Fields & Waves

Electric fields and force

In RF Power and Impedance & Reactance, we used voltage and current to describe energy transfer at circuit connections. For a radio link, we also need to describe the space between the antennas. Electric and magnetic fields have a strength and direction at each point in space. Both can change with time.

This module introduces the basic electromagnetic concepts used in RF engineering. We start with fields around metal plates and a wire, then study how travelling waves carry energy to a receiver. A more detailed treatment of electromagnetism is planned for a future Academy course.

Voltage and electric field

A voltage source connected to two metal plates places opposite charges on them. A small positive charge in the gap experiences a force toward the negative plate. A negative charge experiences a force in the opposite direction. The electric field E is force per unit charge. Its units, newtons per coulomb (N/C), are equivalent to volts per metre (V/m).

With 10 V across 10 mm, E is approximately 1000 V/m. Launch a charge horizontally in the illustration and watch its path bend toward a plate. Compare positive and negative launches, then change the voltage, gap or charge magnitude. Every launch uses the same entry speed and mass.

This force helps explain how an antenna receives a signal. The arriving electric field moves charges along the receiving conductor. The voltage produced at its terminals also depends on the conductor's shape, orientation and connection.

Magnetic fields and flux

A current produces a magnetic field around the conductor. For a long straight wire, point your right thumb in the direction of conventional current: your curled fingers show the magnetic field direction. Reversing the current reverses the field. The circles below represent magnetic field lines, not particles moving around the wire.

Magnetic field intensity H is measured in amperes per metre (A/m). For this wire, $H=|I|/(2\pi r)$, where I is current [A] and r is distance from the wire axis [m]. At a fixed distance, doubling the current doubles H; at a fixed current, doubling the distance halves H. The wire ends and return conductor must be far from the observation point for this estimate.

From the field at a point to flux through an area

We calculated H using the current and the distance from the wire. To calculate magnetic flux through an area, we use magnetic flux density B, measured in teslas (T). H and B are both local field quantities: they describe the field at a point. Their relationship depends on the material, so the same H need not correspond to the same B in different materials. In vacuum, and to a good approximation in air:

To obtain B in this case, multiply H by the vacuum permeability, $\mu_0\approx1.257\times10^{-6}\,\mathrm{H/m}$. The H in this unit means henry; magnetic field intensity H has units of A/m. No area enters this conversion because H and B describe the field at one point.

Now imagine a flat wire loop in a magnetic field. Magnetic flux $\Phi_B$ measures the magnetic field passing through the whole opening. It adds up the contribution of B perpendicular to the opening across its area. If B is uniform across the opening and perpendicular to it, the flux magnitude is:

Place two flat wire loops in the same uniform magnetic field in air, both facing the field directly. The field is equally strong at both loops, so H and B are the same. If one loop encloses twice the area, it has twice the magnetic flux through it.

With this orientation, B tells us how much magnetic flux passes through each square metre. $\Phi_B$ tells us how much passes through the whole area inside the loop. This is why B is called flux density: $1\,\mathrm T=1\,\mathrm{Wb/m^2}$.

Orientation matters too. Tilt the loop away from this position and less field crosses its opening, so the flux decreases even though B and the area inside the loop stay the same. When the field lies in the plane of the loop, the flux is zero. The simple product BA applies to the uniform, perpendicular case.

Electric flux through an area

We can also add up the electric field crossing an area. This is electric flux $\Phi_E$. Return to the plates and choose a flat imaginary surface in the gap, facing the field directly. For a uniform field perpendicular to that surface, the flux magnitude is:

Twice the area gives twice the electric flux while E stays the same. Tilting the surface reduces the flux; it becomes zero when the field lies along the surface. The surface is only a way to choose where we measure the field. No sheet or wire is required, and electric flux does not mean that charge flows through it.

Changing fields act on circuits

When the magnetic flux through a wire loop changes, a voltage appears. This is called electromagnetic induction. If the loop forms a closed circuit, that voltage can drive a current.

Move a magnet toward a stationary wire loop connected to a voltmeter: the changing field through its opening changes the flux. A voltage appears while the magnet moves and falls to zero when it stops. Pulling it away reverses the voltage. Watch the green flux bar and the meter: the meter responds to the change in flux.

Faraday's law relates changing flux to induced voltage. For one stationary turn with fixed area, consider a uniform perpendicular field whose magnitude changes steadily while its direction stays the same. The flux change is the area multiplied by the field change, so:

A 1 cm² loop has an area of 10⁻⁴ m². If B rises by 20 µT in 2 ms under these conditions, first calculate the flux change, then divide by the elapsed time:

One weber per second corresponds to one volt. Halving the time for the same flux change doubles the voltage. Keeping B constant produces no induced voltage in this stationary loop. Keeping a signal conductor close to its return conductor reduces the loop area and can reduce unwanted induced voltage.

The changing magnetic field produces an electric field around the loop. If a resistor closes the loop, this electric field drives a current. The current produces a magnetic field that opposes the change in flux. This is Lenz's law: the response opposes the change in flux, which may be an increase or a decrease.

A changing electric field also produces a magnetic field

Now consider the capacitor. The source below increases the voltage, keeps it constant, then decreases it. Watch charge arrive through the wires and the electric field increase between the plates. The copper-coloured loop represents the magnetic field around the gap. The voltage changes over a much longer time than a field disturbance takes to cross this small capacitor. We therefore ignore propagation delay. This ideal model also neglects edge effects and radiation.

While the voltage is constant, the electric field remains but the magnetic field is zero. During discharge, E decreases without changing direction, while the magnetic field reverses.

Changing the voltage changes the electric flux through the gap. In vacuum, the rate of this flux change determines the displacement current, measured in amperes. The Ampère–Maxwell law includes this contribution alongside ordinary charge current when describing the magnetic field.

For this fixed capacitor, raising the voltage twice as quickly requires twice the wire current to charge the plates. The electric flux also changes twice as quickly, giving twice the displacement current. Through the full plate area, it equals the wire current, although no charge crosses the gap. Once the voltage and flux stop changing, both currents are zero.

How electromagnetic waves travel

The capacitor showed how electric and magnetic fields change around a component. Now consider an antenna connected to a transmitter. The transmitter moves charge back and forth in the antenna, producing an alternating current. The fields change, and these changes reach more distant positions later. Part of the supplied energy travels outward in an electromagnetic wave.

First, separate two observations. At one fixed position, E and H change with time. Along the direction of travel, the same pattern reaches successive positions later. A delay between positions does not mean a delay between E and H at the same position.

The illustration shows a plane wave from an angle. At any instant, each field has the same value everywhere on a plane perpendicular to the direction of travel. Green E arrows and copper H arrows share an origin at every displayed position. The fields are perpendicular to each other and to the direction of travel. Watch each pair grow, shrink and reverse together while the pattern travels forward.

Follow one pair of arrows through a cycle. E and H grow, reach their peaks and reverse together. At a peak, each field is strong but momentarily stops changing. At a zero crossing, each field is zero but changing most rapidly. Field strength and rate of change are different quantities.

Why this agrees with induction

Faraday's law and the Ampère–Maxwell law are part of Maxwell's equations. They apply in empty space too: we can choose an imaginary loop without putting a wire there. There are two different measurements to make. Flux adds up the field through the opening. For the rectangles below, measure the field around each loop by counting the field along each edge multiplied by that edge's length. Contributions with the chosen path direction add; those against it subtract.

Faraday's law relates this around-the-loop electric field to changing magnetic flux through the opening. In empty space, the Ampère–Maxwell law relates the around-the-loop magnetic field to changing electric flux. Compare the two vertical sides of each loop below: E is perpendicular to the horizontal edges, so those edges add nothing to the around-the-loop measurement.

The laws therefore connect differences between nearby positions with changes over time. A zero rate of change does not mean a zero field: at the crest, H can be strong but momentarily not changing. This allows E and H to reach their peaks together.

The capacitor has a different field arrangement: its magnetic field depends on the rate of change of electric flux between the plates. Its H becoming zero during a constant-voltage interval does not require H to vanish at a travelling wave's peak.

In RF Signals, we used $\lambda=v/f$. In vacuum, $v=c\approx3\times10^8\,\mathrm{m/s}$. At 1 GHz, a cycle lasts 1 ns and the wavelength is 0.30 m. A wave crest takes a quarter of a period to travel a quarter of a wavelength.

This plane-wave model describes a travelling field. It does not describe all fields near a component or antenna. Before estimating field strength from antenna power, we will check when this model applies.

Fields carry power

The arriving fields carry energy across space. Power density S is power crossing unit area, measured in watts per square metre (W/m²). The Poynting vector, $\mathbf S=\mathbf E\times\mathbf H$, gives its direction as well as its magnitude. Here × is the vector cross product: it combines the fields' directions as well as their strengths. For the perpendicular E and H in this wave, S has magnitude EH and points perpendicular to both.

Watch the large field arrows at the fixed point below. They form two sides of a rectangle: its area represents the product of their magnitudes, and the right-hand rule gives the power direction. As both fields reverse, the rectangle grows on the opposite side of the point, while power still flows forward.

In a single vacuum plane wave, the ratio of electric field strength to magnetic field strength is the wave impedance, $\eta_0\approx376.7\,\Omega$. This is an E/H ratio in space, rather than a voltage/current ratio at a circuit port.

RMS works for fields just as it did for voltage in RF Power: at a fixed position, square the field values, average over a cycle, then take the square root. E and H are in phase in this travelling wave, so their RMS product gives average power density:

If $E_{\mathrm{rms}}=1\,\mathrm{V/m}$, then $S_{\mathrm{avg}}\approx2.65\,\mathrm{mW/m^2}$. Doubling E quadruples power density in the same medium, just as doubling voltage across the same resistor quadrupled power in RF Power.

Fields also carry power through a cable

A coaxial cable has an inner conductor surrounded by an insulating gap and an outer conductor. Voltage between the conductors produces an electric field across the gap, just as voltage produced a field between the plates. Current along the inner conductor produces a magnetic field around it, as in the wire example.

Assume an ideal lossless cable: none of the travelling energy is absorbed by its materials. For a single forward-travelling wave, E points across the gap toward or away from the inner conductor, and H circles it. Their cross product points along the cable. The energy travels through the insulating region between the conductors, guided by the conductors. This connects the field description to the circuit power you already know.

For example, a sinusoidal signal with 2 V RMS and 40 mA RMS carries $P_{\mathrm{avg}}=V_{\mathrm{rms}}I_{\mathrm{rms}}=80\,\mathrm{mW}$. Adding the field power flow across the cable's insulating cross-section gives the same 80 mW. Here voltage and current are in phase; reflections and transmission-line impedance are developed in Transmission Lines & Matching.

Fields can therefore carry power through a cable or through open space. In both cases, material in their path can change the fields. Next, we examine that material response before returning to the receiving antenna.

Materials and wave speed

The insulator in a cable is a dielectric. Its charges remain bound to atoms or molecules, but an electric field can shift the positive and negative charges slightly relative to one another. This small separation of bound charge is called dielectric polarization. It does not require charge to flow through the material as it does through a wire.

Return to the plates with 10 V across a 10 mm gap. Keep the voltage source connected and fill the gap with an ideal dielectric. Away from the edges, E remains approximately 1000 V/m because the voltage and gap are unchanged. The polarized material contributes a field opposing the plate charges' field, so the source supplies more charge to the plates to maintain the same voltage. More stored charge at the same voltage means greater capacitance.

Permittivity ε describes this electric response. Relative permittivity $\varepsilon_r=\varepsilon/\varepsilon_0$ compares it with vacuum permittivity ε₀. Assume a uniform, linear dielectric: its polarization is proportional to the electric field. Filling the gap with εr = 4 then gives four times the plate charge and capacitance compared with a vacuum gap, at the same voltage and plate spacing.

What changes for a travelling wave?

As a wave's electric field alternates, the bound charges move back and forth. This motion produces a polarization current, which contributes to the magnetic field through the Ampère–Maxwell law. The material response changes how the wave travels.

For the example below, use a uniform, ideal lossless dielectric. Its response is linear and isotropic (the same in every direction), and its properties do not vary with frequency.

Permeability μ describes magnetic response. In a linear, isotropic material, $B=\mu H$, extending the vacuum relationship we used earlier. Relative permeability $\mu_r=\mu/\mu_0$ compares it with vacuum. For our example, μr = 1: the material has the same B/H ratio as vacuum. Then:

At a stationary boundary, the frequency stays the same. With εr = 4, speed and wavelength are half their vacuum values. At 1 GHz, wavelength is 0.15 m instead of 0.30 m: each cycle advances half as far. This shorter wavelength does not mean that energy has been absorbed.

Reflection, absorption and skin depth

The ideal dielectric from the previous section, with μr = 1, changes the E/H ratio as well as wave speed: $\eta=\eta_0/\sqrt{\varepsilon_r}$. With εr = 4, wave impedance is half the vacuum value.

Consider a wave arriving perpendicular to the boundary between two lossless dielectrics, with no conducting sheet at the surface. The total electric and magnetic fields parallel to that surface must each match immediately on its two sides. These matching rules are boundary conditions.

If the wave impedances differ, a forward wave alone cannot match both E and H. A reflected wave travels back into the first material. For the same E direction, a backward wave has the opposite H direction, so its power flows back. Adding it to the arriving wave lets the total E and H match the transmitted wave travelling into the second material.

Absorption is a separate effect. In a lossy material, currents and other material responses convert wave energy into heat or other forms of internal energy. Compare the three effects below.

For example, an ideal lossless boundary that reflects 20% of the arriving power transmits the remaining 80%. If the transmitted wave later enters a lossy region, that region can absorb some of the transmitted power.

Why conductor surfaces matter at RF

Current density J is current per unit cross-sectional area, measured in A/m². It tells us how concentrated the current is. In copper, J is proportional to the local electric field: $J=\sigma E$. Conductivity σ, measured in siemens per metre (S/m), tells us how much current density a given field produces.

At RF, fields and current density become much weaker with depth inside a good conductor. Skin depth δ is the depth at which their amplitudes fall to about 37% of their surface values. E and J fall by the same fraction because conductivity is constant in this model. Current continues beyond δ; it is not a sharp cutoff.

Increase the frequency below and follow the dashed skin-depth marker toward the surface. The depth ruler stays fixed so you can compare penetration. For the same conductor and the same total RMS current, concentrating current into a thinner layer generally raises resistance and heating loss. The later transmission-line lesson develops that calculation.

For a real radio path, reflection and absorption can occur alongside spreading. The propagation course examines these effects in actual environments.

Receiver orientation and polarization

We have followed the wave through its path. Now consider the receiving antenna. The power it delivers depends on its orientation, directional response, losses and load, as well as the arriving power density. Here, we change only its orientation.

Wave polarization describes how the electric-field direction changes over time at one fixed point. This differs from dielectric polarization, which describes the displacement of bound charge in a material. In a linearly polarized wave, E reverses along one fixed axis. A linear receiving antenna has a polarization axis: the direction of electric field to which it responds. It responds to the part of E parallel to that axis, called the field's projection.

If the angle between these axes is θ, the projected field is $E_{\parallel}=E\cos\theta$. With the receiver's other properties fixed, received power is proportional to this field squared. That is why the power fraction below is cos²θ.

The diagram below shows the fields as viewed along the wave's direction of travel. Rotate only the receiver: the copper-coloured projection becomes smaller while the arriving, or incident, green field remains unchanged.

At 60°, the field projection is one half, so received power is one quarter of the aligned value. Rotating the receiver changes its response, not the incident wave's total field strength.

The antenna course extends this to circular and elliptical polarization and practical mismatch.

From antenna power to field strength

Before calculating field strength from transmitter power, check the observation point's position relative to the antenna. Recall how a capacitor stores energy in its electric field and returns it during discharge. Close to an antenna, electric and magnetic fields can also store energy and return some of it to the source. Power can flow toward the antenna during part of a cycle, while net energy still travels outward over the full cycle. This near field cannot generally be described by a single outward-travelling plane wave.

Far enough away, outward radiation dominates. In this far field, the radiation pattern—how strongly the antenna radiates in different directions—changes little with distance. A small region of the outward-travelling wave can be treated as a plane wave. In free space, field amplitudes decrease as 1/r and power density as 1/r². The transition is gradual; it depends on antenna size, wavelength and the required accuracy. Distance alone is not enough to decide.

Account for the antenna's direction

For the calculations below, assume that free-space and far-field conditions have been confirmed. The antenna's linear gain G is a power ratio, rather than a value in decibels. It compares radiation in the chosen direction with an ideal antenna that radiates equally in all directions, for the same accepted input power—the power entering the antenna after input reflection. Gain includes radiation efficiency: the fraction of accepted power that becomes radiation.

Multiplying accepted power by gain gives the equivalent isotropically radiated power (EIRP). This is the power an ideal source radiating equally in every direction would need to produce the same power density at the same distance in the chosen direction.

A sphere of radius r has area 4πr². Dividing the equivalent source's power by this area gives the power crossing each square metre at that distance:

An antenna accepting 50 mW with gain 2 toward the receiver has 100 mW EIRP, or +20 dBm. Under established free-space far-field conditions at 1 m, $S_{\mathrm{avg}}\approx7.96\,\mathrm{mW/m^2}$ and $E_{\mathrm{rms}}=\sqrt{\eta_0S_{\mathrm{avg}}}\approx1.73\,\mathrm{V/m}$. Use S in W/m² inside the square root.

Each moving patch below marks part of a wavefront: places at the same point in the wave cycle, such as a crest. Compare r with 2r: the same radiation spreads over four times the area, so power density becomes one quarter and field strength one half.

For antenna design and complete link calculations, continue to gain, directivity and efficiency and free-space links. Here, the goal is to calculate an estimate and identify the conditions required for it to be valid.

Apply it: explain a receiver's signal

A 900 MHz transmitter feeds an antenna that accepts 10 mW and has linear gain 3 toward a receiver. Free-space far-field conditions are established at 3 m and 6 m along that direction. At 3 m, an aligned linear receiver delivers 12 µW to its load. Use $c\approx3\times10^8\,\mathrm{m/s}$ and $\eta_0\approx376.7\,\Omega$.

1. Find wavelength, directional EIRP, average power density and RMS E at 3 m.
2. Predict how power density and E change at 6 m, without repeating all the arithmetic.
3. At 3 m, rotate only the receiver's polarization axis by 45°. Keep its directional response, losses and load unchanged. Find the received power and explain whether the incident field changed.
4. Could you apply the same spreading formula at 5 cm without checking whether that point is in the far field? State what you would need to know.

You have used circuit power, fields and wave propagation to explain a receiver's response. The next module, Noise, examines the unwanted fluctuations present with the received signal.