Electromagnetic Waves Previous Year Questions — JEE Main Physics
31 past-year Electromagnetic Waves questions from JEE Main (Physics) — page 6.
Q28 (2026)
Match List-I with List-II.
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;" colspan="2">List-I (Relation)</th><th style="border: 1px solid #888; padding: 8px;" colspan="2">List-II (Law)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A.</td><td style="border: 1px solid #888; padding: 8px;">$\oint\vec{E}\cdot\vec{dl}=-\frac{d}{dt}\oint\vec{B}\cdot\vec{da}$</td><td style="border: 1px solid #888; padding: 8px;">I.</td><td style="border: 1px solid #888; padding: 8px;">Ampere's circuital law.</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B.</td><td style="border: 1px solid #888; padding: 8px;">$\oint \vec{B}\cdot\vec{dl}=\mu_{0}\left(1+\epsilon_{0}\frac{d\phi_{E}}{dt}\right)$</td><td style="border: 1px solid #888; padding: 8px;">II.</td><td style="border: 1px solid #888; padding: 8px;">Faraday's laws of electromagnetic induction.</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C)</td><td style="border: 1px solid #888; padding: 8px;">$\oint\vec{E}\cdot\vec{da}=\frac{1}{\epsilon_{0}}\int\rho dv$</td><td style="border: 1px solid #888; padding: 8px;">III.</td><td style="border: 1px solid #888; padding: 8px;">Ampere-Maxwell law</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D)</td><td style="border: 1px solid #888; padding: 8px;">$\oint\vec{B}\cdot\vec{dl}=\mu_{0}I$</td><td style="border: 1px solid #888; padding: 8px;">IV.</td><td style="border: 1px solid #888; padding: 8px;">Gauss's law of electrostatics</td></tr></tbody></table></div>
Choose the correct answer from the options given below :
- A-II, B-III, C-I, D-IV
- A-II, B-III, C-IV, D-I
- A-I, B-IV, C-III, D-II
- A-IV, B-I, C-II, D-III
### Core Logic
Let's systematically identify each integral equation with its corresponding physical law:
(A) $\oint\vec{E}\cdot\vec{dl}=-\frac{d\phi_{B}}{dt}$ corresponds to the line integral of the electric field around a closed loop being equal to the negative rate of change of magnetic flux, which is Faraday's Law of Electromagnetic Induction (II).
(B) $\oint \vec{B}\cdot\vec{dl}=\mu_{0}\left(I+\epsilon_{0}\frac{d\phi_{E}}{dt}\right)$ defines the Ampere-Maxwell Law (III), incorporating the displacement current.
(C) $\oint\vec{E}\cdot\vec{da}=\frac{Q_{\text{enc}}}{\epsilon_{0}}$ relates the electric flux through a closed surface to the enclosed charge, mapping to Gauss's Law of Electrostatics (IV).
(D) $\oint\vec{B}\cdot\vec{dl}=\mu_{0}I$ is the original Ampere's Circuital Law (I) without Maxwell's correction.
### Step 1: Final Conclusion
The correct matches are A-II, B-III, C-IV, D-I.
### Pattern Recognition
Sees: Surface integral of E mapped to volume integral of charge → Gauss's Law. Circulation of E mapped to time derivative of B flux → Faraday's Law.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q47 (2026)
The equation of the electric field of an electromagnetic wave propagating through free space is given by :
$$E = \sqrt{377} \sin(6.27 \times 10^{3} t - 2.09 \times 10^{-5} x) \text{ N/C}$$
The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right)\text{W}/\text{m}^{2}$. The value of $\alpha$ is ____
$$\left(\text{Take } \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}} = 377 \text{ in SI units}\right)$$
### Related Formula
$$I = \frac{1}{2}\varepsilon_{0} E_{0}^{2} c$$
$$c = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$$
### Core Logic
The average power per unit area (Intensity) of an electromagnetic wave in vacuum is $I = \frac{1}{2}\varepsilon_{0} E_{0}^{2} c$. We can reorganize the formula using the intrinsic impedance of free space $Z_{0} = \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}} = 377\Omega$.
### Step 1: Rewrite Intensity Formula
$$I = \frac{1}{2} \varepsilon_{0} E_{0}^{2} \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$$
$$I = \frac{1}{2} E_{0}^{2} \sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}$$
We are given the intrinsic impedance $\sqrt{\frac{\mu_{0}}{\varepsilon_{0}}} = 377 \implies \sqrt{\frac{\varepsilon_{0}}{\mu_{0}}} = \frac{1}{377}$.
### Step 2: Extract Amplitude and Calculate
From the wave equation, $E_{0} = \sqrt{377} \text{ N/C}$.
Substitute into the intensity formula:
$$I = \frac{1}{2} (\sqrt{377})^{2} \times \frac{1}{377}$$
$$I = \frac{1}{2} \times 377 \times \frac{1}{377} = \frac{1}{2} \text{ W/m}^{2}$$
### Step 3: Compare with given format
$$I = \frac{1}{\alpha} \implies \alpha = 2$$
### Pattern Recognition
Sees: "average power... W/m^2" → the problem is asking for Intensity. When given intrinsic impedance ($377\Omega$), use the form $I = E_{\text{rms}}^{2} / Z_0 = E_{0}^{2} / (2Z_{0})$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q43 (2026)
The ratio of speeds of electromagnetic waves in vacuum and a medium, having dielectric constant $k = 3$ and permeability of $\mu = 2\mu_0$ , is ($\mu_0 = \text{permeability of vacuum}$)
- $36 : 1$
- $3 : 2$
- $6 : 1$
- $\sqrt{6} : 1$
### Related Formula
$$v = \frac{1}{\sqrt{\mu \epsilon}}$$
$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$
$$n = \frac{c}{v} = \sqrt{\mu_r \epsilon_r}$$
### Core Logic
The speed of light in a medium is slower than in vacuum by a factor equal to the refractive index $n$.
The relative permittivity (dielectric constant) $\epsilon_r = k = 3$.
The relative permeability $\mu_r = \frac{\mu}{\mu_0} = 2$.
### Step 1: Calculate Ratio
$$\frac{c}{v} = \sqrt{\mu_r \epsilon_r}$$
$$\frac{c}{v} = \sqrt{2 \times 3} = \sqrt{6}$$
Thus, the ratio is $\sqrt{6} : 1$.
### Pattern Recognition
Refractive index is geometrically derived from electrical properties: $n = \sqrt{\mu_r \epsilon_r}$. Just plug in the constants relative to vacuum values.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q35 (2026)
Match the List-I with List-II
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I</th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Radio-wave</td><td style="border: 1px solid #888; padding: 8px;">I. is produced by Magnetron value</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Micro-wave</td><td style="border: 1px solid #888; padding: 8px;">II. Due to change in the vibrational modes of atoms</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Infrared-wave</td><td style="border: 1px solid #888; padding: 8px;">III. Due to inner shell electrons moving from higher energy level to lower energy level</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. X-ray</td><td style="border: 1px solid #888; padding: 8px;">IV. Due to rapid acceleration of electrons</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- $\text{A-II, B-IV, C-III, D-I}$
- $\text{A-IV, B-III, C-I, D-II}$
- $\text{A-IV, B-I, C-II, D-III}$
- $\text{A-IV, B-II, C-I, D-III}$
### Core Logic
Mapping production sources to electromagnetic waves:
Radio waves: Produced by rapid acceleration and deceleration of electrons in aerials. (A -> IV)
Microwaves: Produced by special vacuum tubes like klystrons, magnetrons, and Gunn diodes. (B -> I)
Infrared waves: Produced by hot bodies and molecules. They originate due to the change in vibrational modes of atoms/molecules. (C -> II)
X-rays: Produced when high-energy electrons are stopped by a metal target, or by transitions of inner shell electrons from a higher to a lower energy level. (D -> III)
### Step 1: Final Match
A - IV
B - I
C - II
D - III
This corresponds to Option (3).
### Pattern Recognition
Memorize fundamental EM wave sources: Radio (accelerating charges), Microwave (magnetrons), IR (vibrational modes/heat), X-ray (inner shell transitions).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q32 (2026)
The electric field of an electromagnetic wave travelling through a medium is given by $\vec{E}(x,t) = 25\sin (2.0\times 10^{15}t - 10^{7}x)\hat{n}$ then the refractive index of the medium is ____. (All given measurement are in SI units)
- $1.2$
- $2$
- $1.5$
- $1.7$
### Related Formula
$$v = \frac{\omega}{k}$$
$$\mu = \frac{c}{v}$$
### Core Logic
Compare the given equation with the standard wave equation $E = E_0 \sin(\omega t - kx)$ to extract angular frequency ($\omega$) and wave number ($k$). Find wave velocity $v$ and then refractive index.
### Step 1: Extracting Constants
From the given equation $\vec{E}(x,t) = 25\sin (2.0\times 10^{15}t - 10^{7}x)\hat{n}$:
$$\omega = 2 \times 10^{15} \mathrm{~rad / s}$$
$$k = 10^{7} \mathrm{~m}^{-1}$$
### Step 2: Calculating Wave Velocity
$$v = \frac{\omega}{k} = \frac{2 \times 10^{15}}{10^7} = 2 \times 10^8 \mathrm{~m/s}$$
### Step 3: Calculating Refractive Index
$$v = \frac{c}{\mu} \Rightarrow 2 \times 10^8 = \frac{3 \times 10^8}{\mu}$$
$$\mu = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$$
### Pattern Recognition
For standard wave format $\omega t \pm kx$, the ratio $\omega/k$ immediately yields the phase velocity in the medium.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
