Displacement current

A puzzle at the capacitor

Narrated video. Transcript:

Faraday showed that a changing magnetic field produces an electric field. Maxwell asked the reverse question: can a changing electric field produce a magnetic field? The answer came from a puzzle about a capacitor.

Here is a parallel plate capacitor being charged by a steady current i. Charge flows along the wire onto the left plate, and the same amount leaves the right plate. But no charge crosses the gap between the plates.

Let us find the magnetic field at a point P, a distance r from the wire, just outside the capacitor. Draw a circular loop of radius r through P, centred on the wire.

Ampère’s circuital law says: the line integral of B around the loop equals μ₀ times the current through a surface bounded by the loop. Take a flat disc as that surface. The wire pierces it, so B times 2πr equals μ₀ i.

Now keep the same loop, but stretch the surface into a pot whose bottom lies between the plates. No wire crosses this pot. There is no current through it, so Ampère’s law now gives B equals zero.

Same loop, same point P, two different answers. Both cannot be right. Ampère’s law, as it stands, is missing something.

Look at what does cross the pot: the electric field between the plates. With charge Q on plates of area A, the field is E equals Q over ε₀A, straight across the gap.

So the electric flux through the pot is E times A, which is Q over ε₀. As the capacitor charges, Q grows, and this flux grows too: dΦ_E by dt equals 1 over ε₀ times dQ by dt.

But dQ by dt is just the current i in the wire. So ε₀ times dΦ_E by dt equals i. The changing electric flux, multiplied by ε₀, is exactly the current that was missing.

The Ampère–Maxwell law

Narrated video. Transcript:

Maxwell called ε₀ dΦ_E by dt the displacement current, i_d. No charge flows in it. It is a changing electric flux, but it produces a magnetic field just as an ordinary current does.

The ordinary current, carried by moving charges, is the conduction current, i_c. The total current is the sum: i equals i_c plus i_d.

So Ampère’s law becomes the Ampère–Maxwell law: the line integral of B dot dl around a closed loop equals μ₀ i_c plus μ₀ ε₀ dΦ_E by dt.

Check it at our capacitor. Through the flat disc, i_c is i and there is no electric flux. Through the pot, i_c is zero, but i_d is i. Both surfaces now give the same B. The puzzle is solved.

In the wires the current is all conduction current. Between the plates it is all displacement current. Together the total current is continuous, so Kirchhoff’s junction rule holds for the total current.

A worked example. Circular plates of radius 12 cm are charged by a steady current of 0.15 A. Find B between the plates, 6 cm from the axis.

The plate area is π × 0.12², which is 0.0452 m². So E between the plates grows at dE by dt equals i over ε₀A: 0.15 over 8.85 × 10⁻¹² × 0.0452, about 3.7 × 10¹¹ volt per metre per second.

A loop of radius 6 cm encloses only part of this flux: (6 over 12)², one quarter. So the displacement current through the loop is 0.15 over 4, which is 0.0375 A.

Then B × 2π × 0.06 equals μ₀ × 0.0375. That gives B equals 2 × 10⁻⁷ × 0.0375 over 0.06, which is 1.25 × 10⁻⁷ tesla. Tiny, but real.

The big idea: a changing magnetic field makes an electric field, and a changing electric field makes a magnetic field. With Gauss’s two laws, Faraday’s law and the Ampère–Maxwell law, Maxwell had four equations for all of electricity and magnetism.

Try it: two surfaces, one loop

Interactive simulation. Things to try:

• The loop sits around the wire. Switch from “Flat disc” to “Bag into gap”: the conduction current becomes displacement current, but B stays the same.
• Untick Maxwell’s term with the bag selected. Ampère’s original law now gives B = 0 on the same loop: the contradiction.
• Drag the loop between the plates. Make r smaller than R: only the fraction r²/R² of the flux passes through the loop, and B grows in proportion to r.
• Make r bigger than R between the plates: B is the same as around the wire, μ₀i/2πr. Compare the two curves in the graph.

Current in the gap

A capacitor is being charged by a steady current of 0.15 A in the connecting wires. What is the displacement current between its plates?

1. Zero, because no charge crosses the gap
2. 0.15 A
3. 0.075 A, half the conduction current
4. It depends on the area of the plates

Field between the plates

A capacitor with circular plates of radius 10 cm is charged by a steady current of 2 A. What is the magnetic field between the plates at 5 cm from the axis? (Ignore edge effects; μ₀/2π = 2 × 10⁻⁷ T m A⁻¹.)

1. Zero
2. 2 × 10⁻⁶ T
3. 4 × 10⁻⁶ T
4. 8 × 10⁻⁶ T

Summary

• Displacement current: i_d = ε₀ dΦ_E/dt
• Ampère–Maxwell law: ∮ B · dl = μ₀ i_c + μ₀ε₀ dΦ_E/dt
• Charging capacitor (Φ_E = Q/ε₀): i_d = dQ/dt = i

• Ampère’s original law gives different answers for different surfaces bounded by the same loop near a charging capacitor.
• Maxwell’s fix: a changing electric flux acts as a current, the displacement current. It produces a magnetic field just like a conduction current.
• Total current i_c + i_d is continuous: conduction current in the wires, displacement current between the plates.
• A changing magnetic field produces an electric field (Faraday); a changing electric field produces a magnetic field (Maxwell). Together they make electromagnetic waves possible.
• For steady fields dΦ_E/dt = 0 and the law reduces to Ampère’s circuital law.