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This MCQ module is based on: Displacement Current
TOPIC 34 OF 38
Displacement Current
🎓 Class 12
Physics
CBSE
Theory
Ch 8 – Electromagnetic Waves
⏱ ~14 min
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Displacement Current
8.1 Introduction
Faraday discovered that a changing magnetic flux produces an EMF and hence an electric field. In the 1860s, James Clerk Maxwell asked the symmetric question: can a changing electric flux produce a magnetic field? Answering this required modifying Ampere's law, introducing a new term called the displacement current. The breakthrough led directly to the prediction - and the existence - of electromagnetic waves, including light itself.
8.2 Displacement Current
Recall Ampere's circuital law in its original form:
\(\oint \vec{B}\cdot d\vec{l} = \mu_0 I\)
where I is the conduction current threading the Amperian loop. The law works perfectly for steady currents in wires - but consider what happens when we charge a parallel-plate capacitor.
8.2.1 The Capacitor Paradox
Imagine an Amperian loop drawn around the connecting wire (Fig. 8.1). Two surfaces are bounded by the same loop:
- Surface 1 — a flat disc pierced by the wire. Conduction current through it = I, so \(\oint B\cdot dl = \mu_0 I\).
- Surface 2 — a curved bowl that bulges between the capacitor plates without touching the wire. Conduction current through it = 0, so \(\oint B\cdot dl = 0\).
But the line integral on the LHS is a property of the loop only, not the surface! The two surfaces give contradictory answers - Ampere's law as stated is incomplete.
8.2.2 Maxwell's Resolution
Between the plates the conduction current vanishes, but the electric field E (and hence the electric flux \(\Phi_E\)) is rapidly increasing as charge accumulates. Maxwell defined a new "current":
Displacement Current: The quantity
\(i_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}\)
where \(\Phi_E = \int \vec{E}\cdot d\vec{A}\) is the electric flux through the surface. \(\varepsilon_0\) is the permittivity of free space.For a parallel-plate capacitor with plate area A: \(E = q/(\varepsilon_0 A)\) so \(\Phi_E = EA = q/\varepsilon_0\), giving:
\(i_d = \varepsilon_0\,\dfrac{d}{dt}\!\left(\dfrac{q}{\varepsilon_0}\right) = \dfrac{dq}{dt} = i_c\)
The displacement current between the plates exactly equals the conduction current in the wire. With the new term added Ampere's law becomes:
\(\oint \vec{B}\cdot d\vec{l} = \mu_0 (i_c + i_d) = \mu_0 i_c + \mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}\)
This Ampere-Maxwell law is the corrected version. Whether the Amperian surface cuts a wire or runs through a vacuum gap, the right-hand side gives the same answer.
Symmetry: Faraday's law tells us a changing B produces an E. The Maxwell term tells us a changing E produces a B. The two together make self-sustaining electromagnetic waves possible.
8.2.3 Total Current Is Continuous
Conduction current in the wires and displacement current between the plates together form one continuous "current loop". Even though no charges cross the dielectric gap, the displacement current preserves the continuity of the magnetic field around the loop.
| Quantity | Conduction current ic | Displacement current id |
|---|---|---|
| Source | flow of real charges | changing electric flux |
| Formula | ic = dq/dt | id = ε₀ dΦE/dt |
| Medium | conductor | vacuum / dielectric |
| Heat dissipated? | Yes (I²R) | No |
| Produces B-field? | Yes (Ampere) | Yes (Maxwell) |
8.3 Maxwell's Equations
All of classical electromagnetism can be summarised in four equations, the Maxwell equations:
8.3.1 Why They Lead to EM Waves
Equations (3) and (4) form a self-coupling pair. A changing B field generates a curling E field; that changing E field generates a curling B field; which in turn generates E ... a self-propagating disturbance moves outward through space at speed:
\(c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}\)
Numerically: \(c = 1/\sqrt{(4\pi\times10^{-7})(8.854\times10^{-12})} = 3.00\times10^8\) m/s — the speed of light. This is one of the most beautiful results in physics.
Example 8.1 — Displacement current in a charging capacitor
A parallel-plate capacitor with circular plates of radius 12 cm is being charged at a constant rate dq/dt = 0.15 C/s. (a) What is the displacement current between the plates? (b) What is the magnetic field at a point 6 cm from the central axis between the plates?
(a) For a parallel-plate capacitor id = ic = 0.15 A.
(b) Apply Ampere-Maxwell to a circular loop of radius r = 6 cm. The displacement current is uniformly distributed over the plate area, so the fraction enclosed is (r/R)² = (6/12)² = 0.25. Enclosed displacement current = 0.25 × 0.15 = 0.0375 A. Then B(2πr) = μ₀ × 0.0375 ⇒ B = μ₀ × 0.0375 / (2π × 0.06) = (4π×10⁻⁷ × 0.0375)/(2π × 0.06) = 1.25 × 10⁻⁷ T.
Example 8.2 — Rate of change of E required
What must be the rate of change of the electric field between the plates of a parallel-plate capacitor of plate area 100 cm² to produce a displacement current of 1 A?
id = ε₀ A (dE/dt). So dE/dt = id/(ε₀ A) = 1/(8.854×10⁻¹² × 100×10⁻⁴) = 1.13 × 10¹³ V/m/s.
An astonishingly large rate - which is why displacement-current effects are negligible in most ordinary circuits and become important only at radio frequencies and above.
Simulation: Displacement Current Calculator
Set the plate area and the rate of change of E-field. Watch the displacement current update live.
| Displacement current id | 88.5 nA |
| Rate of change of flux dΦE/dt | 10000 V·m/s |
Activity 8.1 — Continuity of current at a capacitor
Connect a low-voltage AC source (e.g. 6 V, 50 Hz) in series with a 1 μF capacitor and a tiny AC ammeter or galvanometer-bridge. Measure the current.
Predict: will any current be detected, given that no charges cross the gap between the plates?
An AC current is detected because the alternating electric flux between the plates constitutes a displacement current that flows through the gap continuously. Conduction current in the wires and displacement current in the gap match exactly - this is Maxwell's insight in action.
Competency-Based Questions L1L2L3L4L6
A parallel-plate capacitor of plate area 25 cm² is being charged. The conduction current in the connecting wire is 0.5 mA.
1. The displacement current between the plates is: L1
(c) 0.5 mA. id = ic = dq/dt at all times during charging.
2. State the physical meaning of displacement current. L2
It is the "current equivalent" of a changing electric flux - it produces the same magnetic field that a real conduction current of equal magnitude would produce, even though no charges flow.
3. Calculate the rate of change of E-field between the plates. L3
id = ε₀ A dE/dt. dE/dt = id/(ε₀A) = 5×10⁻⁴/(8.854×10⁻¹² × 25×10⁻⁴) = 2.26 × 10¹³ V/m/s.
4. Compare and contrast Ampere's original law with the Ampere-Maxwell law. L4
Both relate the line integral of B around a closed loop to the current threading it. Ampere's original law counts only conduction current; the Ampere-Maxwell law adds the displacement current term ε₀ dΦE/dt. The corrected law is consistent for any choice of bounded surface and predicts EM waves.
5. Propose what would happen to the prediction of EM waves if the displacement-current term were absent. L6
Without the displacement-current term, only Faraday's law would couple E and B (B → E by induction). The reverse coupling (E → B) would be absent. The fields would not be able to sustain each other in vacuum, and electromagnetic waves would not exist. Light, radio, X-rays - all would be unexplained.
Assertion-Reason Questions
Assertion: Magnetic monopoles have never been observed.
Reason: Gauss's law for magnetism states the net B-flux through any closed surface is zero.
(A). Both true; R is the mathematical expression of A.
Assertion: Displacement current produces a magnetic field just like conduction current.
Reason: Maxwell's correction to Ampere's law treats them symmetrically inside the closed loop integral.
(A). Both true; R explains A.
Assertion: A DC current cannot pass through a capacitor.
Reason: A capacitor stores energy in the electric field between its plates.
(B). Both true but R does not explain A. The real reason DC cannot pass continuously is that once the capacitor is fully charged, dΦE/dt = 0 ⇒ no displacement current.
Frequently Asked Questions - Displacement Current
What is the main concept covered in Displacement Current?
In NCERT Class 12 Physics Chapter 8 (Electromagnetic Waves), "Displacement Current" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Displacement Current useful in real-life applications?
Real-life applications of "Displacement Current" from NCERT Class 12 Physics Chapter 8 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Displacement Current?
Key formulas in "Displacement Current" (NCERT Class 12 Physics Chapter 8 Electromagnetic Waves) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 8?
NCERT Class 12 Physics Chapter 8 (Electromagnetic Waves) is structured so each part builds on the previous one. "Displacement Current" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Displacement Current?
CBSE board questions from "Displacement Current" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Displacement Current" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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