Unit 9

AC Circuits & Electromagnetic Waves

The generator of Unit 8 makes a voltage that swings back and forth sixty times a second, and nearly every wall outlet, motor, radio and charger on Earth runs on that kind of signal. This unit first learns to handle it: phasors turn the calculus of inductors and capacitors into the algebra of arrows, impedance lets every DC rule you know work for AC, and resonance explains how a radio picks one station out of hundreds. Then it closes the course: one missing term in Ampère's law — Maxwell's displacement current — makes the four field equations predict waves that travel at the speed of light. That is radio, Wi-Fi, and light itself. The lab lets you drive a series RLC circuit and watch its phasors, impedance and resonance peak move as you turn the knobs.

1 · AC sources and phasors

A coil spinning in a magnetic field (Unit 8) produces a sinusoidal emf. We write an AC source as

\[ v(t) = V_0\cos(\omega t),\qquad \omega = 2\pi f,\qquad T = \frac{1}{f} \] \(V_0\) is the amplitude (peak voltage, in volts); \(f\) is the frequency in hertz (cycles per second); \(\omega\) is the angular frequency in rad/s; \(T\) is the period in seconds. A North American outlet is \(f = 60\) Hz, \(\omega = 2\pi(60) = 377\) rad/s and \(V_0 = 170\) V (it is labelled “120 V” — the RMS value of Section 5). Europe uses 50 Hz and \(V_0 = 325\) V (“230 V”).

In a circuit driven by one sinusoidal source, once the start-up transients of Units 5 and 8 have died away, every current and voltage is a sinusoid at the same \(\omega\). They differ only in two numbers: their amplitude and their phase (how far ahead or behind in time they are). Adding shifted cosines with trig identities is miserable. The fix is a picture.

The phasor picture

Draw an arrow of length \(V_0\) from the origin, at angle \(\omega t + \phi\) from the horizontal axis, and let it spin counter-clockwise at \(\omega\). Its horizontal shadow is \(V_0\cos(\omega t + \phi)\) — exactly the signal. That spinning arrow is a phasor. Because every phasor in the circuit spins at the same rate, you can freeze the picture at any instant: the lengths are the amplitudes and the angles between arrows are the phase differences. Adding two sinusoids becomes adding two arrows tip-to-tail.

A phasor is a spinning arrow whose shadow is the signal. Here the current phasor \(I\) trails the voltage phasor \(V\) by \(50^\circ\), so on the time graph the current reaches each peak a little later: the current lags the voltage by \(50^\circ\). Both arrows spin together, so that angle never changes.

Phasors as complex numbers

\[ v(t) = \operatorname{Re}\!\left\{\tilde V e^{j\omega t}\right\},\qquad \tilde V = V_0 e^{j\phi} = V_0\angle\phi,\qquad j = \sqrt{-1} \] Engineers write \(j\), not \(i\), because \(i\) is already the current. The tilde marks a phasor — a complex number holding amplitude and phase — not a spatial vector. Euler's formula \(e^{j\theta} = \cos\theta + j\sin\theta\) is what makes the arrow picture and the complex number the same thing.

\[ \frac{d}{dt}\;\longrightarrow\;\times\, j\omega \qquad\qquad \int dt\;\longrightarrow\;\times\,\frac{1}{j\omega} \qquad\qquad \times\, j = \text{rotate } 90^\circ \text{ counter-clockwise} \] Differentiating \(e^{j\omega t}\) just multiplies it by \(j\omega\). So the derivative in \(v = L\,di/dt\) and the integral in \(v = q/C\) become multiplications — phasors turn differential equations into algebra. That is the entire reason they exist.

Common mistakes

“Leads” means earlier in time, which on the time graph is to the left and on the phasor diagram is counter-clockwise ahead. Students often read a curve that is further right as “ahead”. It is behind.
Phasors only describe steady-state sinusoids at one frequency. They do not describe the switch-on transient, and you cannot add phasors of different frequencies.
Use radians for \(\omega t\) (\(\omega\) is in rad/s), but phase angles are usually quoted in degrees. Check your calculator mode every time.

2 · Resistors, inductors and capacitors in AC: reactance

Push the same current \(i(t) = I_0\cos\omega t\) through each element alone and see what voltage appears across it.

\[ \text{Resistor: } v_R = iR = I_0R\cos\omega t \qquad\Longrightarrow\qquad V_R = I_0R,\ \text{in phase} \] \[ \text{Inductor: } v_L = L\frac{di}{dt} = -\omega L I_0\sin\omega t = \omega L I_0\cos(\omega t + 90^\circ) \qquad\Longrightarrow\qquad V_L = I_0X_L,\ \ X_L = \omega L \] \[ \text{Capacitor: } v_C = \frac{q}{C} = \frac{1}{C}\int i\,dt = \frac{I_0}{\omega C}\sin\omega t = \frac{I_0}{\omega C}\cos(\omega t - 90^\circ) \qquad\Longrightarrow\qquad V_C = I_0X_C,\ \ X_C = \frac{1}{\omega C} \] \(X_L\) and \(X_C\) are the reactances, measured in ohms (check: \(\omega L\) has units \(\text{s}^{-1}\cdot\text{V}\cdot\text{s}/\text{A} = \text{V}/\text{A}\)). They play the role of resistance — amplitude of voltage over amplitude of current — but come with a \(\pm 90^\circ\) phase shift.

Why the shifts make sense.

Inductor: it fights changes in current (Lenz). Its voltage is largest when the current is changing fastest — when the current crosses zero — and zero when the current is at its peak. So the voltage peaks a quarter cycle before the current: \(v_L\) leads \(i\) by \(90^\circ\).
Capacitor: its voltage is set by the charge piled up. Charge has to flow in first, then the voltage builds; the voltage peaks just as the current reverses. So \(v_C\) lags \(i\) by \(90^\circ\) — equivalently the current leads.
Frequency dependence: faster wiggling means larger \(di/dt\), so \(X_L = \omega L\) grows with frequency; and less time to pile up charge each half-cycle, so \(X_C = 1/(\omega C)\) shrinks.

The same cosine voltage across each element. In the inductor the current reaches its peak a quarter period after the voltage; in the capacitor, a quarter period before. Mnemonic: ELI — in an inductor (L), voltage (E) comes before current (I); ICE — in a capacitor (C), current comes before voltage.

Complex impedance

Impedance is the phasor version of Ohm's law: \(\tilde V = Z\,\tilde I\). The factor \(j\) is the \(90^\circ\) rotation.
Element	Impedance \(Z\)	\(\|Z\|\)	Phase of \(v\) relative to \(i\)	At DC (\(\omega\to0\))	At very high \(\omega\)
Resistor	\(R\)	\(R\)	\(0\)	\(R\)	\(R\)
Inductor	\(j\omega L\)	\(X_L = \omega L\)	\(+90^\circ\) (leads)	short circuit	open circuit
Capacitor	\(\dfrac{1}{j\omega C} = \dfrac{-j}{\omega C}\)	\(X_C = \dfrac{1}{\omega C}\)	\(-90^\circ\) (lags)	open circuit	short circuit

Common mistakes

Using \(f\) where \(\omega\) belongs: \(X_L = 2\pi fL\), not \(fL\). That is a factor of 6.28 lost.
Treating reactance like resistance for energy. An ideal L or C dissipates nothing: it stores energy for a quarter cycle and hands it back the next quarter.
Writing \(X_C = \omega C\). Big capacitors pass AC easily, so \(X_C\) must get smaller as \(C\) grows: \(X_C = 1/(\omega C)\).

EE connection

Frequency-dependent impedance is how every filter works. An RC low-pass filter takes its output across the capacitor: at low frequency the capacitor is nearly open and the output follows the input; at high frequency it is nearly a short and the output dies. The crossover, where \(X_C = R\), is \(\omega_c = 1/(RC)\) — the same \(RC\) as the time constant of Unit 5. A loudspeaker crossover puts an inductor in series with the woofer (blocks the highs) and a capacitor in series with the tweeter (blocks the lows). And the small “decoupling” capacitor next to every chip's power pin is a short circuit for high-frequency noise.

3 · Series RLC: impedance and phase

Now put \(R\), \(L\) and \(C\) in series across a source. In series the same current flows through all three, so take the current phasor as the reference (pointing along \(+x\)). Then:

\(\tilde V_R\) points along the current (length \(IR\)),
\(\tilde V_L\) points \(90^\circ\) ahead, straight up (length \(IX_L\)),
\(\tilde V_C\) points \(90^\circ\) behind, straight down (length \(IX_C\)).

Kirchhoff's loop rule says the source voltage is their phasor sum:

\[ \tilde V = \tilde I\left[R + j\left(X_L - X_C\right)\right] = \tilde I\,Z \] \[ |Z| = \sqrt{R^2 + \left(X_L - X_C\right)^2},\qquad \tan\phi = \frac{X_L - X_C}{R},\qquad I = \frac{V}{|Z|} \] \(|Z|\) is the impedance (ohms); \(\phi\) is the angle by which the source voltage leads the current. \(\phi > 0\): the circuit is inductive (current lags). \(\phi < 0\): capacitive (current leads). \(I\) and \(V\) may both be amplitudes or both RMS — never mix them.

A running example

A series circuit with \(R = 40\ \Omega\), \(L = 0.25\) H, \(C = 40\ \mu\)F on a 120 V (RMS), 60 Hz outlet. With \(\omega = 377\) rad/s:

\[ X_L = (377)(0.25) = 94.2\ \Omega,\qquad X_C = \frac{1}{(377)(40\times10^{-6})} = 66.3\ \Omega \] \[ |Z| = \sqrt{40^2 + (94.2 - 66.3)^2} = \sqrt{1600 + 780} = 48.8\ \Omega,\qquad \phi = \tan^{-1}\frac{27.9}{40} = 34.9^\circ \] \[ I_{\mathrm{rms}} = \frac{120}{48.8} = 2.46\ \text{A};\qquad V_R = 98.4\ \text{V},\quad V_L = 231.8\ \text{V},\quad V_C = 163.1\ \text{V} \]

Phasor diagram (left, volts) and impedance triangle (right, ohms) for the running example. \(\tilde V_L\) and \(\tilde V_C\) point in opposite directions and mostly cancel; only their difference, \(68.7\) V, adds to \(V_R\). Check: \(\sqrt{98.4^2 + 68.7^2} = 120.0\) V. The source voltage leads the current by \(34.9^\circ\): an inductive circuit.

Common mistake: adding magnitudes

\(98.4 + 231.8 + 163.1 = 493\) V, far more than the 120 V source. That is not a violation of Kirchhoff's law: the loop rule holds for instantaneous voltages and for phasors, never for amplitudes or RMS readings. A voltmeter across the inductor really does read 232 V — almost twice the outlet. The capacitor's voltage is out of phase and cancels most of it.

EE connection

With \(Z\) in hand, every DC technique carries over to AC: impedances in series add, in parallel \(1/Z = \sum 1/Z_k\), voltage dividers, Thévenin equivalents, node and mesh analysis — all with complex numbers in place of resistances. Your first circuits course spends weeks on exactly this, and it all rests on the \(d/dt\to j\omega\) trick of Section 1.

4 · Resonance and quality factor

Sweep the source frequency. \(X_L = \omega L\) rises, \(X_C = 1/(\omega C)\) falls, so at one special frequency they are equal and cancel completely:

\[ \omega_0 L = \frac{1}{\omega_0 C}\quad\Longrightarrow\quad \omega_0 = \frac{1}{\sqrt{LC}},\qquad f_0 = \frac{1}{2\pi\sqrt{LC}} \] The resonant frequency — the same frequency at which an LC circuit oscillates on its own (Unit 8). Driving a system at its natural frequency gives the biggest response, like pushing a swing in time with it.

At resonance:

\(|Z| = R\), its minimum value, so the current is maximum: \(I_{\max} = V/R\).
\(\phi = 0\): voltage and current are in phase; the circuit looks purely resistive.
\(V_L = V_C\), equal and opposite. Each can be much larger than the source voltage.

For the running example, \(\omega_0 = 1/\sqrt{(0.25)(40\times10^{-6})} = 316\) rad/s, \(f_0 = 50.3\) Hz. Driven at 60 Hz it sits above resonance, where \(X_L > X_C\) — which is why it came out inductive.

How sharp is the peak? The quality factor

\[ Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 RC} = \frac{1}{R}\sqrt{\frac{L}{C}},\qquad \Delta\omega = \frac{R}{L} = \frac{\omega_0}{Q},\qquad \Delta f = \frac{f_0}{Q} \] \(Q\) is dimensionless. \(\Delta\omega\) is the bandwidth: the width of the resonance curve between the two frequencies where the current falls to \(1/\sqrt2\) of its peak (where the power falls to half). Those points are exactly where \(|X_L - X_C| = R\); solving gives \(\omega_2 - \omega_1 = R/L\) exactly. Small \(R\) → high \(Q\) → tall, narrow peak.

What \(Q\) means physically: \(Q = 2\pi\times\dfrac{\text{energy stored}}{\text{energy dissipated per cycle}}\). At resonance the stored energy is \(\tfrac12LI_0^2\) and the loss per cycle is \(\tfrac12I_0^2R\cdot\tfrac{2\pi}{\omega_0}\); their ratio times \(2\pi\) is \(\omega_0L/R\). Also, at resonance \(V_L = V_C = Q\,V\): the running example (\(Q = 1.98\)) would put 237 V across the inductor at 50.3 Hz.

Resonance curves for the same \(L\) and \(C\) with two resistances. Both peak at \(\omega_0 = 1/\sqrt{LC}\) — \(R\) does not move the peak. Smaller \(R\) gives a taller peak (\(I_{\max} = V/R\)) and a narrower one (\(\Delta\omega = R/L\)). A high-\(Q\) circuit responds strongly to one narrow band of frequencies and ignores the rest: that is selectivity.

Interactive lab Series RLC: resonance & phasors Open it in Interactive Laboratories →

Common mistakes

\(\omega_0\) versus \(f_0\): \(1/\sqrt{LC}\) is in rad/s. Divide by \(2\pi\) for hertz.
Three different things are called \(Q\) in this course: charge, quality factor, and (Section 5) reactive power. Context tells you which.
“At resonance the inductor and capacitor voltages are zero.” No — they are large; it is their sum that is zero.

EE connection

Turning a radio's tuning knob changes \(C\) (a variable capacitor, or today a varactor diode) so that \(f_0\) lands on the station you want; the circuit's \(Q\) sets how well it rejects the neighbours (Worked example 3). Quartz crystals in every clock and microcontroller are mechanical resonators with \(Q\) of \(10^4\) to \(10^6\), which is why they keep time so well. Wireless phone chargers use a pair of coils tuned to resonance (around 100–200 kHz) so energy transfers efficiently. And resonance can be a hazard: a power-factor capacitor bank accidentally resonating with line inductance can produce damaging overvoltages.

5 · Power in AC circuits: RMS, power factor and transformers

Instantaneous power is \(p(t) = v(t)\,i(t)\), as in DC. For a resistor with \(i = I_0\cos\omega t\):

\[ p(t) = I_0^2R\cos^2\omega t = \tfrac12 I_0^2R\left(1 + \cos 2\omega t\right) \quad\Longrightarrow\quad P_{\mathrm{avg}} = \tfrac12 I_0^2R \] The power pulses at twice the line frequency between 0 and \(I_0^2R\), never negative: a resistor always takes energy. Its average is half the peak.

To keep the familiar \(P = I^2R\), define the root-mean-square value: square the signal, average over a cycle, take the square root.

\[ I_{\mathrm{rms}} = \sqrt{\langle i^2\rangle} = \frac{I_0}{\sqrt2},\qquad V_{\mathrm{rms}} = \frac{V_0}{\sqrt2},\qquad P_{\mathrm{avg}} = I_{\mathrm{rms}}^2R = \frac{V_{R,\mathrm{rms}}^2}{R} \] \(\langle\ \rangle\) means time average. An RMS value is the DC value that would heat a resistor equally. Outlet and meter ratings are RMS: \(120\ \text{V}_{\mathrm{rms}}\) means a peak of \(120\sqrt2 = 170\) V.

Average power and power factor

When current and voltage are out of phase by \(\phi\), the average of \(V_0\cos\omega t\cdot I_0\cos(\omega t - \phi)\) is \(\tfrac12V_0I_0\cos\phi\):

\[ P_{\mathrm{avg}} = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi,\qquad \text{power factor} = \cos\phi = \frac{R}{|Z|}\ \ \text{(series circuit)} \] Only the part of the current in phase with the voltage delivers energy. For a pure inductor or capacitor \(\phi = \pm90^\circ\), \(\cos\phi = 0\), and the average power is zero. In any circuit, all the average power ends up in the resistances: \(P_{\mathrm{avg}} = \sum_k I_{\mathrm{rms},k}^2R_k\) (for a single series loop, \(I_{\mathrm{rms}}^2R\)).

Running example: \(\cos 34.9^\circ = 40/48.8 = 0.820\), so \(P = (120)(2.46)(0.820) = 242\) W. Check: \(I_{\mathrm{rms}}^2R = (2.46)^2(40) = 242\) W. ✓

Power engineers also name the product without the \(\cos\phi\): the apparent power \(S = V_{\mathrm{rms}}I_{\mathrm{rms}}\), in volt-amperes (VA), and the reactive power \(Q_{\mathrm r} = V_{\mathrm{rms}}I_{\mathrm{rms}}\sin\phi\), in var — energy that sloshes into and out of the L's and C's each cycle without being used. They form a right triangle: \(S^2 = P^2 + Q_{\mathrm r}^2\). In the running example \(S = 295\) VA, \(P = 242\) W, \(Q_{\mathrm r} = 169\) var.

Common mistakes

\(P = V_0I_0\) is twice too big. Use \(\tfrac12V_0I_0\cos\phi\) or \(V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi\).
\(V_{\mathrm{rms}}I_{\mathrm{rms}}\) is the real power only when \(\cos\phi = 1\).
\(V_0/\sqrt2\) is the RMS of a sine wave only. A square wave's RMS equals its peak; a triangle wave's is \(V_0/\sqrt3\). (“True-RMS” multimeters compute it properly for any waveform.)

EE connection

Motors are inductive, so factories run at power factors like 0.7–0.8 lagging. The utility must push extra current through its wires (losses \(\propto I^2\)) to deliver the same real power, so it charges industrial customers for poor power factor. The fix is power-factor correction: capacitors in parallel with the load supply the reactive current locally (Worked example 4).

Transformers

From Unit 8: two coils share one iron core, so the same changing flux \(d\Phi_B/dt\) threads every turn of both. Each turn sees the same emf, so voltage is proportional to the number of turns:

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p},\qquad V_pI_p = V_sI_s\ \ \Longrightarrow\ \ \frac{I_s}{I_p} = \frac{N_p}{N_s},\qquad R_{\mathrm{eff}} = \left(\frac{N_p}{N_s}\right)^2R_{\mathrm{load}} \] Subscripts \(p\) = primary (input), \(s\) = secondary (output); \(N\) = number of turns. For an ideal (lossless) transformer power in equals power out, so stepping voltage up steps current down by the same factor. A load \(R_{\mathrm{load}}\) on the secondary looks like \(R_{\mathrm{eff}}\) from the primary side — the basis of impedance matching.

A transformer needs a changing flux. On steady DC it does nothing (and its primary, being just a low-resistance coil, would draw a huge current). This is the historical reason the grid is AC: in the 1880s only AC could be stepped up and down efficiently.

Why the grid uses high voltage

A power plant must deliver power \(P = VI\) through lines with resistance \(R_{\mathrm{line}}\). The current in the line is \(I = P/V\), so the power wasted heating the line is

\[ P_{\mathrm{loss}} = I^2R_{\mathrm{line}} = \frac{P^2R_{\mathrm{line}}}{V^2} \] For a fixed delivered power, raising the transmission voltage by a factor \(n\) cuts the loss by \(n^2\). Going from 20 kV to 400 kV (factor 20) cuts losses 400-fold (Worked example 5).

The grid steps voltage up for the long haul and down for safe use, with transformers at each stage. The long-distance section carries the highest voltage and therefore the smallest current. (Today, power electronics also make very long or undersea HVDC links practical — but the local grid is still AC.)

Common mistake: “\(P = V^2/R\), so higher voltage means more loss”

In \(P_{\mathrm{loss}} = V^2/R_{\mathrm{line}}\), \(V\) would have to be the voltage across the line resistance — the small drop \(IR_{\mathrm{line}}\) along the wire — not the 400 kV between the line and ground. Using the transmission voltage there gives absurd answers. Always compute the line current first, then \(I^2R_{\mathrm{line}}\).

6 · Displacement current and Maxwell's equations

Ampère's law (Unit 7) says \(\oint\mathbf B\cdot d\boldsymbol\ell = \mu_0 I_{\mathrm{enc}}\), where \(I_{\mathrm{enc}}\) is the current through any surface bounded by the loop. Apply it to a wire charging a capacitor, with a loop around the wire:

Surface 1, a flat disk, is pierced by the wire: \(I_{\mathrm{enc}} = I\).
Surface 2, a bag with the same rim that bulges out between the plates, is pierced by no wire: \(I_{\mathrm{enc}} = 0\).

Same loop, same \(\oint\mathbf B\cdot d\boldsymbol\ell\), two different answers. Ampère's law, as written, is broken whenever charge is piling up somewhere.

The same Amperian loop bounds a flat surface pierced by the current and a bulging surface that slips between the plates. Between the plates there is no moving charge — but there is an electric field that is growing. Maxwell's fix counts that changing electric flux as a current.

Between the plates \(E = \sigma/\varepsilon_0 = q/(\varepsilon_0A)\) (Unit 4), so the electric flux is \(\Phi_E = EA = q/\varepsilon_0\). Then \(\varepsilon_0\,d\Phi_E/dt = dq/dt = I\): the changing flux carries exactly the current that the wire delivers. Maxwell (1861–65) added it to Ampère's law:

\[ I_d = \varepsilon_0\frac{d\Phi_E}{dt},\qquad \oint\mathbf B\cdot d\boldsymbol\ell = \mu_0\left(I_{\mathrm{enc}} + \varepsilon_0\frac{d\Phi_E}{dt}\right) \] \(I_d\) is the displacement current (amperes). It is not charges moving; it is a changing electric field, and it makes a magnetic field exactly as a real current would. With it, both surfaces give \(\mu_0I\). The name is historical and unfortunate.

Magnetic field between capacitor plates

Circular plates of radius \(R = 5.0\) cm charge with \(I = 2.0\) A. Between the plates \(dE/dt = I/(\varepsilon_0\pi R^2) = 2.88\times10^{13}\ \text{V}\,\text{m}^{-1}\text{s}^{-1}\). An Amperian circle of radius \(r < R\) encloses the fraction \(r^2/R^2\) of the displacement current, so \(B(2\pi r) = \mu_0I\,r^2/R^2\), giving \(B = \mu_0Ir/(2\pi R^2)\). At \(r = 3.0\) cm: \(B = 4.8\ \mu\)T. At the rim (\(r = 5.0\) cm): \(8.0\ \mu\)T. Outside, at \(r = 10\) cm: \(B = \mu_0I/(2\pi r) = 4.0\ \mu\)T — identical to the field around the wire itself. The gap is magnetically invisible.

Maxwell's equations

All of electromagnetism in four lines (integral form, in vacuum):

\[ \text{(1) Gauss:}\quad \oint\mathbf E\cdot d\mathbf A = \frac{Q_{\mathrm{enc}}}{\varepsilon_0} \qquad\qquad \text{(2) Gauss for }\mathbf B\text{:}\quad \oint\mathbf B\cdot d\mathbf A = 0 \] \[ \text{(3) Faraday:}\quad \oint\mathbf E\cdot d\boldsymbol\ell = -\frac{d\Phi_B}{dt} \qquad\qquad \text{(4) Amp\`ere–Maxwell:}\quad \oint\mathbf B\cdot d\boldsymbol\ell = \mu_0I_{\mathrm{enc}} + \mu_0\varepsilon_0\frac{d\Phi_E}{dt} \] (1) Charges are sources of \(\mathbf E\): field lines start on \(+\) and end on \(-\). (2) There are no magnetic charges: \(\mathbf B\) lines always close on themselves. (3) A changing magnetic flux makes a circulating \(\mathbf E\). (4) Currents and changing electric flux make a circulating \(\mathbf B\). Surface integrals \(\oint d\mathbf A\) are over closed surfaces; line integrals \(\oint d\boldsymbol\ell\) are around the closed rim of the surface used for the flux.

Add the Lorentz force \(\mathbf F = q(\mathbf E + \mathbf v\times\mathbf B)\), which says what the fields do to charges, and you have the complete theory. Notice the symmetry between (3) and (4): a changing \(\mathbf B\) makes \(\mathbf E\), and a changing \(\mathbf E\) makes \(\mathbf B\). Each can regenerate the other with no charges or currents anywhere. That is a wave.

Common mistakes

Displacement current is not a current of charges and does not flow through the dielectric or vacuum “as electrons”. Nothing crosses the gap.
Losing the minus sign in Faraday's law (it is Lenz's law) — or adding one to the Ampère–Maxwell term. The two equations are not perfectly symmetric in sign, and that sign difference is what makes waves travel instead of grow.
Equation (2) does not say \(\mathbf B = 0\); it says as much \(\mathbf B\) flux leaves any closed surface as enters it.

7 · Electromagnetic waves

Wiggle a charge (or drive an AC current up and down an antenna) and the changing fields near it start the chain: changing \(\mathbf E\) → \(\mathbf B\) → changing \(\mathbf B\) → \(\mathbf E\) → … The disturbance detaches and travels outward. Far from the source it is a plane wave:

A plane wave moving along \(+x\). \(\mathbf E\) oscillates along \(y\), \(\mathbf B\) along \(z\) (drawn in perspective, toward you); they rise and fall together (in phase), are perpendicular to each other and to the direction of travel, and \(\mathbf E\times\mathbf B\) points along the travel direction: \(\hat{\mathbf y}\times\hat{\mathbf z} = \hat{\mathbf x}\).

\[ E_y(x,t) = E_0\cos(kx - \omega t),\qquad B_z(x,t) = B_0\cos(kx - \omega t),\qquad k = \frac{2\pi}{\lambda},\quad \frac{\omega}{k} = f\lambda = c \] \(E_0, B_0\) are the field amplitudes; \(\lambda\) the wavelength (m); \(k\) the wave number (rad/m). A crest moves at speed \(\omega/k\).

Where \(c\) and \(E = cB\) come from

Apply Faraday's law to a thin rectangle in the \(xy\)-plane, and the Ampère–Maxwell law (no currents) to a thin rectangle in the \(xz\)-plane. In the limit of tiny rectangles they become

\[ \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{\partial t},\qquad -\frac{\partial B_z}{\partial x} = \mu_0\varepsilon_0\frac{\partial E_y}{\partial t} \] \[ \text{Substitute the waves:}\quad kE_0 = \omega B_0,\qquad kB_0 = \mu_0\varepsilon_0\,\omega E_0 \] \[ \Longrightarrow\quad \frac{E_0}{B_0} = \frac{\omega}{k} = c,\qquad c^2 = \frac{1}{\mu_0\varepsilon_0},\qquad c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = \frac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}} = 3.00\times10^8\ \text{m/s} \] Two constants measured with charges and magnets on a lab bench — \(\varepsilon_0\) from capacitors, \(\mu_0\) from currents — predict the speed of light. Maxwell did this calculation and concluded that light is an electromagnetic wave. At every point and instant, \(E = cB\).

Energy flow: the Poynting vector and intensity

\[ \mathbf S = \frac{1}{\mu_0}\,\mathbf E\times\mathbf B,\qquad u = \tfrac12\varepsilon_0E^2 + \frac{B^2}{2\mu_0} = \varepsilon_0E^2 \] \(\mathbf S\) (the Poynting vector, W/m²) is the power crossing unit area, pointing the way energy flows. \(u\) is the energy density (J/m³) from Units 4 and 8; in a wave the electric and magnetic halves are equal (use \(B = E/c\) and \(c^2 = 1/(\mu_0\varepsilon_0)\)).

\[ I = \langle S\rangle = \frac{E_0B_0}{2\mu_0} = \frac{E_0^2}{2\mu_0c} = \tfrac12c\,\varepsilon_0E_0^2 = \frac{E_{\mathrm{rms}}^2}{\mu_0c},\qquad I = \frac{P}{4\pi r^2}\ \ \text{(point source)} \] The intensity \(I\) (W/m² — not a current!) is the time-averaged \(S\). The \(\tfrac12\) is the average of \(\cos^2\), exactly as in RMS power. A source radiating power \(P\) equally in all directions spreads it over a sphere of area \(4\pi r^2\): the inverse-square law.

Example — sunlight. At the ground, \(I\approx1000\) W/m². Then \(E_0 = \sqrt{2I/(c\varepsilon_0)} = \sqrt{2(1000)/[(3.00\times10^8)(8.85\times10^{-12})]} = 868\) V/m and \(B_0 = E_0/c = 2.89\ \mu\)T — only a few percent of the Earth's field (about \(50\ \mu\)T).

Radiation pressure (briefly)

\[ p_{\mathrm{rad}} = \frac{I}{c}\ \ \text{(absorbed)},\qquad p_{\mathrm{rad}} = \frac{2I}{c}\ \ \text{(perfectly reflected)} \] Waves carry momentum \(U/c\) along with energy \(U\). Sunlight on a black surface pushes with \(1000/(3\times10^8) = 3.3\ \mu\)Pa — tiny, but enough to steer spacecraft and propel solar sails.

The electromagnetic spectrum

One theory, one speed \(c\) in vacuum, frequencies spanning more than 20 orders of magnitude. Band edges are conventions, not physics.
Band	Typical frequency	Wavelength \(\lambda = c/f\)	Where an EE meets it
Power line	50 / 60 Hz	6000 / 5000 km	The grid; so long that the wiring in a house is “lumped” circuit theory
AM radio	540–1700 kHz	556–176 m	Tuned LC circuits (Worked example 3)
FM radio, TV	88–108 MHz (FM)	3.4–2.8 m	Car and rooftop antennas
Microwave	0.6–6 GHz (cell, Wi-Fi 2.4/5 GHz); up to ~300 GHz	12.5 cm at 2.4 GHz; 6 cm at 5 GHz	Wi-Fi, Bluetooth, phones, radar, microwave ovens (2.45 GHz)
Infrared	~0.3–430 THz	1 mm – 700 nm	Remote controls, thermal cameras, fibre optics at 1550 nm (193 THz)
Visible	430–750 THz	700–400 nm	LEDs, lasers, displays, solar cells
Ultraviolet	750 THz – 30 PHz	400–10 nm	Photolithography for chips (EUV at 13.5 nm)
X-rays, gamma rays	above ~30 PHz	below ~10 nm	Medical imaging, inspection of solder joints

EE connection: antennas

An antenna is a conductor in which an AC current sloshes charge back and forth, launching exactly the waves of this section; receiving is the reverse — the wave's \(\mathbf E\) pushes electrons in the wire. Antennas work best when their size is comparable to \(\lambda\): a half-wave dipole for 2.4 GHz Wi-Fi is \(\lambda/2 = 6.25\) cm long, a quarter-wave whip is about 3.1 cm — which is why the antennas inside a phone are a few centimetres, while AM broadcast towers are tens of metres tall. And it is why the 60 Hz grid barely radiates at all: a wire would have to be thousands of kilometres long.

Common mistakes

Thinking \(B\) is unimportant because \(B_0 = E_0/c\) is numerically tiny. The magnetic energy density equals the electric one; \(c\) just converts units.
Forgetting the \(\tfrac12\) in \(I = \tfrac12c\varepsilon_0E_0^2\) (it is there because \(E_0\) is the peak; with \(E_{\mathrm{rms}}\) there is no \(\tfrac12\)).
Getting the direction wrong: travel is along \(\mathbf E\times\mathbf B\), in that order.
Mixing up intensity \(I\) and current \(I\). Same letter, different quantity.

8 · The whole course on one page

Everything in Physics II is the four Maxwell equations plus the Lorentz force, applied to different situations. Here is where each one came from in this course.

Plus \(\mathbf F = q(\mathbf E + \mathbf v\times\mathbf B)\) (Units 1 and 6), charge conservation and \(\mathbf J = \sigma\mathbf E\) (Unit 5).
Equation	In words	Units that built it
\(\oint\mathbf E\cdot d\mathbf A = \dfrac{Q_{\mathrm{enc}}}{\varepsilon_0}\)	Charge is the source of \(\mathbf E\).	1 Coulomb's law and fields (Gauss's law with spherical symmetry); 2 Gauss's law itself; 3 potential \(V\); 4 capacitance and field energy \(\tfrac12\varepsilon_0E^2\).
\(\oint\mathbf B\cdot d\mathbf A = 0\)	No magnetic monopoles; \(\mathbf B\) lines close.	6 magnetic fields and dipoles; 7 every Biot–Savart field loops around its current.
\(\oint\mathbf E\cdot d\boldsymbol\ell = -\dfrac{d\Phi_B}{dt}\)	Changing \(\mathbf B\) makes a circulating \(\mathbf E\).	3 and 5 its static case \(\oint\mathbf E\cdot d\boldsymbol\ell = 0\) — a potential exists and Kirchhoff's loop rule works; 8 induction, generators, inductors, transformers.
\(\oint\mathbf B\cdot d\boldsymbol\ell = \mu_0I_{\mathrm{enc}} + \mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}\)	Currents and changing \(\mathbf E\) make a circulating \(\mathbf B\).	7 Biot–Savart and Ampère; 9 Maxwell's displacement current, which turns the set into waves.

Circuits are the same physics, lumped. A capacitor is a place where energy is stored in \(\mathbf E\) (Unit 4); an inductor stores it in \(\mathbf B\) (Unit 8); a resistor turns it into heat (Unit 5); and an antenna lets it escape as radiation (Unit 9). Circuit theory works when the circuit is much smaller than \(\lambda\) — at 60 Hz, always; at 5 GHz, only for a board a few millimetres across. Beyond that you need the fields themselves, which is the subject of the electromagnetics course in an EE degree (transmission lines, waveguides, antennas), alongside circuits, signals and systems, and power.

9 · Worked examples

1. Reactance versus frequency

Find the reactances of a 10 mH inductor and a 1.0 µF capacitor at 100 Hz, 1 kHz and 10 kHz, and the frequency at which they are equal.

\[ X_L = 2\pi fL:\quad 6.28\ \Omega,\ \ 62.8\ \Omega,\ \ 628\ \Omega \] \[ X_C = \frac{1}{2\pi fC}:\quad 1592\ \Omega,\ \ 159\ \Omega,\ \ 15.9\ \Omega \] \[ X_L = X_C\ \Longrightarrow\ f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(0.010)(1.0\times10^{-6})}} = \frac{1}{2\pi(1.0\times10^{-4})} = 1.59\ \text{kHz} \]

Each factor of 10 in frequency multiplies \(X_L\) by 10 and divides \(X_C\) by 10. The crossover is the resonant frequency \(f_0\) — previewing Section 4. In a circuit, below 1.59 kHz the capacitor dominates; above it, the inductor does.

2. A capacitive series RLC circuit, from source to \(i(t)\)

\(v(t) = 10\cos(1000t)\) V drives \(R = 100\ \Omega\), \(L = 50\) mH, \(C = 5.0\ \mu\)F in series. Find \(i(t)\), the element voltage amplitudes and the average power.

\[ X_L = (1000)(0.050) = 50\ \Omega,\qquad X_C = \frac{1}{(1000)(5.0\times10^{-6})} = 200\ \Omega,\qquad X_L - X_C = -150\ \Omega \] \[ |Z| = \sqrt{100^2 + 150^2} = \sqrt{32\,500} = 180.3\ \Omega,\qquad \phi = \tan^{-1}\!\left(\frac{-150}{100}\right) = -56.3^\circ \] \[ I_0 = \frac{10}{180.3} = 55.5\ \text{mA},\qquad \boxed{i(t) = 55.5\cos(1000t + 56.3^\circ)\ \text{mA}} \]

\(\phi < 0\): the voltage lags the current, so the current leads by \(56.3^\circ\) — hence the \(+\) sign in \(i(t)\). The circuit is capacitive because it is driven below resonance (\(\omega_0 = 1/\sqrt{LC} = 2000\) rad/s).

\[ V_{R0} = (0.0555)(100) = 5.55\ \text{V},\quad V_{L0} = (0.0555)(50) = 2.77\ \text{V},\quad V_{C0} = (0.0555)(200) = 11.1\ \text{V} \] \[ \text{Check: } \sqrt{5.55^2 + (2.77 - 11.1)^2} = \sqrt{30.8 + 69.2} = 10.0\ \text{V}\ \checkmark \] \[ P_{\mathrm{avg}} = \tfrac12I_0^2R = \tfrac12(0.0555)^2(100) = 0.154\ \text{W} = \tfrac12V_0I_0\cos\phi = \tfrac12(10)(0.0555)(0.555)\ \checkmark \]

Note \(V_{C0} = 11.1\) V exceeds the 10 V source — fine, as in Section 3.

3. Tuning an AM radio

A radio's tuned circuit has \(L = 240\ \mu\)H and resistance \(R = 15\ \Omega\). (a) What \(C\) tunes it to a station at 1000 kHz? (b) What range of \(C\) covers the AM band, 540–1700 kHz? (c) Find \(Q\) and the bandwidth at 1000 kHz.

\[ \text{(a)}\quad C = \frac{1}{(2\pi f_0)^2L} = \frac{1}{(2\pi\cdot10^6)^2(240\times10^{-6})} = 1.06\times10^{-10}\ \text{F} = \boxed{106\ \text{pF}} \] \[ \text{(b)}\quad C(540\ \text{kHz}) = 362\ \text{pF},\qquad C(1700\ \text{kHz}) = 36.5\ \text{pF} \]

A 10-to-1 capacitance range covers a 3.1-to-1 frequency range, because \(f_0\propto1/\sqrt C\): \((1700/540)^2 = 9.9\).

\[ \text{(c)}\quad Q = \frac{\omega_0L}{R} = \frac{(2\pi\cdot10^6)(240\times10^{-6})}{15} = \frac{1508}{15} = 101,\qquad \Delta f = \frac{f_0}{Q} = \frac{1000\ \text{kHz}}{101} \approx 10\ \text{kHz} \]

A 10 kHz passband is just right: AM stations in the Americas are spaced 10 kHz apart, so the circuit passes one station's audio sidebands and attenuates its neighbours.

4. Power-factor correction (EE application)

A workshop's motors draw 5.0 kW of real power at power factor 0.70 (lagging) from a 240 V rms, 60 Hz supply. (a) What current do they draw? (b) What capacitor in parallel raises the power factor to 1.00? (c) What is the new current?

\[ \text{(a)}\quad I = \frac{P}{V\cos\phi} = \frac{5000}{(240)(0.70)} = 29.8\ \text{A} \] \[ \phi = \cos^{-1}0.70 = 45.6^\circ,\qquad Q_{\mathrm r} = P\tan\phi = (5000)(1.020) = 5.10\ \text{kvar} \]

(b) A capacitor across the supply draws a current that leads by \(90^\circ\), supplying reactive power \(V^2/X_C = V^2\omega C\). Make it cancel the motor's:

\[ V^2\omega C = Q_{\mathrm r}\ \Longrightarrow\ C = \frac{5101}{(240)^2(377)} = \frac{5101}{2.17\times10^7} = \boxed{235\ \mu\text{F}} \] \[ \text{(c)}\quad I = \frac{P}{V} = \frac{5000}{240} = 20.8\ \text{A} \]

The motors get exactly the same 5.0 kW, but the supply current drops from 29.8 A to 20.8 A, and the wiring losses drop by \((20.8/29.8)^2 = 0.49\) — roughly half. The capacitor and the motor's inductance now trade their reactive energy back and forth locally instead of through the utility's lines.

5. Why transmit at 400 kV?

A plant delivers 20 MW through lines with total resistance \(4.0\ \Omega\). Compare the line losses at 20 kV and 400 kV (treat as a single-phase line, power factor 1), and give the step-up turns ratio.

\[ 20\ \text{kV}:\quad I = \frac{20\times10^6}{20\times10^3} = 1000\ \text{A},\qquad P_{\mathrm{loss}} = (1000)^2(4.0) = 4.0\ \text{MW}\ \ (20\%) \] \[ 400\ \text{kV}:\quad I = \frac{20\times10^6}{400\times10^3} = 50\ \text{A},\qquad P_{\mathrm{loss}} = (50)^2(4.0) = 10\ \text{kW}\ \ (0.05\%) \]

20 times the voltage, 1/400 the loss. The step-up transformer needs \(N_s/N_p = 400/20 = 20\). Sanity check on the trap from Section 5: the voltage drop along the line at 400 kV is only \(IR = (50)(4.0) = 200\) V, and \(200^2/4.0 = 10\) kW — consistent. Using \((400\ \text{kV})^2/4.0\ \Omega = 40\) GW would be nonsense.

6. The fields of a Wi-Fi signal

A Wi-Fi router radiates 0.10 W at 2.4 GHz, assume equally in all directions. At 3.0 m, find the intensity, \(E_0\), \(B_0\), and the wavelength.

\[ I = \frac{P}{4\pi r^2} = \frac{0.10}{4\pi(3.0)^2} = 8.84\times10^{-4}\ \text{W/m}^2 \] \[ E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} = \sqrt{\frac{2(8.84\times10^{-4})}{(3.00\times10^8)(8.85\times10^{-12})}} = \boxed{0.82\ \text{V/m}},\qquad B_0 = \frac{E_0}{c} = 2.7\times10^{-9}\ \text{T} \] \[ \lambda = \frac{c}{f} = \frac{3.00\times10^8}{2.4\times10^9} = 0.125\ \text{m} = 12.5\ \text{cm} \]

Less than one volt per metre, yet a receiver decodes hundreds of megabits per second from it. The router's antennas are typically a half-wave long, about 6 cm. Compare sunlight's \(E_0\approx 870\) V/m: the Sun delivers about a million times the intensity.

Checklist

I can draw a phasor diagram and say which quantity leads, and I know “leads” means earlier in time.
I know \(X_L = \omega L\), \(X_C = 1/(\omega C)\), which way each shifts the phase, and how each behaves at DC and at high frequency.
I can find \(|Z|\), \(\phi\), \(I\) and every element voltage in a series RLC circuit, and check them with a phasor sum — not by adding magnitudes.
I can find \(\omega_0 = 1/\sqrt{LC}\), \(Q\) and the bandwidth \(\Delta\omega = R/L\), and I know what happens at resonance.
I use RMS values, \(P = V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi = I_{\mathrm{rms}}^2R\), and I can correct a power factor.
I can use the transformer ratios and explain, with \(I^2R_{\mathrm{line}}\), why the grid transmits at high voltage.
I can explain displacement current with the charging capacitor and write all four Maxwell equations, saying what each means.
I know \(c = 1/\sqrt{\mu_0\varepsilon_0}\), \(E = cB\), the direction \(\mathbf E\times\mathbf B\), and \(I = \tfrac12c\varepsilon_0E_0^2\).