,\mathrm{V}/\mathrm{m}; U=2.1 VU = 2.1\,\mathrm{V}; P=UI=34 WP = UI = 34\,\mathrm{W}.

Exercise 10.2 ★

Current densities: a lightning stroke (30 kA30\,\mathrm{kA}, radius 1 cm1\,\mathrm{cm}); an electron beam (2 mA2\,\mathrm{mA}, 0.5 mm20.5\,\mathrm{mm}^{2}); a nerve axon (1 nA1\,\mathrm{nA}, radius 2 µm2\,\text{µ}\mathrm{m}); the Earth’s magnetosphere ring current (1 MA1\,\mathrm{MA} through about 1×1014 m21 \times 10^{14}\,\mathrm{m}^{2}). Which exceed the wire of the previous exercise?

Solution

Solution of Exercise 10.2.

Lightning 3×104/π10−4=1×108 A/m23 \times 10^4/\pi10^{-4} = 1 \times 10^{8}\,\mathrm{A}/\mathrm{m}^{2}; beam 4×103 A/m24 \times 10^{3}\,\mathrm{A}/\mathrm{m}^{2}; axon 10−9/1.3×10−11=80 A/m210^{-9}/1.3 \times 10^{-11} = 80\,\mathrm{A}/\mathrm{m}^{2}; ring current 1×10−8 A/m21 \times 10^{-8}\,\mathrm{A}/\mathrm{m}^{2}. Only the lightning exceeds the wire’s 6×1066 \times 10^6.

Exercise 10.3 ★

Resistivities at 20∘C20{}^{\circ}\mathrm{C}: copper 1.7×10−8 Ω m1.7 \times 10^{-8}\,\Omega\,\mathrm{m}, aluminium 2.8×10−8 Ω m2.8 \times 10^{-8}\,\Omega\,\mathrm{m}, iron 1.0×10−7 Ω m1.0 \times 10^{-7}\,\Omega\,\mathrm{m}, nichrome 1.1×10−6 Ω m1.1 \times 10^{-6}\,\Omega\,\mathrm{m}. Resistance of 100 m100\,\mathrm{m} of 2.5 mm22.5\,\mathrm{mm}^{2} in each; section of aluminium equivalent to 2.5 mm22.5\,\mathrm{mm}^{2} of copper, and the mass ratio (densities 89008900 and 2700 kg/m32700\,\mathrm{kg}/\mathrm{m}^{3}); length of 0.5 mm0.5\,\mathrm{mm} nichrome wire for a 1 kW1\,\mathrm{kW}, 230 V230\,\mathrm{V} heater.

Solution

Solution of Exercise 10.3.

R=ρeL/SR = \rho_eL/S: 0.680.68, 1.11.1, 4.04.0, 44 Ω44\,\Omega. Aluminium: 2.5×2.8/1.7=4.1 mm22.5 \times 2.8/1.7 = 4.1\,\mathrm{mm}^{2}, mass ratio (4.1×2700)/(2.5×8900)=0.50(4.1 \times 2700)/(2.5 \times 8900) = 0.50 — half the mass. Heater: R=2302/1000=53 ΩR = 230^2/1000 = 53\,\Omega, S=0.196 mm2S = 0.196\,\mathrm{mm}^{2}, L=RS/ρe=9.4 mL = RS/\rho_e = 9.4\,\mathrm{m}.

Exercise 10.4 ★

Drude: τ\tau, mobility and mean free path (random speed 1.6×106 m/s1.6 \times 10^{6}\,\mathrm{m}/\mathrm{s}) for copper (γ=6.0×107 S/m\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}, n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}) and silver (γ=6.3×107 S/m\gamma = 6.3 \times 10^{7}\,\mathrm{S}/\mathrm{m}, n=5.9×1028 m−3n = 5.9 \times 10^{28}\,\mathrm{m}^{-3}); conductivity of a silicon sample with n=1×1022 m−3n = 1 \times 10^{22}\,\mathrm{m}^{-3} and μ=0.14 m2/(V s)\mu = 0.14\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s}); drift speed in that sample at 1 kV/m1\,\mathrm{kV}/\mathrm{m}.

Solution

Solution of Exercise 10.4.

τ=mγ/ne2\tau = m\gamma/ne^2: copper 2.5×10−14 s2.5 \times 10^{-14}\,\mathrm{s}, μ=eτ/m=4.4×10−3 m2/(V s)\mu = e\tau/m = 4.4 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s}), ℓ=40 nm\ell = 40\,\mathrm{nm}; silver 3.8×10−14 s3.8 \times 10^{-14}\,\mathrm{s}, 6.7×10−3 m2/(V s)6.7 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s}), 61 nm61\,\mathrm{nm}. Silicon: γ=neμ=1022×1.6×10−19×0.14=220 S/m\gamma = ne\mu = 10^{22} \times 1.6 \times 10^{-19} \times 0.14 = 220\,\mathrm{S}/\mathrm{m}; v=μE=140 m/sv = \mu E = 140\,\mathrm{m}/\mathrm{s}.

Exercise 10.5 ★★

A capacitor of plates SS is charged through a wire by I(t)I(t). (a) Apply the integral conservation law to a closed surface enclosing one plate: relate II and the plate charge QQ. (b) Why is div⁡j→≠0\operatorname{div}\vect j \ne 0 at the plate? (c) A spherical electrode of radius aa in a conducting medium emits a radial current II: j→(r)\vect j(r); check div⁡j→=0\operatorname{div} \vect j = 0 for r>ar > a (use div⁡j→=1r2∂r(r2jr)\operatorname{div}\vect j = \frac1{r^2}\partial_r(r^2j_r) for a radial field). (d) Where, then, does the charge conservation "break"?

Solution

Solution of Exercise 10.5.

(a) Current in, none out:  ⁣dQ/ ⁣dt=I\dd Q/\dd t = I. (b) Charge accumulates: ∂tρ≠0\partial_t \rho \ne 0. (c) j→=(I/4πr2)e→r\vect j = (I/4\pi r^2)\vect e_r: div⁡j→=1r2∂r(I/4π)=0\operatorname{div}\vect j = \frac1{r^2} \partial_r(I/4\pi) = 0. (d) Only at the electrode, where the current is injected from the wire — a source of the flux.

Exercise 10.6 ★★

Leakage of a coaxial cable. The insulator between the conductors (radii a=1 mma = 1\,\mathrm{mm}, b=3.5 mmb = 3.5\,\mathrm{mm}) has γ=1×10−13 S/m\gamma = 1 \times 10^{-13}\,\mathrm{S}/\mathrm{m}; a voltage UU is applied between them, a radial leakage current II flows per length LL. (a) j→(r)\vect j(r) and E→(r)\vect E(r). (b) Show that the resistance of a length LL is R=ln⁡(b/a)/2πγLR = \ln(b/a)/2\pi\gamma L; value per kilometre. (c) Leakage current and power per kilometre at 1 kV1\,\mathrm{kV}. (d) Compare with the capacitance per length 2πε0εr/ln⁡(b/a)2\pi\varepsilon_0\varepsilon_r/\ln(b/a): show RC=ε0εr/γRC = \varepsilon_0\varepsilon_r/\gamma and interpret with the relaxation time.

Solution

Solution of Exercise 10.6.

(a) j=I/2πrLj = I/2\pi rL, E=j/γE = j/\gamma, radial. (b) U=∫abE  ⁣dr=Iln⁡(b/a)/2πγLU = \int_a^bE\,\dd r = I\ln(b/a)/ 2\pi\gamma L: R=ln⁡3.5/(2π×10−13×103)=2×109 ΩR = \ln3.5/(2\pi \times 10^{-13} \times 10^3) = 2 \times 10^{9}\,\Omega per kilometre. (c) 0.5 µA0.5\,\text{µ}\mathrm{A}, 0.5 mW0.5\,\mathrm{mW}. (d) RC=ε0εr/γ=200 sRC = \varepsilon_0\varepsilon_r/\gamma = 200\,\mathrm{s} for εr=2.3\varepsilon_r = 2.3: the cable, charged and isolated, discharges through its own insulator with the relaxation time of that insulator.

Exercise 10.7 ★★

Complex conductivity of copper (τ=2.5×10−14 s\tau = 2.5 \times 10^{-14}\,\mathrm{s}). (a) ωτ\omega\tau at 50 Hz50\,\mathrm{Hz}, 10 GHz10\,\mathrm{GHz} (microwaves), 30 THz30\,\mathrm{THz} (infrared), 5×1014 Hz5 \times 10^{14}\,\mathrm{Hz} (green light). (b) Modulus and phase of γ‾/γ0\underline\gamma/\gamma_0 in each case. (c) Up to what frequency is Ohm’s law good to 1%1\%? (d) In the visible, write γ‾\underline\gamma and show that j→‾\underline{\vect j} is in quadrature with E→‾\underline{\vect E}: mean Joule power?

Solution

Solution of Exercise 10.7.

(a) ωτ\omega\tau: 8×10−128 \times 10^{-12}, 1.6×10−31.6 \times 10^{-3}, 4.74.7, 7979. (b) ∣γ‾∣/γ0=1/1+ω2τ2|\underline\gamma|/\gamma_0 = 1/\sqrt{1 + \omega^2\tau^2}: 11, 11, 0.210.21, 0.0130.013; phase −arctan⁡ωτ-\arctan\omega\tau: 00, −0.09∘-0.09{}^{\circ}, −78∘-78{}^{\circ}, −89∘-89{}^{\circ}. (c) ωτ≤0.14\omega\tau \le 0.14: f≤0.9 THzf \le 0.9\,\mathrm{THz}. (d) γ‾≈ne2/imω\underline\gamma \approx ne^2/\iu m\omega: j→‾\underline{\vect j} lags E→‾\underline{\vect E} by 90∘90{}^{\circ}, ⟨j→⋅E→⟩=0\langle\vect j\cdot\vect E\rangle = 0: no Joule heating — the electrons oscillate freely, as in a plasma.

Exercise 10.8 ★★

A 0.20 mm0.20\,\mathrm{mm} copper fuse wire (copper: γ=6.0×107 S/m\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}, ρ=8900 kg/m3\rho = 8900\,\mathrm{kg}/\mathrm{m}^{3}, c=385 J/(kg K)c = 385\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K}), melts at 1085∘C1085{}^{\circ}\mathrm{C}, latent heat 205 kJ/kg205\,\mathrm{kJ}/\mathrm{kg}). (a) Current density and Joule power per unit volume at 10 A10\,\mathrm{A}. (b) Time to melt from 20∘C20{}^{\circ}\mathrm{C} if no heat escapes (ignore the rise of resistivity with temperature). (c) Same at 30 A30\,\mathrm{A}. (d) In reality the wire loses heat to the air: how does that set the rated current of a fuse, and why are fuses sealed in a sand-filled cartridge?

Solution

Solution of Exercise 10.8.

(a) S=3.1×10−8 m2S = 3.1 \times 10^{-8}\,\mathrm{m}^{2}, j=3.2×108 A/m2j = 3.2 \times 10^{8}\,\mathrm{A}/\mathrm{m}^{2}, p=j2/γ=1.7×109 W/m3p = j^2/\gamma = 1.7 \times 10^{9}\,\mathrm{W}/\mathrm{m}^{3}. (b) Energy to melt per unit volume ρ(cΔT+Lf)=8900(4.1×105+2.05×105)=5.5×109 J/m3\rho(c\Delta T + L_f) = 8900(4.1 \times 10^5 + 2.05 \times 10^5) = 5.5 \times 10^{9}\,\mathrm{J}/\mathrm{m}^{3}: 3.2 s3.2\,\mathrm{s}. (c) Nine times faster: 0.36 s0.36\,\mathrm{s}. (d) The rated current is the one whose Joule heating the air just evacuates below the melting point; the sand quenches the arc that would otherwise keep conducting across the gap after the wire melts.

Exercise 10.9 ★★

Relaxation. (a) Derive ∂tρ=−(γ/ε0)ρ\partial_t\rho = -(\gamma/\varepsilon_0)\rho from charge conservation, Ohm’s law and div⁡E→=ρ/ε0\operatorname{div}\vect E = \rho/\varepsilon_0. (b) τr\tau_r for copper, sea water (γ=5 S/m\gamma = 5\,\mathrm{S}/\mathrm{m}), tap water (5×10−2 S/m5 \times 10^{-2}\,\mathrm{S}/\mathrm{m}), glass (1×10−12 S/m1 \times 10^{-12}\,\mathrm{S}/\mathrm{m}), PTFE (1×10−16 S/m1 \times 10^{-16}\,\mathrm{S}/\mathrm{m}). (c) Up to what frequency is each a "good conductor" (charge relaxing within a period)? (d) A charged glass rod keeps its charge for minutes: is that consistent?

Solution

Solution of Exercise 10.9.

(a) ∂tρ=−div⁡j→=−γdiv⁡E→=−γρ/ε0\partial_t\rho = -\operatorname{div}\vect j = -\gamma\operatorname{div}\vect E = -\gamma\rho/ \varepsilon_0. (b) ε0/γ\varepsilon_0/\gamma: 1.5×10−19 s1.5 \times 10^{-19}\,\mathrm{s}, 1.8×10−12 s1.8 \times 10^{-12}\,\mathrm{s}, 1.8×10−10 s1.8 \times 10^{-10}\,\mathrm{s}, 9 s9\,\mathrm{s}, 9×1049 \times 10^4 s. (c) f≲1/2πτrf \lesssim 1/2\pi\tau_r: 101810^{18} Hz (the Drude model fails long before), 90 GHz90\,\mathrm{GHz}, 0.9 GHz0.9\,\mathrm{GHz}, 0.02 Hz0.02\,\mathrm{Hz}, 2×10−6 Hz2 \times 10^{-6}\,\mathrm{Hz}. (d) A glass rod would lose its charge in a minute through its volume; real dry glass is nearer 10−1410^{-14} S/m (hours), and it is surface humidity that usually discharges it — consistent in order of magnitude.

Exercise 10.10 ★★★

Grounding. A hemispherical electrode of radius a=0.50 ma = 0.50\,\mathrm{m} is buried flush with the surface of soil of conductivity γ=0.010 S/m\gamma = 0.010\,\mathrm{S}/\mathrm{m}; a current II enters it and spreads radially into the half-space. (a) j→(r)\vect j(r), E→(r)\vect E(r), potential V(r)V(r) with V(∞)=0V(\infty) = 0. (b) Resistance of the grounding, R=1/2πγaR = 1/2\pi\gamma a; value. (c) A lightning stroke of 20 kA20\,\mathrm{kA} hits the rod: potential of the rod; potential at r=10 mr = 10\,\mathrm{m} and 10.7 m10.7\,\mathrm{m}; the "step voltage" between the feet of a person standing there (feet 0.7 m0.7\,\mathrm{m} apart, radially). (d) Why are cattle, with their four legs far apart, killed by nearby strikes more often than people?

Solution

Solution of Exercise 10.10.

(a) j→=(I/2πr2)e→r\vect j = (I/2\pi r^2)\vect e_r, E=I/2πγr2E = I/2\pi\gamma r^2, V=I/2πγrV = I/2\pi\gamma r. (b) R=V(a)/I=1/2πγa=32 ΩR = V(a) /I = 1/2\pi\gamma a = 32\,\Omega. (c) Vrod=640 kVV_{\text{rod}} = 640\,\mathrm{kV}; V(10)=32 kVV(10) = 32\,\mathrm{kV}, V(10.7)=30 kVV(10.7) = 30\,\mathrm{kV}: 2.1 kV2.1\,\mathrm{kV} between the feet. (d) Front and hind legs 1.5 m1.5\,\mathrm{m} apart along the radius: 44–5 kV5\,\mathrm{kV}, and the current path crosses the heart.

Exercise 10.11 ★★★

Hall sensor. A strip of thickness tt and width ww carries II along xx in a field BB along zz. (a) From E→H=−j→∧B→/nq\vect E_H = -\vect j\wedge\vect B /nq, show that the Hall voltage across the width is VH=IB/nqtV_H = IB/nqt. (b) Copper strip, t=0.10 mmt = 0.10\,\mathrm{mm}, 1 A1\,\mathrm{A}, 1 T1\,\mathrm{T}: VHV_H. (c) A doped semiconductor with n=1×1022 m−3n = 1 \times 10^{22}\,\mathrm{m}^{-3}, same geometry: VHV_H; why sensors use semiconductors. (d) A sensor with n=1×1022 m−3n = 1 \times 10^{22}\,\mathrm{m}^{-3}, t=20 µmt = 20\,\text{µ}\mathrm{m}, fed with 1 mA1\,\mathrm{mA}, reads 10 µV10\,\text{µ}\mathrm{V}: field measured. What does the sign of VHV_H tell?

Solution

Solution of Exercise 10.11.

(a) EH=jB/nqE_H = jB/nq with j=I/wtj = I/wt: VH=EHw=IB/nqtV_H = E_Hw = IB/nqt. (b) 1/(8.5×1028×1.6×10−19×10−4)=0.7 µV1/(8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-4}) = 0.7\,\text{µ}\mathrm{V}. (c) 6.3 mV6.3\,\mathrm{mV}: VH∝1/nV_H \propto 1/n, and semiconductors have 10610^6 times fewer carriers. (d) B=VHnqt/I=10−5×1022×1.6×10−19×2×10−5/10−3=0.32 TB = V_Hnqt/I = 10^{-5} \times 10^{22} \times 1.6 \times 10^{-19} \times 2 \times 10^{-5}/10^{-3} = 0.32\,\mathrm{T}; the sign of VHV_H gives the sign of the carriers (for a known B→\vect B), or the direction of B→\vect B (for a known sensor).

Exercise 10.12 ★★★

Electrolyte. In water, Na+^+ and Cl−^- ions have mobilities μ+=5.2×10−8 m2/(V s)\mu_+ = 5.2 \times 10^{-8}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s}) and μ−=7.9×10−8 m2/(V s)\mu_- = 7.9 \times 10^{-8}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s}). (a) Show that γ=ne(μ++μ−)\gamma = ne(\mu_+ + \mu_-) for nn ion pairs per unit volume. (b) Sea water, 0.5 mol/L0.5\,\mathrm{mol}/\mathrm{L} of NaCl: γ\gamma; compare with the measured 5 S/m5\,\mathrm{S}/\mathrm{m}. (c) Drift speeds at 10 V/m10\,\mathrm{V}/\mathrm{m}; which ion carries more of the current? (d) A column of sea water, 1 cm21\,\mathrm{cm}^{2} by 10 cm10\,\mathrm{cm}, carries 1 A1\,\mathrm{A}: voltage, power, and the time to warm it by 10 K10\,\mathrm{K} (c=4.0 kJ/(kg K)c = 4.0\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})); why electrophoresis gels are cooled.

Solution

Solution of Exercise 10.12.

(a) j→=neμ+E→+(−ne)(−μ−E→)\vect j = ne\mu_+\vect E + (-ne)(-\mu_-\vect E): both species drift in opposite directions and carry current in the same direction. (b) n=0.5×103×6.02×1023=3×1026 m−3n = 0.5 \times 10^3 \times 6.02 \times 10^{23} = 3 \times 10^{26}\,\mathrm{m}^{-3}: γ=3×1026×1.6×10−19×1.31×10−7=6.3 S/m\gamma = 3 \times 10^{26} \times 1.6 \times 10^{-19} \times 1.31 \times 10^{-7} = 6.3\,\mathrm{S}/\mathrm{m} (the ions hinder each other at this concentration: 5 S/m5\,\mathrm{S}/\mathrm{m} measured). (c) 0.52 µm/s0.52\,\text{µ}\mathrm{m}/\mathrm{s} and 0.79 µm/s0.79\,\text{µ}\mathrm{m}/\mathrm{s}: Cl−^- carries 60%60\%. (d) R=L/γS=200 ΩR = L/\gamma S = 200\,\Omega, U=200 VU = 200\,\mathrm{V}, P=200 WP = 200\,\mathrm{W} into 10 g10\,\mathrm{g} of water: 10 K10\,\mathrm{K} in 2 s2\,\mathrm{s} — hence the cooling plates of electrophoresis tanks.

10.6 Problem: The copper wire, from the atom to the grid

Problem 10.1

Weekend problem — one metre of household wire examined at four scales: the electron, the local laws, the circuit, and the power line

Copper: molar mass 63.5 g/mol63.5\,\mathrm{g}/\mathrm{mol}, density 8900 kg/m38900\,\mathrm{kg}/\mathrm{m}^{3}, one conduction electron per atom, conductivity γ=6.0×107 S/m\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m} at 20∘C20{}^{\circ}\mathrm{C}, rising resistivity ρe(T)=ρe(20)[1+α(T−20)]\rho_e(T) = \rho_e(20)[1 + \alpha(T - 20)] with α=4.0×10−3 K−1\alpha = 4.0 \times 10^{-3}\,\mathrm{K}^{-1}; c=385 J/(kg K)c = 385\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K}). Wire: section S=2.5 mm2S = 2.5\,\mathrm{mm}^{2}, length 1 m1\,\mathrm{m}, current 16 A16\,\mathrm{A}.

Part I — The electrons.

  1. Number density nn of conduction electrons.
  2. Current density and drift velocity; how long an electron needs to cross the metre.
  3. Drude’s τ\tau and the mobility; mean free path with a random speed of 1.6×106 m/s1.6 \times 10^{6}\,\mathrm{m}/\mathrm{s}; how many atomic spacings (0.26 nm0.26\,\mathrm{nm}) is that, and what does it say about what the electrons collide with?
  4. The thermal vibrations grow with TT, so τ∝1/T\tau \propto 1/T roughly: check against α\alpha near room temperature ( ⁣dρe/ρe= ⁣dT/T\dd\rho_e/\rho_e = \dd T/T would give α=1/293\alpha = 1/293).
  5. Electric field in the wire; total voltage over the metre; the kinetic energy of the drift per electron compared with kBTk_BT — is the gas of electrons disturbed by the current?
  6. Number of electrons crossing a section of the wire per second.
  7. Charge of the conduction electrons in the metre of wire; the force the field exerts on all of them together; compare with the weight of the wire. Where does that force end up (think of the collisions)?

Part II — The local laws.

  1. Joule power per unit volume and per metre.
  2. The wire, in air, loses heat at a rate h(T−Ta)h(T - T_a) per unit surface with h=10 W/(m2 K)h = 10\,\mathrm{W}/(\mathrm{m}^{2}\,\mathrm{K}) (convection): equilibrium temperature rise; check whether the resistivity correction matters.
  3. The same wire buried in insulation that lets only h=2 W/(m2 K)h = 2\,\mathrm{W}/(\mathrm{m}^{2}\,\mathrm{K}) through: new temperature rise, with the resistivity correction included (solve the balance).
  4. Left without any cooling, rate of temperature rise at 16 A16\,\mathrm{A} and at 160 A160\,\mathrm{A} (a short circuit); time to reach the melting point (1085∘C1085{}^{\circ}\mathrm{C}) in the second case; what protects the installation?
  5. At 50 Hz50\,\mathrm{Hz}, ωτ=?\omega\tau = ? Is Ohm’s law affected? (The skin effect of Chapter 14 is a different matter.)
  6. Write the relaxation time ε0/γ\varepsilon_0/\gamma of copper; what does it imply for the charge density inside the wire, and where does the charge that produces the field EE inside the wire sit?

Part III — Conservation of charge in the circuit. The wire feeds a capacitor C=100 µFC = 100\,\text{µ}\mathrm{F} through a switch; the current is I(t)=I0e−t/τcI(t) = I_0\eu^{-t/\tau_c} with I0=16 AI_0 = 16\,\mathrm{A}, τc=1 ms\tau_c = 1\,\mathrm{ms}.

  1. Charge that reaches the plate; final voltage.
  2. Apply  ⁣dQ/ ⁣dt=−Iout\dd Q/\dd t = -I_{\text{out}} to a closed surface around the plate: which side carries a conduction current, which does not? What quantity must therefore vary in the gap?
  3. At t=0.5 mst = 0.5\,\mathrm{ms}, rate of change of the plate charge; if the plates are 1 cm21\,\mathrm{cm}^{2} apart by 0.1 mm0.1\,\mathrm{mm} in air, rate of change of the field between them.
  4. Show that ε0  ⁣dE/ ⁣dt×\varepsilon_0\,\dd E/\dd t \times plate area equals the current in the wire — the quantity that Chapter 11 adds to Ampère’s law.
  5. The wire is cut and the two ends are 1 mm1\,\mathrm{mm} apart in air: what current flows (air γ≈1×10−14 S/m\gamma \approx 1 \times 10^{-14}\,\mathrm{S}/\mathrm{m})? At what field does air break down (3 MV/m3\,\mathrm{MV}/\mathrm{m}): what voltage across the gap makes a spark?

Part IV — The line. A 230 V230\,\mathrm{V} line of 10 mm210\,\mathrm{mm}^{2} copper, 500 m500\,\mathrm{m} long (go and return: 1 km1\,\mathrm{km} of wire), feeds a farm drawing 40 A40\,\mathrm{A}.

  1. Resistance, voltage drop, and power lost in the line; fraction of the delivered power.
  2. The same power at 20 kV20\,\mathrm{kV} with a transformer at each end: current, drop and loss; conclude why transmission is done at high voltage.
  3. Aluminium (γ=3.6×107 S/m\gamma = 3.6 \times 10^{7}\,\mathrm{S}/\mathrm{m}, 2700 kg/m32700\,\mathrm{kg}/\mathrm{m}^{3}) is used on pylons: section for the same resistance as 10 mm210\,\mathrm{mm}^{2} of copper; mass per kilometre for both; why aluminium wins up there and copper in the wall.
  4. A Hall sensor clamped around the line reads its field: for a clamp meter with n=1×1022 m−3n = 1 \times 10^{22}\,\mathrm{m}^{-3}, t=0.1 mmt = 0.1\,\mathrm{mm}, fed with 5 mA5\,\mathrm{mA} in a 0.1 T0.1\,\mathrm{T} field concentrated by an iron core, the Hall voltage.
  5. After a hot day’s load the wire sits at 80∘C80{}^{\circ}\mathrm{C}: its resistance then, and the extra loss.
  6. Lightning hits the line: 20 kA20\,\mathrm{kA} for 100 µs100\,\text{µ}\mathrm{s}. Energy dissipated in the 1 km1\,\mathrm{km} of wire; temperature rise if none escapes; current density — compare with Part I.
  7. Sum up: the five quantities (nn, τ\tau, γ\gamma, jj, EE) that describe the wire at each scale, and the law that links each pair.
Solution

Solution of Problem 10.1.

1. n=ρNA/M=8.4×1028 m−3n = \rho N_A/M = 8.4 \times 10^{28}\,\mathrm{m}^{-3}.

2. j=6.4×106 A/m2j = 6.4 \times 10^{6}\,\mathrm{A}/\mathrm{m}^{2}, v=j/ne=0.47 mm/sv = j/ne = 0.47\,\mathrm{mm}/\mathrm{s}; 35 min35\,\mathrm{min} per metre.

3. τ=mγ/ne2=2.5×10−14 s\tau = m\gamma/ne^2 = 2.5 \times 10^{-14}\,\mathrm{s}, μ=4.4×10−3 m2/(V s)\mu = 4.4 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s}), ℓ=40 nm\ell = 40\,\mathrm{nm}, 150150 spacings: the electrons collide with defects and thermal vibrations, not with the atoms.

4. ρe∝T\rho_e \propto T gives α=1/293=3.4×10−3 K−1\alpha = 1/293 = 3.4 \times 10^{-3}\,\mathrm{K}^{-1}, close to the measured 4×10−34 \times 10^{-3}.

5. E=0.11 V/mE = 0.11\,\mathrm{V}/\mathrm{m}, U=0.11 VU = 0.11\,\mathrm{V}; 12mv2=1×10−37 J\tfrac12mv^2 = 1 \times 10^{-37}\,\mathrm{J} against kBT=4×10−21 Jk_BT = 4 \times 10^{-21}\,\mathrm{J}: the electron gas does not notice the current.

6. I/e=1×1020 I/e = 1 \times 10^{20}\, electrons per second.

7. Q=neSL=3.4×104 CQ = neSL = 3.4 \times 10^{4}\,\mathrm{C}; QE=3.6 kNQE = 3.6\,\mathrm{kN}, against a weight of 0.2 N0.2\,\mathrm{N}. The field pulls the positive lattice with the opposite force; the electrons hand their momentum to the lattice by collisions: the wire as a whole feels nothing.

8. p=j2/γ=6.8×105 W/m3p = j^2/\gamma = 6.8 \times 10^{5}\,\mathrm{W}/\mathrm{m}^{3}, 1.7 W/m1.7\,\mathrm{W}/\mathrm{m}.

9. Surface πd=5.6×10−3 m2\pi d = 5.6 \times 10^{-3}\,\mathrm{m}^{2} per metre: ΔT=1.7/(10×5.6×10−3)=30 K\Delta T = 1.7/(10 \times 5.6 \times 10^{-3}) = 30\,\mathrm{K}; the resistivity rises 12%12\%, the power to 1.9 W1.9\,\mathrm{W}: ΔT≈35 K\Delta T \approx 35\,\mathrm{K}.

10. ΔT=1.7(1+0.004ΔT)/(2×5.6×10−3)\Delta T = 1.7(1 + 0.004\Delta T)/(2 \times 5.6 \times 10^{-3}): ΔT(1−0.6)=150\Delta T(1 - 0.6) = 150, ΔT≈400 K\Delta T \approx 400\,\mathrm{K} — near thermal runaway: buried cables must be derated.

11. p/ρc=0.2 K/sp/\rho c = 0.2\,\mathrm{K}/\mathrm{s}; at 160 A160\,\mathrm{A}, 20 K/s20\,\mathrm{K}/\mathrm{s}: melting in about a minute (less as ρe\rho_e rises); the breaker opens in milliseconds.

12. ωτ=8×10−12\omega\tau = 8 \times 10^{-12}: Ohm’s law is untouched.

13. ε0/γ=1.5×10−19 s\varepsilon_0/\gamma = 1.5 \times 10^{-19}\,\mathrm{s}: no charge inside; the field in the wire is made by charges on its surface, whose density varies along the wire.

14. Q=I0τc=16 mCQ = I_0\tau_c = 16\,\mathrm{mC}, V=Q/C=160 VV = Q/C = 160\,\mathrm{V}.

15. The wire side carries II, the gap side no conduction current: the charge grows, and with it the electric field in the gap.

16. I=16e−0.5=9.7 AI = 16\eu^{-0.5} = 9.7\,\mathrm{A}; E=Q/ε0AE = Q/\varepsilon_0A,  ⁣dE/ ⁣dt=I/ε0A=9.7/(8.85×10−12×10−4)=1.1×1016 V m−1 s−1\dd E/\dd t = I/\varepsilon_0A = 9.7/(8.85 \times 10^{-12} \times 10^{-4}) = 1.1 \times 10^{16}\,\mathrm{V}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}.

17. ε0A  ⁣dE/ ⁣dt=I\varepsilon_0A\,\dd E/\dd t = I: the displacement current.

18. j=γE=10−14×2.3×105=2×10−9 A/m2j = \gamma E = 10^{-14} \times 2.3 \times 10^5 = 2 \times 10^{-9}\,\mathrm{A}/\mathrm{m}^{2}, 10−1410^{-14} A through the section: nothing; breakdown at 3 kV3\,\mathrm{kV} across the millimetre.

19. R=1000/(6×107×10−5)=1.7 ΩR = 1000/(6 \times 10^7 \times 10^{-5}) = 1.7\,\Omega; ΔU=67 V\Delta U = 67\,\mathrm{V}; P=2.7 kWP = 2.7\,\mathrm{kW}, 29%29\% of the 9.2 kW9.2\,\mathrm{kW} delivered.

20. I=0.46 AI = 0.46\,\mathrm{A}, ΔU=0.8 V\Delta U = 0.8\,\mathrm{V}, 0.35 W0.35\,\mathrm{W}: transmission at high voltage divides the loss by (U2/U1)2(U_2/U_1)^2.

21. 10×6/3.6=17 mm210 \times 6/3.6 = 17\,\mathrm{mm}^{2}; 89 kg/km89\,\mathrm{kg}/\mathrm{km} of copper against 45 kg/km45\,\mathrm{kg}/\mathrm{km} of aluminium: lighter and cheaper on pylons; in the wall copper’s smaller section fits the conduits and terminals.

22. VH=IB/nqt=5×10−3×0.1/(1022×1.6×10−19×10−4)=3.1 mVV_H = IB/nqt = 5 \times 10^{-3} \times 0.1/(10^{22} \times 1.6 \times 10^{-19} \times 10^{-4}) = 3.1\,\mathrm{mV}.

23. +24%+24\%: 2.1 Ω2.1\,\Omega, 3.3 kW3.3\,\mathrm{kW} lost.

24. RI2t=1.7×4×108×10−4=67 kJRI^2t = 1.7 \times 4 \times 10^8 \times 10^{-4} = 67\,\mathrm{kJ} into 89 kg89\,\mathrm{kg}: 2 K2\,\mathrm{K}. j=2×109 A/m2j = 2 \times 10^{9}\,\mathrm{A}/\mathrm{m}^{2}, three hundred times the household value.

25. nn (matter) and τ\tau (collisions) give γ=ne2τ/m\gamma = ne^2\tau/m; γ\gamma links jj and EE (Ohm); jj integrates to II, EE to UU (the circuit); j2/γj^2/\gamma is the heat.

Terms defined in this chapter

See all 393 terms in the glossary