electromagnetic field as a unified model

Before Maxwell, electric and magnetic phenomena were described by separate laws. Faraday's experiments showed that a changing magnetic flux creates a circulating electric field, so a changing $\vec{B}$ produces an $\vec{E}$. Maxwell added the symmetric statement: a changing electric field is equivalent to a displacement current $I_d=\varepsilon_0\,d\Phi_E/dt$ and itself produces a magnetic field. Therefore $\vec{E}$ and $\vec{B}$ cannot be treated separately: they are two aspects of one electromagnetic field, and this mutual generation makes a free wave possible without any charges or currents. In vacuum such a wave travels at $c=1/\sqrt{\mu_0\varepsilon_0}\approx 3{,}00\times10^{8}$ m/s, which matched the measured speed of light and became the main argument for the unified model. In this model $\vec{E}$ and $\vec{B}$ turn into each other when you change inertial frame, so the split into “electric” and “magnetic” depends on the observer.

Estimate the speed of an electromagnetic wave in vacuum from $c=1/\sqrt{\mu_0\varepsilon_0}$, taking $\mu_0=4\pi\times10^{-7}$ H/m and $\varepsilon_0=8.85\times10^{-12}$ F/m. Compare the result with the tabulated value $c=2.998\times10^{8}$ m/s.

A plane electromagnetic wave travels along the $z$ axis in vacuum with $E_x=E_0\sin(kz-\omega t)$, where $E_0=300$ V/m. Find the amplitude and direction of the magnetic field of the wave and explain why such a wave does not die out in empty space.

A circular capacitor of radius $R=5.0$ cm is being charged by a current $I=2.0$ A. Assuming the field between the plates is uniform, find the magnetic field at a distance $r=R/2$ from the axis and at the edge of the plates $r=R$, where outside the field is given by Ampère's law for a wire carrying current $I$. Explain what the unified field model demonstrates here.

Self-check with a simulation (safe, no equipment). Open a single-charge simulation, for example PhET “Radio Waves & Electromagnetic Fields”.

1. Place the charge at rest. Note what you see: only $\vec{E}$, no magnetic field.
2. Make the charge oscillate. Describe what appears around it: what the $\vec{E}$ and $\vec{B}$ lines look like and where the wave travels.
3. Estimate from the picture the angle between $\vec{E}$ and $\vec{B}$ and between each of them and the direction of propagation; check that they are perpendicular.
4. Double the oscillation frequency and note how the wavelength changes; compare with $\lambda=c/f$.
5. Write a one-sentence conclusion: why the field far from the charge keeps existing.

Optional (safe). Build a circuit from a 1.5 V cell, a piece of insulated wire, and a compass: when you close the circuit for 2–3 s the needle deflects. Use battery power only, never mains; the wire may warm up, so open the circuit immediately.

Show that the electric and magnetic fields are not two independent objects but two mutually perpendicular components of one travelling wave that sustain each other.

An axonometric sketch: the $z$ axis is the direction of propagation; a sine curve for $\vec{E}$ in the $xz$ plane with amplitude $E_0$; a sine curve for $\vec{B}$ in the $yz$ plane with amplitude $B_0=E_0/c$; the arrows $\vec{E}$, $\vec{B}$, and $\vec{v}$ are mutually perpendicular; labels “$E/B=c$” and one wavelength $\lambda$ between two neighbouring maxima.

A phone lies $3\ \text{m}$ from a Wi‑Fi router transmitting at $2.4\ \text{GHz}$. The router's antenna current oscillates $2.4\times10^9$ times per second, so one period is $T = 1/(2.4\times10^9\ \text{Hz}) \approx 4.2\times10^{-10}\ \text{s}$. The wave travels the $3\ \text{m}$ to the phone at $c = 3.0\times10^8\ \text{m/s}$, arriving after $t = 3.0\ \text{m} / (3.0\times10^8\ \text{m/s}) = 1.0\times10^{-8}\ \text{s}$. The wavelength in air is $\lambda = c/f = (3.0\times10^8\ \text{m/s})/(2.4\times10^9\ \text{Hz}) = 0.125\ \text{m} = 12.5\ \text{cm}$. The phone detects the oscillating electric and magnetic fields of the wave.

This is Maxwell's unified electromagnetic field in action. The oscillating current in the router antenna produces a changing electric field $E$ around the antenna, which produces a changing magnetic field $B$; that changing $B$ in turn produces a changing $E$, so the pair propagates as a self-sustaining wave without any medium. The energy density carried by the wave is $u = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2$, showing that $E$ and $B$ are two inseparable components of one field. The phone's antenna responds to the local $E$ and $B$ of this unified field.

A bicycle moves at $18\ \text{km/h} = 5.0\ \text{m/s}$. The wheel radius is $0.35\ \text{m}$, so the wheel turns at about $5.0/(2\pi \times 0.35) \approx 2.3$ revolutions per second. A sidewall dynamo steps this up so its permanent magnet rotates at $20\ \text{Hz}$ next to a coil of $N = 8000$ turns with area $A = 1.5\ \text{cm}^2 = 1.5\times10^{-4}\ \text{m}^2$. The magnetic field through the coil has peak value $B_0 = 0.20\ \text{T}$. The peak induced emf is $\mathcal{E}_0 = N A B_0 \omega$, with $\omega = 2\pi f = 2\pi(20\ \text{Hz}) \approx 126\ \text{rad/s}$, giving $\mathcal{E}_0 \approx 8000 \times 1.5\times10^{-4} \times 0.20 \times 126 \approx 6.0\ \text{V}$. The dynamo delivers about $3\ \text{W}$ to the headlamp.

The rotating magnet makes the magnetic flux through the coil change. According to the unified electromagnetic field, a changing magnetic field $B$ is accompanied by a circulating electric field $E$; this induced $E$ drives the current that lights the lamp. The same coupled field explains why a dynamo and an electric motor are one device run in opposite directions: a current creates $B$, and a changing $B$ creates $E$. Here the unified model replaces separate magnetic and electric effects with a single field whose components transform into each other.

An induction hob has a coil under the glass. The coil carries an alternating current at $25\ \text{kHz}$ and the hob delivers $2000\ \text{W}$ to a pot. The rapidly changing magnetic field $B$ (peak about $0.01\ \text{T}$ at the pot bottom) induces eddy currents in the ferromagnetic base of the pot. To heat $1.0\ \text{L}$ of water from $20\ ^\circ\text{C}$ to $100\ ^\circ\text{C}$ requires $Q = mc\Delta T = (1.0\ \text{kg})(4186\ \text{J kg}^{-1}\text{K}^{-1})(80\ \text{K}) = 3.35\times10^5\ \text{J}$. At $2000\ \text{W}$ this takes $t = Q/P = 3.35\times10^5/2000 \approx 168\ \text{s}$, about $2.8\ \text{min}$, matching a real hob. The glass stays cool because it is not electrically conducting.

The coil's alternating current produces a time-varying magnetic field $B$. Because $B$ reverses $25\,000$ times per second, $dB/dt$ is large, and the unified electromagnetic field requires an accompanying electric field $E$ inside the pot. That $E$ drives eddy currents, which dissipate energy as heat. This is a direct use of the unified model: an electric field appears where there is no battery and no wire, purely because the magnetic field is changing. It also shows that $E$ and $B$ are not independent but two aspects of one field.