The Workbook

Mutual induction: the alternating current in the primary makes an alternating flux in the core. This changing flux induces an emf in every turn of the secondary. A steady direct current gives a constant flux, so no emf is induced in the secondary.

1. The same flux threads every turn, so the emf per turn is the same in both coils: \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\). \(V_s = 220 \times \dfrac{2000}{200} = 2200\) V.

2. \(N_s > N_p\), so the voltage is raised: step-up.

Ideal: \(V_pI_p = V_sI_s\), so \(I_p = \dfrac{2200 \times 1}{220} = 10\) A. The voltage goes up ten times and the current comes down ten times.

\(\dfrac{N_p}{N_s} = \dfrac{V_p}{V_s} = \dfrac{240}{12} = 20\). With \(N_p = 1000\), \(N_s = 50\) turns.

Soft iron has a high permeability, so it carries almost all the flux from the primary to the secondary. It also magnetises and demagnetises easily, with little hysteresis loss. Laminating the core breaks up the paths of eddy currents. Insulation stops the windings shorting through the conducting core.

Copper loss, \(I^2R\) heating in the windings: thick wire. Eddy current loss in the core: laminations. Hysteresis loss: soft iron or special alloys with a thin loop. Flux leakage, flux that misses the secondary: winding the coils one over the other on a closed core.

Efficiency \(= \dfrac{900}{1000} = 90\%\). Loss \(= 1000 - 900 = 100\) W, mostly as heat.

1. \(I = \dfrac{P}{V} = \dfrac{10^{5}}{10^{3}} = 100\) A. Loss \(= I^2R = 10^{4} \times 2 = 20\) kW.

2. \(I = \dfrac{10^{5}}{10^{4}} = 10\) A. Loss \(= 100 \times 2 = 200\) W.

3. \(\dfrac{20}{100} = 20\%\) at \(1\) kV but only \(\dfrac{0.2}{100} = 0.2\%\) at \(10\) kV. Ten times the voltage cuts the loss a hundred times.

For a given power, a higher voltage means a smaller current, and the \(I^2R\) loss in the lines falls as the square of the current. High voltages are dangerous and useless for appliances, so transformers step the voltage down again near homes and factories.

Input \(= 220 \times 2 = 440\) W. Output \(= 0.9 \times 440 = 396\) W. \(I_s = \dfrac{396}{11} = 36\) A.

With no secondary current, the primary is just a coil of large inductance across the mains. Its back emf almost equals the supply voltage, so only a small magnetising current flows, enough to set up the flux. The primary current grows only when the secondary draws power.

\(V_pI_p = V_sI_s\) with \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\) gives \(\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}\). Power cannot be created, so if the voltage is multiplied by a factor the current must be divided by the same factor.

1. \(\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{0.1 \times 10^{-5}}} = 1000\) rad/s. \(f = \dfrac{\omega}{2\pi} \approx 159\) Hz.

2. \(T = \dfrac{2\pi}{\omega} \approx 6.3 \times 10^{-3}\) s.

\(U_0 = \dfrac{q_0^2}{2C} = \dfrac{(10^{-4})^2}{2 \times 10^{-5}} = 5 \times 10^{-4}\) J. \(I_0 = q_0\omega = 10^{-4} \times 1000 = 0.1\) A. Check: \(\dfrac{1}{2}LI_0^2 = \dfrac{1}{2} \times 0.1 \times 0.01 = 5 \times 10^{-4}\) J.

\(q = q_0\cos\omega t\) and \(i = \dfrac{dq}{dt} = -q_0\omega\sin\omega t\). The current is largest when \(\sin\omega t = \pm 1\), at \(t = \dfrac{T}{4}, \dfrac{3T}{4}, \dots\); the charge is then zero and all the energy is in the inductor.

1. \(U_E = \dfrac{q_0^2}{2C}\cos^2\omega t\) and \(U_B = \dfrac{q_0^2}{2C}\sin^2\omega t\) each complete a cycle in half a period, so the energy shuttles at \(2f \approx 318\) Hz.

2. \(\cos^2 + \sin^2 = 1\), so the sum is always \(\dfrac{q_0^2}{2C} = 5 \times 10^{-4}\) J.

Round the loop the capacitor voltage equals the inductor's emf: \(\dfrac{q}{C} = -L\dfrac{di}{dt}\). With \(i = \dfrac{dq}{dt}\): \(L\dfrac{d^2q}{dt^2} + \dfrac{q}{C} = 0\). This has the same form as \(m\dfrac{d^2x}{dt^2} + kx = 0\). Here \(q\) stands in place of \(x\), \(L\) in place of \(m\) and \(\dfrac{1}{C}\) in place of \(k\). Its solution is simple harmonic, \(q = q_0\cos\omega t\) with \(\omega = \dfrac{1}{\sqrt{LC}}\), just as \(\omega = \sqrt{\dfrac{k}{m}}\).

Displacement \(x \leftrightarrow\) charge \(q\); velocity \(v \leftrightarrow\) current \(i\); mass \(m \leftrightarrow\) inductance \(L\). Spring constant \(k \leftrightarrow \dfrac{1}{C}\); kinetic energy \(\dfrac{1}{2}mv^2 \leftrightarrow \dfrac{1}{2}Li^2\); potential energy \(\dfrac{1}{2}kx^2 \leftrightarrow \dfrac{q^2}{2C}\).

\(C\) is four times larger, so \(\omega = \dfrac{1}{\sqrt{LC}}\) is halved: \(500\) rad/s. \(T = \dfrac{2\pi}{500} \approx 12.6\) ms, twice as long.

Real coils and wires have resistance, so part of the energy is turned into heat in every cycle and the amplitude decays. An oscillating circuit also radiates a little energy as electromagnetic waves. To keep it going, energy must be fed in, as in a radio transmitter.

1. A starts at its maximum, like the charge \(q_0\); B starts at zero, like the current. A is \(q\) and B is \(i\).

2. All the energy is magnetic when \(q = 0\) and \(i\) is largest, first at \(t = \dfrac{T}{4} \approx 1.6\) ms.

\(\omega = 2\pi \times 10^{6}\) rad/s. \(C = \dfrac{1}{\omega^2L} = \dfrac{1}{4\pi^2 \times 10^{12} \times 10^{-3}} \approx 2.5 \times 10^{-11}\) F \(= 25\) pF.

\(\dfrac{q^2}{2C} = \dfrac{1}{2}\cdot\dfrac{q_0^2}{2C}\), so \(q = \dfrac{q_0}{\sqrt2} = \dfrac{100}{1.41} \approx 71\) μC.

\(q = q_0\cos\omega t\) and \(i = -q_0\omega\sin\omega t\). \(\dfrac{q^2}{2C} = \dfrac{q_0^2}{2C}\cos^2\omega t\) and \(\dfrac{1}{2}Li^2 = \dfrac{1}{2}Lq_0^2\omega^2\sin^2\omega t = \dfrac{q_0^2}{2C}\sin^2\omega t\), using \(\omega^2 = \dfrac{1}{LC}\). Their sum is \(\dfrac{q_0^2}{2C}(\cos^2\omega t + \sin^2\omega t) = \dfrac{q_0^2}{2C}\), constant.

Ampère's law says \(\oint\vec B\cdot d\vec l\) equals \(\mu_0\) times the current through any surface bounded by the loop. A flat surface cut by the wire has current \(I\) through it. A bulging surface passing between the plates has no current through it. The same loop then gives \(\mu_0I\) and \(0\), so the law as it stood cannot be right.

The displacement current is \(I_d = \epsilon_0\dfrac{d\Phi_E}{dt}\), where \(\Phi_E\) is the electric flux through the surface. Maxwell added it to the conduction current: \(\oint\vec B\cdot d\vec l = \mu_0(I + I_d)\). Between the plates the growing flux gives an \(I_d\) equal to the wire current, so both surfaces now agree.

1. \(\Phi_E = EA\), so \(I_d = \epsilon_0A\dfrac{dE}{dt} = 8.85 \times 10^{-12} \times 0.1 \times 10^{12} = 0.885\) A.

2. Also \(0.885\) A. The charge on the plates is \(q = \epsilon_0AE\), so the wire current \(\dfrac{dq}{dt}\) is exactly the displacement current. The two together form a continuous current around the circuit.

Between the plates \(E = \dfrac{\sigma}{\epsilon_0} = \dfrac{q}{\epsilon_0A}\), so \(q = \epsilon_0AE = \epsilon_0\Phi_E\). Differentiating: \(\dfrac{dq}{dt} = \epsilon_0\dfrac{d\Phi_E}{dt} = I_d\). The conduction current in the wires equals the displacement current between the plates at every instant.

\(q = CV\), so \(I_d = \dfrac{dq}{dt} = C\dfrac{dV}{dt} = 2 \times 10^{-6} \times 10^{6} = 2\) A.

Gauss's law for electricity: \(\oint\vec E\cdot d\vec A = \dfrac{q}{\epsilon_0}\). Gauss's law for magnetism: \(\oint\vec B\cdot d\vec A = 0\) (no magnetic monopoles). Faraday's law: \(\oint\vec E\cdot d\vec l = -\dfrac{d\Phi_B}{dt}\). Ampère-Maxwell law: \(\oint\vec B\cdot d\vec l = \mu_0I + \mu_0\epsilon_0\dfrac{d\Phi_E}{dt}\).

1. Along \(+x\). The direction of travel is that of \(\vec E \times \vec B\): here \(\hat j \times \hat k = \hat i\).

2. Zero. \(\vec E\) and \(\vec B\) reach their maxima at the same points and vanish together.

3. Both fields oscillate perpendicular to the direction in which the wave moves, and perpendicular to each other.

\(\mu_0\epsilon_0 = 4\pi \times 10^{-7} \times 8.85 \times 10^{-12} = 1.11 \times 10^{-17}\) s\(^2\)/m\(^2\). \(c = \dfrac{1}{\sqrt{1.11 \times 10^{-17}}} = 3.0 \times 10^{8}\) m/s.

\(B_0 = \dfrac{E_0}{c} = \dfrac{300}{3 \times 10^{8}} = 10^{-6}\) T \(= 1\) μT. The magnetic field of light is very weak in tesla, but it carries as much energy as the electric field.

\(k = 0.5\) rad/m, so \(\lambda = \dfrac{2\pi}{k} = 4\pi \approx 12.6\) m. \(\omega = 1.5 \times 10^{8}\) rad/s, so \(f = \dfrac{\omega}{2\pi} \approx 2.4 \times 10^{7}\) Hz. \(v = \dfrac{\omega}{k} = 3 \times 10^{8}\) m/s. \(B_0 = \dfrac{E_0}{c} = 2 \times 10^{-7}\) T. With \(\vec E\) along \(y\) and travel along \(+x\), \(\vec B\) is along \(z\) so that \(\vec E \times \vec B\) points along \(x\).

\(v = \dfrac{1}{\sqrt{\mu\epsilon}} = \dfrac{c}{\sqrt{\mu_r\epsilon_r}} = \dfrac{3 \times 10^{8}}{\sqrt{2.25}} = 2 \times 10^{8}\) m/s. \(n = \dfrac{c}{v} = \sqrt{\epsilon_r} = 1.5\).

\(u = \dfrac{1}{2}\epsilon_0E^2 + \dfrac{B^2}{2\mu_0}\). With \(B = \dfrac{E}{c}\) and \(c^2 = \dfrac{1}{\mu_0\epsilon_0}\): \(\dfrac{B^2}{2\mu_0} = \dfrac{E^2}{2\mu_0c^2} = \dfrac{1}{2}\epsilon_0E^2\), equal to the electric part. Over a cycle the average of \(E^2\) is \(\dfrac{E_0^2}{2}\), so \(\bar u = \dfrac{1}{2}\epsilon_0E_0^2 = \dfrac{1}{2} \times 8.85 \times 10^{-12} \times 9 \times 10^{4} \approx 4 \times 10^{-7}\) J/m\(^3\).

\(I = \bar u\,c = 4 \times 10^{-7} \times 3 \times 10^{8} \approx 120\) W/m\(^2\). Equivalently \(I = \dfrac{1}{2}\epsilon_0cE_0^2\).

1. Energy per second on \(1\) m\(^2\) is \(1000\) J; its momentum is \(\dfrac{U}{c} = \dfrac{1000}{3 \times 10^{8}} \approx 3.3 \times 10^{-6}\) kg m/s, delivered each second.

2. Pressure \(= \dfrac{I}{c} = 3.3 \times 10^{-6}\) Pa. Force \(= pA = 3.3 \times 10^{-6}\) N.

3. Reflection reverses the momentum, so twice as much is transferred: \(p = \dfrac{2I}{c} \approx 6.7 \times 10^{-6}\) Pa.

Accelerated charges radiate. In an antenna the charges oscillate up and down. The changing electric field creates a changing magnetic field and so on, sending out a wave. The wave has the same frequency as the oscillation of the charges, set by the driving circuit.

\(I = \dfrac{P}{4\pi r^2} = \dfrac{100}{4\pi} \approx 8\) W/m\(^2\). \(E_0 = \sqrt{\dfrac{2I}{\epsilon_0c}} = \sqrt{\dfrac{2 \times 8}{8.85 \times 10^{-12} \times 3 \times 10^{8}}} = \sqrt{6.0 \times 10^{3}} \approx 77\) V/m.

1. \(I \propto \dfrac{1}{r^2}\): at \(2\) m the intensity is a quarter, \(2\) W/m\(^2\).

2. \(E_0 \propto \sqrt I \propto \dfrac{1}{r}\): half the value at \(1\) m, about \(39\) V/m.

Sunlight gives about \(3 \times 10^{-6}\) Pa, while the atmosphere presses with \(10^{5}\) Pa, some \(3 \times 10^{10}\) times more. The force on a hand is well below a micronewton, far too small to feel. Only on huge, light surfaces in space (solar sails) does it matter.

1. Radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays. Wavelength falls along the same order.

2. Gamma rays: photon energy \(hf\) grows with frequency, and they have the highest frequency.

(i) \(f = \dfrac{c}{\lambda} = \dfrac{3 \times 10^{8}}{0.03} = 10^{10}\) Hz. (ii) \(\lambda = \dfrac{3 \times 10^{8}}{10^{8}} = 3\) m. (iii) \(f = \dfrac{3 \times 10^{8}}{10^{-9}} = 3 \times 10^{17}\) Hz.

Radio: oscillating currents in antennas. Microwaves: magnetrons and klystrons. Infrared: hot bodies, vibrating molecules. X-rays: fast electrons stopped by a metal target. Gamma rays: radioactive nuclei.

Microwaves: cooking (water molecules absorb \(2.45\) GHz), radar. Infrared: remote controls, night-vision cameras. Ultraviolet: sterilising water and instruments. X-rays: medical imaging. Gamma rays: destroying cancer cells.

(i) \(500\) nm: visible (green). (ii) \(1\) mm: microwave, at the border with far infrared. (iii) \(10\) pm: gamma rays, overlapping hard X-rays. (iv) \(300\) m: radio (medium wave).

\(\dfrac{\partial E}{\partial x} = kE_0\cos(kx - \omega t)\) and \(\dfrac{\partial B}{\partial t} = -\omega B_0\cos(kx - \omega t)\). Faraday's law gives \(kE_0\cos(kx - \omega t) = \omega B_0\cos(kx - \omega t)\), so \(kE_0 = \omega B_0\) and \(\dfrac{E_0}{B_0} = \dfrac{\omega}{k} = c\).