PhysicsClass 12Electromagnetic Waves

Electromagnetic Waves | Class 12 Physics Notes

By ConceptScroll Team · Published on 17 July 2026 · 6 min read

Electromagnetic Waves | Class 12 Physics Notes

Electromagnetic Waves – this guide gives you a concise, exam-ready overview of Electromagnetic Waves from Class 12 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

8.2 DISPLACEMENT CURRENT

Ampere's circuital law states that the line integral of the magnetic field B around a closed loop equals μ₀ times the conduction current passing through any surface bounded by the loop. However, when applied to a charging capacitor, this law leads to a contradiction. Consider a parallel plate capacitor connected to a time-varying current i(t). Outside the capacitor, the conduction current flows in the wires, producing a magnetic field B at a point P. Using Ampere's law with a loop encircling the wire, the magnetic field is given by B(2πr) = μ₀ i(t), where r is the radius of the circular loop.

Now, consider a different surface bounded by the same loop but passing between the capacitor plates where no conduction current flows. Ampere's law then predicts zero magnetic field, contradicting the previous result. The resolution lies in recognizing that the changing electric field between the capacitor plates acts like a current, called the displacement current.

The electric field E between the plates is related to the charge Q on the plates as E = Q/(ε₀ A), where A is the plate area and ε₀ is the permittivity of free space. The electric flux through the surface between the plates is Φ_E = E × A = Q/ε₀. Differentiating with respect to time, dΦ_E/dt = (1/ε₀) dQ/dt = i/ε₀. Multiplying both sides by ε₀, we get ε₀ dΦ_E/dt = i, which identifies the displacement current i_d = ε₀ dΦ_E/dt.

Thus, the total current passing through any surface bounded by the loop is the sum of the conduction current i_c and the displacement current i_d. Ampere's law is generalized to the Ampere-Maxwell law:

∮ B · dl = μ₀ i_c + μ₀ ε₀ dΦ_E/dt

This generalization restores consistency and symmetry in Maxwell's equations. The displacement current produces magnetic fields just like conduction current. Inside the capacitor, conduction current is zero but displacement current is non-zero; outside the capacitor, conduction current exists but displacement current is zero.

This concept has profound implications, including the prediction of electromagnetic waves. The symmetry between the laws of induction (changing magnetic fields produce electric fields) and the Ampere-Maxwell law (changing electric fields produce magnetic fields) implies that time-varying electric and magnetic fields sustain each other and propagate through space as waves.

📊 Diagram: See figure_2, figure_3, figure_4: (a) A parallel plate capacitor C, as part of a circuit through which a time dependent current i(t) flows, with a loop of radius r to determine magnetic field at point P; (b) a pot-shaped surface passing through the interior between the capacitor plates with the loop as its rim; (c) a tiffin-shaped surface with the circular loop as its rim and a flat circular bottom S between the capacitor plates. Arrows show uniform electric field between the capacitor plates. See figure_5 and figure_6: (a) The electric and magnetic fields E and B between the capacitor plates at point M; (b) cross-sectional view of the fields.

🧪 Activity: Measuring the magnetic field at a point between capacitor plates to verify the presence of displacement current.

🔗 Connection: This section leads to Maxwell's equations in vacuum, summarizing the fundamental laws of electromagnetism including displacement current.

Frequently asked questions

8.1 Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain.

Given: Radius of plates, r = 12 cm = 0.12 m Separation between plates, d = 5.0 cm = 0.05 m Charging current, I = 0.15 A

(a) Calculate capacitance C: Capacitance of parallel plate capacitor, C = ε₀ A / d Area, A = π r² = π × (0.12)² = π × 0.0144 = 0.04524 m² ε₀ = 8.854 × 10⁻¹² F/m So, C = (8.854 × 10⁻¹²) × 0.04524 / 0.05 = 8.01 × 10⁻¹² F = 8.01 pF

Rate of change of potential difference (dV/dt): Current I = C (dV/dt) ⇒ dV/dt = I / C = 0.15 / (8.01 × 10⁻¹²) = 1.87 × 10¹⁰ V/s

(b) Displacement cur

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s⁻¹. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Given: Radius R = 6.0 cm = 0.06 m Capacitance C = 100 pF = 100 × 10⁻¹² F Voltage amplitude V₀ = 230 V (rms given, so V_rms = 230 V) Angular frequency ω = 300 rad/s

(a) RMS conduction current I_rms: For capacitor, current leads voltage by 90° I = C dV/dt = C ω V₀ cos(ω t) RMS current I_rms = ω C V_rms I_rms = 300 × 100 × 10⁻¹² × 230 = 6.9 × 10⁻⁶ A = 6.9 μA

(b) The conduction current in the wires is equal in magnitude to the displacement current between the plates because the displacement curren

8.3 What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?

The physical quantity that is the same for all electromagnetic waves, regardless of their wavelength, is the speed of the wave in vacuum. All electromagnetic waves travel at the speed of light, c = 3 × 10⁸ m/s.

8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

In a plane electromagnetic wave traveling along the z-direction, the electric field vector (E) and magnetic field vector (B) are perpendicular to each other and both are perpendicular to the direction of propagation (z-axis). Thus, E and B lie in the x-y plane.

Frequency, ν = 30 MHz = 30 × 10⁶ Hz Wavelength, λ = c / ν = (3 × 10⁸ m/s) / (30 × 10⁶ Hz) = 10 m

Ready to ace this chapter?

Get the full Electromagnetic Waves chapter — interactive notes, diagrams, worked solutions, polls and a free practice quiz — in the ConceptScroll app.

Open in ConceptScroll →

Study smarter with ConceptScroll

Daily NCERT-aligned reels, AI doubt solving and chapter quizzes — all free.

Start learning free
#cbse notes#class 12#ncert#physics

Continue reading