Chapter 8 · Electromagnetic Waves

8.1 Displacement Current and Maxwell's Equations

A changing electric field produces a magnetic field exactly as a current does. Adding this displacement current to Ampère's circuital law removes a contradiction at a charging capacitor and completes Maxwell's four equations of electromagnetism.

Touch the screen of a phone and nothing flows through your finger in the ordinary sense: the screen senses the tiny change in capacitance between your fingertip and a grid of transparent electrodes. Every one of those electrodes is a capacitor being charged and discharged millions of times a second. Current flows along the leads to its plates — but not across the gap between them, which is an insulator. Does the magnetic field of that current simply stop at the plates? Measurements say no: the magnetic field around a charging capacitor is just as if the current carried straight on across the gap.

In 1865 James Clerk Maxwell found the reason by a careful look at Ampère's circuital law (Section 4.3). Applied to a capacitor that is being charged, the law gives two different answers for the same loop. Maxwell repaired it by adding a second source of magnetic field: a changing electric field, which acts exactly like a current and which he called the displacement current. In this section we meet the contradiction, define the displacement current, use the corrected law to find the magnetic field between the plates of a capacitor, and put together the four laws of electromagnetism — Maxwell's equations — which in Section 8.2 predict electromagnetic waves.

A Gap in Ampère's Circuital Law

Ampère's circuital law says that for steady currents the circulation of $\mathbf B$ round any closed loop equals $\mu_0$ times the current threading the loop: $\oint\mathbf B\cdot d\mathbf l=\mu_0 i$. "Threading the loop" means crossing any surface whose edge is the loop — and for steady currents it does not matter which surface you choose, because charge cannot pile up anywhere: whatever flows into a region flows out again.

Now connect a parallel-plate capacitor to a source so that a current $i$ flows along the leads and charges it. Draw a circular loop $C$ of radius $r$ round one lead, a little before the capacitor. Near the lead, but far from the plates, the field is that of a long straight wire, so by symmetry $B$ is the same all round the loop and the left-hand side of Ampère's law is $B(2\pi r)$. For the right-hand side we need a surface bounded by $C$:

Surface $S_1$ — a flat disc. The lead pierces it, so the current through it is $i$ and $B(2\pi r)=\mu_0i$, giving $B=\dfrac{\mu_0i}{2\pi r}$.

Surface $S_2$ — a bag. Keep the same edge $C$ but stretch the surface into a pouch that passes between the plates, like a bag hanging into the gap. No wire crosses it: the conduction current through $S_2$ is zero, and Ampère's law would give $B=0$ on the very same loop.

The field on the loop cannot be both. Something crosses $S_2$ that we have not counted. Between the plates there is no current, but there is an electric field, and while the capacitor charges, that field is growing.

The Displacement Current and the Ampère–Maxwell Law

Let the plates have area $A$ and charge $\pm q$ at some instant. The field between them is uniform (we ignore the fringing at the edges), $E=\dfrac{q}{\varepsilon_0A}$ (Section 2.3), so the electric flux through the part of $S_2$ that lies between the plates is $$\Phi_E=EA=\frac{q}{\varepsilon_0}.$$ As the capacitor charges, $q$ grows at the rate $dq/dt=i$, so $$\varepsilon_0\frac{d\Phi_E}{dt}=\frac{dq}{dt}=i.$$ The quantity $\varepsilon_0\,d\Phi_E/dt$ through $S_2$ is exactly equal to the current $i$ through $S_1$. Maxwell's idea was to treat it as a current in its own right.

📘 Definition 8.1 — Displacement current
The displacement current through a surface is $\varepsilon_0$ times the rate of change of the electric flux through that surface: $$i_d=\varepsilon_0\,\frac{d\Phi_E}{dt}.$$ It is measured in amperes. The ordinary current carried by moving charges is then called the conduction current, $i_c$. Despite its name, the displacement current is not a flow of charge: it is a changing electric field, and it is a source of magnetic field just as a conduction current is.
📐 Law 8.1 — The Ampère–Maxwell law

For any closed loop and any surface bounded by it, the circulation of $\mathbf B$ round the loop is $\mu_0$ times the total current — conduction plus displacement — through the surface: $$\oint\mathbf B\cdot d\mathbf l=\mu_0\left(i_c+i_d\right)$$ $$\oint\mathbf B\cdot d\mathbf l=\mu_0i_c+\mu_0\varepsilon_0\,\frac{d\Phi_E}{dt}.$$ The sign of each current is fixed by the right-hand rule: curl the fingers along the direction of travel round the loop; currents and growing flux along the thumb count positive. For steady currents and fields that do not change, $i_d=0$ and the law reduces to Ampère's circuital law.

📐 Result 8.2 — The total current is continuous

For a parallel-plate capacitor being charged or discharged, the displacement current between the plates equals the conduction current in the leads at every instant: $$i_d=\varepsilon_0\frac{d\Phi_E}{dt}=\frac{dq}{dt}=i_c.$$ Conduction current flows along the wires up to a plate, and displacement current carries on across the gap, in the same direction (along $d\mathbf E/dt$). The sum $i_c+i_d$ has the same value through every surface bounded by the same loop, so the Ampère–Maxwell law gives the same $B$ whichever surface you choose. If the capacitor is connected to a voltage $V$ that changes, $q=CV$ and $i_d=C\,\dfrac{dV}{dt}$.

Notice the symmetry this brings into electromagnetism. Faraday's law (Chapter 6) says that a changing magnetic flux produces an electric field. The Ampère–Maxwell law says that a changing electric flux produces a magnetic field. Each kind of field can be produced by changes in the other — the key to electromagnetic waves.

Why did Ampère and everyone before Maxwell miss the extra term? Because it is usually tiny. In an ordinary circuit the electric field inside the wires barely changes, and the factor $\varepsilon_0=8.85\times10^{-12}$ makes $\varepsilon_0\,d\Phi_E/dt$ negligible unless the flux changes extremely fast. In a capacitor the conduction current stops at the plates, and the displacement current is the only thing that carries on.

✏️ Example 1 — The capacitor puzzle in numbers
A parallel-plate capacitor with circular plates of radius $6.0$ cm is charged by a steady current of $0.50$ A. A circular loop of radius $12$ cm surrounds one lead, far from the plates. (a) Use Ampère's law with a flat surface $S_1$ to find $B$ on the loop. (b) What does the original Ampère's law give with a bag-shaped surface $S_2$ passing between the plates? (c) Find the rate of change of the electric flux between the plates and the displacement current through $S_2$, and show that the Ampère–Maxwell law gives the same field as in (a).
(a) Flat disc The lead crosses $S_1$: $B(2\pi r)=\mu_0i$, so $$B=\frac{\mu_0i}{2\pi r}=\frac{(4\pi\times10^{-7})(0.50)}{2\pi(0.12)}$$ $\approx8.3\times10^{-7}$ T $=0.83\ \mu$T, circling the lead (right-hand grip rule).
(b) Bag, old law No wire crosses $S_2$, so $i_c=0$ and the original law gives $B=0$ — a contradiction with (a) for the same loop.
(c) Displacement current $\Phi_E=q/\varepsilon_0$, so $$\frac{d\Phi_E}{dt}=\frac{i}{\varepsilon_0}=\frac{0.50}{8.85\times10^{-12}}$$ $\approx5.6\times10^{10}$ V m/s. Then $i_d=\varepsilon_0\,d\Phi_E/dt=0.50$ A through $S_2$, and $B(2\pi r)=\mu_0i_d$ gives the same $0.83\ \mu$T.
Check Interactive 1 opens with this capacitor and loop: with either surface and the Ampère–Maxwell law it reads $B=0.833\ \mu$T. ✓
✏️ Example 2 — Displacement current from a changing field
The plates of a capacitor each have area $0.020\ \text{m}^2$. At a certain instant the uniform electric field between them is increasing at $3.0\times10^{12}$ V m$^{-1}$ s$^{-1}$. Find the displacement current between the plates and the conduction current in the leads at that instant.
Flux $\Phi_E=EA$ with $A$ fixed, so $\dfrac{d\Phi_E}{dt}=A\dfrac{dE}{dt}$.
Displacement current $$i_d=\varepsilon_0A\frac{dE}{dt}$$ $=(8.85\times10^{-12})(0.020)(3.0\times10^{12})\approx0.53$ A.
Leads By Result 8.2 the conduction current in the leads is also $0.53$ A. Units: $\text{C}^2\text{N}^{-1}\text{m}^{-2}\cdot\text{m}^2\cdot\text{N C}^{-1}\text{s}^{-1}=\text{C/s}=\text{A}$. ✓
Interactive 1 — One Loop, Two Surfaces

A capacitor with circular plates of radius $6.0$ cm is charged by a steady current $i$ (the moving dots). The dashed ellipse is a loop of radius $r$ round the left lead. Choose the surface bounded by that loop: the flat disc $S_1$ is pierced by the wire; the bag $S_2$ hangs into the gap, where only the growing field $\mathbf E$ (red arrows) crosses it. Switch between Ampère's original law and the Ampère–Maxwell law and compare the field each predicts on the loop. The drawing is not to scale: the loop is imagined far enough from the plates for the lead to act as a long straight wire. When the plates are "full" the animation empties them and starts again.

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conduction current through surface—
displacement current $\varepsilon_0\,d\Phi_E/dt$—
$dE/dt$ between the plates—
predicted $B$ on the loop—

The Magnetic Field between the Plates

The Ampère–Maxwell law does more than rescue the field round the leads: it predicts a magnetic field in the empty gap itself, where no charge moves at all. Take a capacitor with circular plates of radius $R$, charged by a current $i$, and ignore fringing, so that $\mathbf E$ is uniform between the plates and zero outside. By symmetry the magnetic field in the gap forms circles round the axis of the plates, with the same magnitude $B(r)$ everywhere on a circle of radius $r$. Apply the law to such a circle, using the flat disc it bounds (in the gap) as the surface. No conduction current crosses it, so only the displacement current counts.

Inside, $r\le R$. The disc catches only the fraction $\pi r^2/\pi R^2$ of the flux, so $i_d=i\,\dfrac{r^2}{R^2}$ and $B(2\pi r)=\mu_0i\dfrac{r^2}{R^2}$.

Outside, $r\ge R$. The disc catches all the flux, $i_d=i$, and $B(2\pi r)=\mu_0i$ — the same field as round the lead itself.

📐 Result 8.3 — Magnetic field between the plates of a charging capacitor

For a capacitor with circular plates of radius $R$ carrying a charging (or discharging) current $i$, with a uniform field between the plates and fringing neglected, the magnetic field in the gap at distance $r$ from the axis circles the axis and has magnitude

$$B=\frac{\mu_0\,i\,r}{2\pi R^2}\quad(r\le R),$$ $$B=\frac{\mu_0\,i}{2\pi r}\quad(r\ge R).$$

$B$ is zero on the axis, grows linearly to its largest value $\mu_0i/(2\pi R)$ at the edge of the plates, and then falls as $1/r$. Its sense is given by the right-hand grip rule with the thumb along $d\mathbf E/dt$ — the same sense as round the leads. Writing $i=\varepsilon_0\pi R^2\,dE/dt$, the inside field is $B=\dfrac{\mu_0\varepsilon_0r}{2}\dfrac{dE}{dt}$.

The pattern is exactly that of the magnetic field inside and outside a thick wire carrying a uniform current (Section 4.3). As far as the magnetic field is concerned, the gap behaves as a fat "wire" of displacement current.

✏️ Example 3 — Inside and outside the plates
For the capacitor of Example 1 (circular plates of radius $6.0$ cm, steady charging current $0.50$ A), find (a) the rate at which the field between the plates is increasing, (b) the magnetic field in the gap at $3.0$ cm, $6.0$ cm and $12$ cm from the axis, and (c) the displacement current through a circle of radius $3.0$ cm in the gap.
(a) dE/dt $A=\pi R^2=\pi(0.060)^2=1.13\times10^{-2}\ \text{m}^2$. $$\frac{dE}{dt}=\frac{i}{\varepsilon_0A}=\frac{0.50}{(8.85\times10^{-12})(1.13\times10^{-2})}$$ $\approx5.0\times10^{12}$ V m$^{-1}$ s$^{-1}$.
(b) r = 3.0 cm Inside: $$B=\frac{\mu_0ir}{2\pi R^2}=\frac{(2\times10^{-7})(0.50)(0.030)}{(0.060)^2}$$ $\approx8.3\times10^{-7}$ T.
r = 6.0 cm At the edge: $B=\dfrac{\mu_0i}{2\pi R}=\dfrac{(2\times10^{-7})(0.50)}{0.060}\approx1.7\times10^{-6}$ T, the largest value.
r = 12 cm Outside: $B=\dfrac{\mu_0i}{2\pi r}=\dfrac{(2\times10^{-7})(0.50)}{0.12}\approx8.3\times10^{-7}$ T — the same as round the lead in Example 1.
(c) Fraction $i_d=i\,\dfrac{r^2}{R^2}=0.50\times\dfrac14=0.125$ A — one quarter of the charging current.
Check Half the edge value occurs at both $r=R/2$ and $r=2R$, as the two $8.3\times10^{-7}$ T results show. Interactive 2 draws this $B(r)$ graph for the same capacitor. ✓
✏️ Example 4 — An alternating voltage across a capacitor
A $2.0$ nF capacitor is connected to a signal generator whose voltage is $v=50\sin(2\pi\nu t)$ volts with $\nu=1.0\times10^{4}$ Hz. (a) Find an expression for the displacement current between the plates and its peak value. (b) What is the displacement current at the instants when $v$ is greatest, and when $v=0$?
(a) Differentiate $q=Cv$, so $i_d=C\dfrac{dv}{dt}=C(50)(2\pi\nu)\cos(2\pi\nu t)$. The peak value is $$i_0=(2.0\times10^{-9})(50)(2\pi\times1.0\times10^4)$$ $\approx6.3\times10^{-3}$ A $=6.3$ mA.
(b) Instants When $v$ is at its peak, $\cos(2\pi\nu t)=0$: the voltage is momentarily not changing, so $i_d=0$. When $v=0$ the voltage changes fastest, and $i_d=\pm6.3$ mA.
Meaning A steady voltage gives no displacement current, so a capacitor blocks a steady current; an alternating voltage keeps the field changing, so alternating current "passes through" the gap as displacement current. The current in the leads is the same $i_d$ at every instant.
Interactive 2 — The Magnetic Field of a Displacement Current

Looking along the axis of the capacitor of Examples 1 and 3 (plates of radius $R=6.0$ cm, shaded), with $\mathbf E$ pointing into the page and growing (⊗). The displacement current is spread uniformly over the plates. The blue circles are magnetic field lines, with arrows sized by $B$. The probe circle of radius $r$ catches the displacement current inside it (shaded green); the graph shows $B(r)$ from Result 8.3. Press Play to sweep the probe, or set it with the slider.

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probe radius $r$—
$i_d$ through the probe circle—
$B$ on the probe circle—
$dE/dt$ between the plates—

Maxwell's Equations

With the displacement current in place, all of electricity and magnetism can be summarised in four laws. You have met each of them already; Maxwell's contribution was the last term of the fourth, and the realisation that together they form a complete and consistent set. Here they are for fields in vacuum, with $Q$ the charge enclosed by a closed surface and $i_c$ the conduction current through a surface bounded by a closed loop.

📐 Law 8.4 — Maxwell's equations

1. Gauss's law for electricity (Section 1.3). The electric flux out of any closed surface is the enclosed charge divided by $\varepsilon_0$:

$$\oint\mathbf E\cdot d\mathbf A=\frac{Q}{\varepsilon_0}.$$

2. Gauss's law for magnetism (Section 5.1). The magnetic flux out of any closed surface is zero — there are no isolated magnetic poles:

$$\oint\mathbf B\cdot d\mathbf A=0.$$

3. Faraday's law of induction (Chapter 6). A changing magnetic flux through a loop produces an electric field circulating round it:

$$\oint\mathbf E\cdot d\mathbf l=-\frac{d\Phi_B}{dt}.$$

4. The Ampère–Maxwell law (Law 8.1). A conduction current or a changing electric flux produces a magnetic field circulating round it:

$$\oint\mathbf B\cdot d\mathbf l=\mu_0i_c+\mu_0\varepsilon_0\frac{d\Phi_E}{dt}.$$

Together with the force on a charge, $\mathbf F=q\left(\mathbf E+\mathbf v\times\mathbf B\right)$ (Section 4.1), these laws describe every electric and magnetic phenomenon outside the atomic scale.

Read the four equations in words and a pattern appears. The first two say where field lines can start: electric field lines begin and end on charges, magnetic field lines never begin or end. The last two say what makes fields circulate: currents and changing electric fields make $\mathbf B$ circulate; changing magnetic fields make $\mathbf E$ circulate. Now imagine a region of empty space — no charges, no currents. Equations 3 and 4 still have one term each: a changing $\mathbf B$ makes an $\mathbf E$, and a changing $\mathbf E$ makes a $\mathbf B$. If each change keeps generating the other, a disturbance of the fields can travel through empty space on its own. Maxwell worked out its speed from equations 3 and 4 and found $1/\sqrt{\mu_0\varepsilon_0}$: $$\frac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}}\approx3.0\times10^8\ \text{m/s}.$$ That is the speed of light. Section 8.2 follows this prediction.

✏️ Example 5 — Which law is at work?
For each situation name the one of Maxwell's equations that describes it best. (a) Breaking a bar magnet in half always gives two smaller magnets, each with a north and a south pole. (b) The electric field lines of a charged metal sphere spread out radially and their flux is the same through any closed surface round it. (c) Pushing a magnet into a coil makes a current flow in the coil. (d) A compass needle placed between the plates of a capacitor connected to a high-frequency AC source is disturbed, although no charge crosses the gap. (e) The units of $\mu_0\varepsilon_0$ are $\text{s}^2/\text{m}^2$.
(a) No isolated poles, so the magnetic flux out of any closed surface is zero: Gauss's law for magnetism, $\oint\mathbf B\cdot d\mathbf A=0$.
(b) Flux through a closed surface fixed by the enclosed charge: Gauss's law, $\oint\mathbf E\cdot d\mathbf A=Q/\varepsilon_0$.
(c) A changing magnetic flux produces a circulating electric field that drives the current: Faraday's law.
(d) A changing electric flux produces a magnetic field: the displacement-current term of the Ampère–Maxwell law. (In practice the field is weak and a compass responds too slowly to follow the rapid alternations; Interactive 2 and the Research–Teaching Nexus below show how small it is.)
(e) Check From the Ampère–Maxwell law, $\mu_0i_c$ and $\mu_0\varepsilon_0\,d\Phi_E/dt$ have the same units (T m). $\Phi_E$ is in V m, so $d\Phi_E/dt$ is in V m/s, and $\mu_0\varepsilon_0$ has units $\dfrac{\text{T m}}{\text{V m/s}}=\dfrac{\text{T s}}{\text{V}}$. With $1\ \text{T}=1\ \text{V s/m}^2$ this is $\text{s}^2/\text{m}^2$ — the units of $1/(\text{speed})^2$. ✓
⚠️ Common mistakes with displacement current

1. Thinking charge flows across the gap. The displacement current is a changing electric field, not moving charge. No electron crosses the gap of an ideal capacitor.

2. Using $\Phi_E$ instead of its rate of change. $i_d=\varepsilon_0\,d\Phi_E/dt$. A large but steady field gives no displacement current; a fully charged capacitor in a DC circuit has $i_d=0$.

3. Forgetting the fraction inside the plates. A circle of radius $r\lt R$ between the plates encloses only $i\,r^2/R^2$ of the displacement current, so $B=\mu_0ir/(2\pi R^2)$, not $\mu_0i/(2\pi r)$.

4. Thinking $B=0$ between the plates "because there is no current there". There is no conduction current, but the displacement current makes a magnetic field everywhere in the gap except on the axis.

5. Getting the direction backwards. $i_d$ points along $d\mathbf E/dt$, which for a charging capacitor is along $\mathbf E$ (from the positive to the negative plate) — the same direction as the conduction current in the leads. While the capacitor discharges, $\mathbf E$ shrinks and $i_d$ points the other way, again matching the leads.

🚀 Beyond the board syllabus (JEE Main · NEET) — displacement current in a charging RC circuit

Displacement current is listed by JEE Main and NEET (CBSE keeps only the basic idea); JEE Advanced does not list it, although it lists the RC circuit used here.

When an uncharged capacitor $C$ is connected through a resistor $R$ to a cell of emf $\varepsilon$ (negligible internal resistance), the charging current is not steady: it starts at $\varepsilon/R$ and dies away as the capacitor fills, $$i=\frac{\varepsilon}{R}\,e^{-t/RC},$$ where $RC$ is the time constant. By Result 8.2 the displacement current between the plates follows the same curve, and so does the magnetic field in the gap: it is largest at the moment of switching on and falls to $1/e\approx37\%$ of that after one time constant. For $\varepsilon=10$ V, $R=1.0\ \text{k}\Omega$ and $C=1.0\ \mu$F: $RC=1.0$ ms, $i_d=10$ mA at $t=0$ and about $3.7$ mA at $t=1.0$ ms. Charging and discharging of RC circuits is in the JEE Advanced syllabus; boards do not ask for it.

Exercises

I
Displacement current. Use $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$.
  • 1 Show that $\varepsilon_0\,d\Phi_E/dt$ has the unit of current.
  • 2 The electric flux through a surface changes at $2.0\times10^{9}$ V m/s. Find the displacement current through it.
  • 3 The uniform field between capacitor plates of area $0.010\ \text{m}^2$ is changing at $2.0\times10^{12}$ V m$^{-1}$ s$^{-1}$. Find the displacement current.
  • 4 The voltage across a $5.0\ \mu$F capacitor is rising at $200$ V/s. Find the displacement current between its plates.
  • 5 A capacitor with plates of area $0.040\ \text{m}^2$ is charged by a current of $2.0$ A. At what rate is the field between the plates changing?
  • 6 A capacitor connected to a battery has been fully charged for a long time. What are the conduction current in the leads and the displacement current between the plates?
  • 7 A copper wire carries a steady current. Is there a displacement current inside the wire? Explain.
  • 8 A closed surface encloses one plate of a capacitor that is being charged. Show that the total current ($i_c+i_d$) out of the closed surface is zero.
1 $\Phi_E=Q/\varepsilon_0$ for a closed surface, so $\varepsilon_0\Phi_E$ has the unit of charge (C) and $\varepsilon_0\,d\Phi_E/dt$ has the unit C/s $=$ A.
2 $i_d=(8.85\times10^{-12})(2.0\times10^9)\approx1.8\times10^{-2}$ A $=18$ mA.
3 $i_d=\varepsilon_0A\,dE/dt$ $=(8.85\times10^{-12})(0.010)(2.0\times10^{12})$ $\approx0.18$ A.
4 $i_d=C\,dV/dt$ $=(5.0\times10^{-6})(200)$ $=1.0\times10^{-3}$ A $=1.0$ mA.
5 $\dfrac{dE}{dt}=\dfrac{i}{\varepsilon_0A}$ $=\dfrac{2.0}{(8.85\times10^{-12})(0.040)}$ $\approx5.6\times10^{12}$ V m$^{-1}$ s$^{-1}$.
6 Both are zero: the charge no longer changes, so $i_c=dq/dt=0$, and the field between the plates is constant, so $i_d=0$.
7 For a steady current the field inside the wire is constant in time ($E=V/l$), so $d\Phi_E/dt=0$ and $i_d=0$. (Even for AC the displacement current inside a metal is negligible compared with the conduction current.)
8 The lead brings conduction current $i$ into the surface. Where the surface passes through the gap, the flux out of it grows at $i/\varepsilon_0$, so the displacement current out is $i$. Net total current out: $-i+i=0$ — total current is never lost, it only changes form.
II
The field between the plates. A capacitor has circular plates of radius $10$ cm and is charged by a steady current of $1.5$ A. Neglect fringing; $\mu_0=4\pi\times10^{-7}$ T m/A.
  • 1 Find the magnetic field in the gap $5.0$ cm from the axis.
  • 2 Find the field at the edge of the plates, $r=10$ cm.
  • 3 Find the field in the plane of the gap $20$ cm from the axis.
  • 4 What is the field on the axis? Explain.
  • 5 At what rate is the electric field between the plates increasing?
  • 6 Find the displacement current through a circle of radius $5.0$ cm, centred on the axis, in the gap.
  • 7 At which two distances from the axis is the field half of its largest value?
  • 8 The plate separation is halved while the same charging current flows. How do the displacement current and the field at the edge change?
1 $B=\dfrac{\mu_0ir}{2\pi R^2}$ $=\dfrac{(2\times10^{-7})(1.5)(0.050)}{(0.10)^2}$ $=1.5\times10^{-6}$ T.
2 $B=\dfrac{\mu_0i}{2\pi R}=\dfrac{(2\times10^{-7})(1.5)}{0.10}=3.0\times10^{-6}$ T.
3 $B=\dfrac{\mu_0i}{2\pi r}=\dfrac{(2\times10^{-7})(1.5)}{0.20}=1.5\times10^{-6}$ T.
4 Zero: a circle of zero radius encloses no displacement current ($B\propto r$ inside), and by symmetry the field there could point in no preferred direction round the axis.
5 $\dfrac{dE}{dt}=\dfrac{i}{\varepsilon_0\pi R^2}$ $=\dfrac{1.5}{(8.85\times10^{-12})(0.0314)}$ $\approx5.4\times10^{12}$ V m$^{-1}$ s$^{-1}$.
6 $i_d=i\,r^2/R^2=1.5\times0.25\approx0.38$ A.
7 Inside, $B\propto r$: half at $r=R/2=5.0$ cm. Outside, $B\propto1/r$: half at $r=2R=20$ cm (parts 1 and 3).
8 Neither changes. $i_d=i$ whatever the gap; halving $d$ doubles $C$ and changes how fast the voltage rises, but with the current fixed, $dq/dt$ — and hence $dE/dt=i/(\varepsilon_0A)$ and $B$ — is unchanged.
III
Maxwell's equations.
  • 1 Write down Maxwell's four equations in integral form and name each one.
  • 2 If isolated magnetic poles (monopoles) were discovered, which equation would have to change first, and how?
  • 3 Which equation shows that a changing magnetic field produces an electric field? Give an everyday device that relies on it.
  • 4 Which equation, and which term in it, gives the magnetic field in the gap of a charging capacitor?
  • 5 In a region of space with no charges and no currents, write what is left of each of the four equations.
  • 6 Explain, using two of Maxwell's equations, why electric field lines can begin and end but magnetic field lines always form closed loops.
  • 7 Show that $\mu_0\varepsilon_0$ has the units $\text{s}^2/\text{m}^2$ and evaluate $1/\sqrt{\mu_0\varepsilon_0}$.
1 Gauss's law $\oint\mathbf E\cdot d\mathbf A=Q/\varepsilon_0$; Gauss's law for magnetism $\oint\mathbf B\cdot d\mathbf A=0$; Faraday's law $\oint\mathbf E\cdot d\mathbf l=-d\Phi_B/dt$; Ampère–Maxwell law $\oint\mathbf B\cdot d\mathbf l=\mu_0i_c+\mu_0\varepsilon_0\,d\Phi_E/dt$.
2 Gauss's law for magnetism: the magnetic flux out of a closed surface would no longer be zero but proportional to the magnetic "charge" enclosed, just as $\oint\mathbf E\cdot d\mathbf A$ is proportional to $Q$.
3 Faraday's law. Examples: a generator, a transformer, an induction cooktop, the pick-up of an electric guitar.
4 The Ampère–Maxwell law, through its displacement-current term $\mu_0\varepsilon_0\,d\Phi_E/dt$ ($i_c=0$ in the gap).
5 $\oint\mathbf E\cdot d\mathbf A=0$; $\oint\mathbf B\cdot d\mathbf A=0$; $\oint\mathbf E\cdot d\mathbf l=-d\Phi_B/dt$; $\oint\mathbf B\cdot d\mathbf l=\mu_0\varepsilon_0\,d\Phi_E/dt$. The last two still couple $\mathbf E$ and $\mathbf B$, which allows electromagnetic waves.
6 Gauss's law: a net electric flux out of a closed surface means there is charge inside, so lines start on positive and end on negative charges. Gauss's law for magnetism: the magnetic flux out of every closed surface is zero, so every line that enters a region leaves it — the lines never end and close on themselves.
7 $\mu_0$: T m/A $=\text{N/A}^2$; $\varepsilon_0$: $\text{C}^2\text{N}^{-1}\text{m}^{-2}$. Product: $\dfrac{\text{N}}{\text{A}^2}\cdot\dfrac{\text{C}^2}{\text{N m}^2}=\dfrac{\text{C}^2}{\text{A}^2\text{m}^2}=\dfrac{\text{s}^2}{\text{m}^2}$. $1/\sqrt{(1.257\times10^{-6})(8.85\times10^{-12})}\approx3.0\times10^8$ m/s.
IV
Applications and longer problems.
  • 1 A $10\ \mu$F capacitor is connected to a supply $v=320\sin(100\pi t)$ volts. Find the peak displacement current between its plates.
  • 2 For the capacitor of part 1, find the displacement current at the instants when $v=320$ V and when $v=0$, and explain the difference.
  • 3 The magnetic field at the edge of the circular plates (radius $5.0$ cm) of a charging capacitor is measured to be $4.0\ \mu$T. Find the charging current.
  • 4 Modelling (illustrative values). Under a thundercloud the vertical electric field near the ground is $1.0\times10^{5}$ V/m over an area of $4.0\ \text{km}^2$. A lightning flash makes this field collapse to almost zero in about $0.50$ s. Estimate the average displacement current through that area during the collapse.
  • 5 For the capacitor of Example 1, find the displacement current per unit area of the plates.
  • 6 Two capacitors with circular plates of radii $R$ and $2R$ are charged by equal currents. Compare the largest magnetic fields in their gaps.
  • 7 A capacitor with square plates of side $10$ cm and separation $1.0$ mm is connected to a source whose voltage rises at $2.0\times10^{4}$ V/s. Find its capacitance and the displacement current.
1 $i_d=C\,dv/dt=C(320)(100\pi)\cos(100\pi t)$. Peak: $(10\times10^{-6})(320)(314)\approx1.0$ A.
2 At $v=320$ V the voltage is at its peak and momentarily not changing: $i_d=0$. At $v=0$ it changes fastest: $i_d=\pm1.0$ A. The displacement current depends on how fast $v$ changes, not on $v$ itself.
3 $B=\dfrac{\mu_0i}{2\pi R}\Rightarrow i=\dfrac{2\pi RB}{\mu_0}$ $=\dfrac{2\pi(0.050)(4.0\times10^{-6})}{4\pi\times10^{-7}}$ $=1.0$ A.
4 $i_d\approx\varepsilon_0A\dfrac{\Delta E}{\Delta t}$ $=(8.85\times10^{-12})(4.0\times10^6)\dfrac{1.0\times10^5}{0.50}$ $\approx7.1$ A — a sizeable current through the air, made of nothing but a changing field.
5 $\dfrac{i_d}{A}=\dfrac{0.50}{\pi(0.060)^2}\approx44\ \text{A/m}^2$ (this equals $\varepsilon_0\,dE/dt$).
6 The largest field is at the edge, $\mu_0i/(2\pi R)$, so the capacitor with radius $2R$ has half the largest field (the same displacement current is spread over a wider area).
7 $C=\dfrac{\varepsilon_0A}{d}$ $=\dfrac{(8.85\times10^{-12})(0.010)}{1.0\times10^{-3}}$ $=8.85\times10^{-11}$ F $=88.5$ pF. $i_d=C\,dV/dt$ $=(8.85\times10^{-11})(2.0\times10^4)$ $\approx1.8\times10^{-6}$ A $=1.8\ \mu$A.
🎯 Exam corner — displacement current and Maxwell's equations BoardJEE MainNEET

E1. Board · 3 marks Explain, with the help of a capacitor being charged, why Ampère's circuital law needs to be modified. Define displacement current and show that, for a parallel-plate capacitor, the displacement current between the plates equals the conduction current in the connecting wires.

Problem For a loop round a lead, a flat surface pierced by the lead gives $\oint\mathbf B\cdot d\mathbf l=\mu_0i$, but a pot-shaped surface through the gap, bounded by the same loop, has no current through it and gives $0$. One loop cannot have two values of $\oint\mathbf B\cdot d\mathbf l$, so the law is incomplete. (Diagram: capacitor, loop, surfaces $S_1$ and $S_2$.)
Definition $i_d=\varepsilon_0\,d\Phi_E/dt$; the corrected law is $\oint\mathbf B\cdot d\mathbf l=\mu_0(i_c+i_d)$.
Proof Between the plates $E=\dfrac{q}{\varepsilon_0A}$, so $\Phi_E=EA=\dfrac{q}{\varepsilon_0}$ and $i_d=\varepsilon_0\dfrac{d}{dt}\left(\dfrac{q}{\varepsilon_0}\right)=\dfrac{dq}{dt}=i_c$.
Exam tip One mark is for the diagram with two surfaces on the same loop; one for the definition with the formula; one for the three-line proof.

E2. Board · 1 mark Assertion (A): A capacitor connected to a battery for a long time has no displacement current between its plates. Reason (R): The displacement current is proportional to the rate of change of the electric flux, not to the flux itself. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Concept Fully charged: $q$, $E$ and $\Phi_E$ are constant, so $i_d=\varepsilon_0\,d\Phi_E/dt=0$, although the field is large. R is true and is exactly the reason.
Answer (A). Exam tip: "large field" never means "large displacement current" — only a changing field counts.

E3. JEE Main A capacitor has circular plates of radius $4.0$ cm. The uniform electric field between them is increasing at $1.5\times10^{12}$ V m$^{-1}$ s$^{-1}$. The magnetic field in the gap at $2.0$ cm from the axis is (take $c=3.0\times10^8$ m/s) (A) $1.7\times10^{-7}$ T (B) $3.3\times10^{-7}$ T (C) $8.3\times10^{-8}$ T (D) zero.

Setup $r\lt R$. Ampère–Maxwell round a circle of radius $r$: $B(2\pi r)=\mu_0\varepsilon_0\,\pi r^2\,\dfrac{dE}{dt}$, so $B=\dfrac{\mu_0\varepsilon_0r}{2}\dfrac{dE}{dt}=\dfrac{r}{2c^2}\dfrac{dE}{dt}$.
Numbers $B=\dfrac{0.020}{2(9.0\times10^{16})}(1.5\times10^{12})$ $\approx1.7\times10^{-7}$ T.
Answer (A). Exam tip: $\mu_0\varepsilon_0=1/c^2$ saves a lot of arithmetic; (B) forgets the factor $\tfrac12$, and the plate radius is not needed at all for $r\lt R$.

E4. NEET The potential difference across a $2.0\ \mu$F capacitor is changing at $5.0$ V per millisecond. The displacement current between its plates is (A) $1.0$ mA (B) $10$ mA (C) $0.10$ A (D) $2.5$ mA.

Formula $i_d=C\,\dfrac{dV}{dt}$ $=(2.0\times10^{-6})\left(\dfrac{5.0}{1.0\times10^{-3}}\right)$ $=1.0\times10^{-2}$ A.
Answer (B) $10$ mA. Exam tip: convert "per millisecond" to per second first — (A) is what you get by forgetting a factor of $10$ in that conversion.

E5. JEE Main (One or more correct.) A parallel-plate capacitor with circular plates of radius $R$ is being charged by a constant current $I$. Neglect fringing. Then (A) the magnetic field in the gap is zero on the axis; (B) at distance $r\lt R$ from the axis the magnetic field is proportional to $r$; (C) the displacement current through the whole gap equals $I$; (D) in the plane of the gap at $r\gt R$ the magnetic field is zero because no current, of either kind, flows there.

Options $B=\mu_0Ir/(2\pi R^2)$ inside: zero at $r=0$ (A true), proportional to $r$ (B true). $i_d=\varepsilon_0\,d\Phi_E/dt=dq/dt=I$ (C true). Outside the plates $B=\mu_0I/(2\pi r)\neq0$: the circle of radius $r$ encloses the whole displacement current even though none flows at the circle itself (D false).
Answer (A), (B), (C). Exam tip: what matters is the current through the surface bounded by the loop, not the current at the loop.

E6. Board · Case study A student builds a large capacitor from two circular aluminium plates of radius $8.0$ cm and charges it with a steady current of $0.80$ A. (i) (2 marks) Find the magnetic field in the gap at $4.0$ cm from the axis and at the edge of the plates. (ii) (1 mark) The student looks along the axis from the positive plate towards the negative plate. Is the magnetic field clockwise or anticlockwise? (iii) (1 mark) What happens to the magnetic field in the gap when the capacitor is fully charged?

(i) Edge: $B=\dfrac{\mu_0i}{2\pi R}=\dfrac{(2\times10^{-7})(0.80)}{0.080}=2.0\times10^{-6}$ T. At $r=R/2$: $B=\dfrac{\mu_0ir}{2\pi R^2}$, half of that, $1.0\times10^{-6}$ T.
(ii) $\mathbf E$ points from the positive to the negative plate, away from the student, and is growing, so $i_d$ points away from the student. Right-hand grip rule with the thumb away: clockwise as seen by the student.
(iii) The current stops, $E$ stops changing, $i_d=0$ and the magnetic field in the gap disappears. Exam tip: the field exists only while the capacitor charges or discharges.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How large is the magnetic field produced by a displacement current in a laboratory capacitor, and what would it take to measure it?

Why it matters. Maxwell added the displacement current in 1865 on theoretical grounds. Its existence was confirmed indirectly when Hertz produced electromagnetic waves in 1887, but measuring the magnetic field inside a capacitor gap directly is a delicate experiment that physics laboratories still use to test instruments. The same calculation matters today in high-frequency electronics, where the fields between the closely spaced layers of a circuit board and its tiny capacitors can couple unwanted signals into neighbouring tracks.

What to do.

  1. Model a demonstration capacitor: two circular plates of radius $10$ cm, $5.0$ mm apart, in air. Find its capacitance.
  2. Connect it (in your model) to a sinusoidal supply of peak voltage $100$ V at $1.0$ kHz. Find the peak displacement current and the peak magnetic field at the edge of the plates (Result 8.3).
  3. Repeat for $1.0$ MHz. How does the field scale with frequency, and why?
  4. Compare both answers with the Earth's magnetic field (about $5\times10^{-5}$ T). Explain why a steady field this small, or a slowly varying one, would be hopeless to detect, and why a high frequency helps.

Tools: Calculator or spreadsheet.

Hand in: A short table of $C$, $i_d$ and $B_\text{edge}$ at the two frequencies, the ratio to the Earth's field, and a paragraph proposing how you would detect the field.

Going further. A small search coil placed in the gap with its plane containing the axis picks up the circulating field. Using Faraday's law (Chapter 6), explain why its emf grows with frequency even faster than $B$ does.

Instructor note: $C=\varepsilon_0\pi R^2/d\approx5.6\times10^{-11}$ F ($56$ pF). At $1.0$ kHz: $i_0=C\omega V_0\approx3.5\times10^{-5}$ A, $B_\text{edge}=\mu_0i_0/(2\pi R)\approx7.0\times10^{-11}$ T. At $1.0$ MHz: $i_0\approx3.5\times10^{-2}$ A, $B\approx7.0\times10^{-8}$ T, still only about $1.4\times10^{-3}$ of the Earth's field. $B\propto\omega$ because $i_d=C\,dV/dt$. A high frequency both raises $B$ and lets you separate the signal from the steady Earth field (tuned detection). A search coil's emf is proportional to $dB/dt\propto\omega B\propto\omega^2$.

🧪 Self-check quiz

0/5 correct

Q1. The displacement current through a surface is

Q2. A capacitor is being charged by a current of $2.0$ A. The displacement current between its plates is

Q3. Between the circular plates of a charging capacitor (radius $R$), the magnetic field at distance $r\lt R$ from the axis

Q4. Which of Maxwell's equations expresses the fact that isolated magnetic poles do not exist?

Q5. A steady current flows through a resistor connected to a battery. The displacement current inside the resistor is

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