PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 8 · JEE Main

Electromagnetic Waves

Chapter-wise practice and revision for JEE Main: concise notes on displacement current and Maxwell's equations, the structure of a plane wave, speed in a medium, energy, intensity and radiation pressure, and the spectrum, concept figures, and 54 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

54 selected questions3 topics3 concept figuresPreview · first 3 of 54 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 207 previous-year JEE questions in the PiTheory bank

Wave properties115
Spectrum15
Displacement current12

Contents

  1. 1Displacement current and Maxwell's equationsQ1–Q8
  2. 2Plane waves: fields, speed, energy and pressureQ9–Q48
  3. 3The electromagnetic spectrumQ49–Q54
  4. ✓Last-day checklist
01

Topic 1 · Q1–Q8

Displacement current and Maxwell's equations

12 of the 207 JEE PYQs in the bank test this as their main idea

Look first

IcIcId = ε₀ dΦE/dtId = Ic = C dV/dt
Current is continuous through the gap.

Quick notes

$$I_d=\varepsilon_0\frac{d\Phi_E}{dt},\qquad \oint\vec B\cdot d\vec l=\mu_0\left(I_c+\varepsilon_0\frac{d\Phi_E}{dt}\right)$$

Between capacitor plates no charge flows, but the changing field carries a displacement current equal to the conduction current in the wires. For a capacitor $I_d=C\dfrac{dV}{dt}$.

LawEquationSays
Gauss (electric)$\oint\vec E\cdot d\vec A=\dfrac{Q}{\varepsilon_0}$charges are sources of $\vec E$
Gauss (magnetic)$\oint\vec B\cdot d\vec A=0$no magnetic monopoles
Faraday$\oint\vec E\cdot d\vec l=-\dfrac{d\Phi_B}{dt}$changing $\vec B$ makes $\vec E$
Ampere–Maxwell$\oint\vec B\cdot d\vec l=\mu_0I_c+\mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$current and changing $\vec E$ make $\vec B$
  • AC on a capacitor: $I_{d,0}=V_0\omega C$; rms values follow the same rule.
  • Uniform field between plates: displacement current through part of the area is shared in proportion to area.
  • Only accelerated charges radiate EM waves.
Trap. $\varepsilon_0\,d\Phi_E/dt$ has the dimensions of current, not of field or charge.

Practice: first 3 of 8 questions

Q1JEE Main 2026 (5 Apr, Shift 1)Easy

A displacement current of $4.0$ A can be set up in the space between two parallel plates of $6\,\mu\text{F}$ capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\alpha \times 10^6$ V/s. The value of $\alpha$ is ______.

  1. A$0.58$
  2. B$0.67$
  3. C$0.82$
  4. D$0.75$
Answer and solution

Answer: Option B: $0.67$

Concept. Between the plates no charge flows, but the changing field carries a displacement current equal to the conduction current in the wires: $I_d=\varepsilon_0\dfrac{d\Phi_E}{dt}=C\dfrac{dV}{dt}$.

Step 1. Rate of change of voltage. $\dfrac{dV}{dt}=\dfrac{I_d}{C}=\dfrac{4.0}{6\times10^{-6}}=0.67\times10^6$ V/s.

Answer. $\alpha\approx0.67$ (Option B).

Point to note. $I_d=C\,dV/dt$.

Q2JEE Main 2025 (23 Jan, Shift 2)Easy

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance $2.5\ \mu\text{F}$. The dielectric constant of the medium between the capacitor plates is 1. It produces an instantaneous displacement current of $0.25\ \text{mA}$ in the intervening space between the capacitor plates. The magnitude of the rate of change of the potential difference will be ______ $\text{V s}^{-1}$.

Numerical answer type

Answer and solution

Answer: 100

Concept. Displacement current in a capacitor: $I_d=C\dfrac{dV}{dt}$ (with dielectric constant 1 the given $C$ already includes $\varepsilon_0$).

Step 1. Solve for the rate. $\dfrac{dV}{dt}=\dfrac{0.25\times10^{-3}}{2.5\times10^{-6}}=100$ V/s.

Answer. $100$ V s$^{-1}$.

Point to note. $I_d=C\,dV/dt$.

Q3JEE Main 2025 (22 Jan, Shift 2)Easy

A parallel plate capacitor of area $A = 16\ \text{cm}^2$ and separation between the plates $10\ \text{cm}$ is charged by a DC current. Consider a hypothetical plane surface of area $A_0 = 3.2\ \text{cm}^2$ inside the capacitor and parallel to the plates. At an instant, the current through the circuit is $6\ \text{A}$. At the same instant the displacement current through $A_0$ is ______ mA.

Numerical answer type

Answer and solution

Answer: 1200

Concept. The field between the plates is uniform, so the displacement current density $J_d=\varepsilon_0\dfrac{dE}{dt}$ is the same everywhere. The whole plate area carries the full $6$ A; a smaller area carries its share.

Step 1. Fraction of the area. $\dfrac{A_0}{A}=\dfrac{3.2}{16}=0.2$.

Step 2. Displacement current through $A_0$. $0.2\times6=1.2$ A $=1200$ mA. The plate separation is not needed.

Answer. $1200$ mA.

Point to note. Share by area.

51 more questions and 2 more topics in the full chapter

  1. 2Plane waves: fields, speed, energy and pressure40 questions with solutions
  2. 3The electromagnetic spectrum6 questions with solutions
  3. ✓Last-day checklist

The complete Electromagnetic Waves chapter is in the Class 12 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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