Electromagnetic Waves Previous Year Questions — JEE Main Physics
31 past-year Electromagnetic Waves questions from JEE Main (Physics) — page 5.
Q40 (2024)
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} \mathrm{~Hz}$ and an amplitude of $50 \mathrm{~Vm}^{-1}$. The total average energy density of the electromagnetic field of the wave is :
[Use $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{Nm}^2$ ]
- $1.106 \times 10^{-8} \, \mathrm{Jm}^{-3}$
- $4.425 \times 10^{-8} \mathrm{~Jm}^{-3}$
- $2.212 \times 10^{-8} \mathrm{~Jm}^{-3}$
- $2.212 \times 10^{-10} \mathrm{~Jm}^{-3}$
### Related Formula
$$U_{\text{total average}} = \frac{1}{2}\epsilon_0 E_0^2$$
### Core Logic
For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so:
$$U_{\text{avg}} = U_E + U_B = 2U_E = 2 \left( \frac{1}{4}\epsilon_0 E_0^2 \right) = \frac{1}{2}\epsilon_0 E_0^2$$
Where $E_0$ is the amplitude of the electric field.
### Step 2: Substitution
Given:
$E_0 = 50 \mathrm{\, V/m}$
$\epsilon_0 = 8.85 \times 10^{-12} \mathrm{\, C^2/(N\cdot m^2)}$
$$U_{\text{avg}} = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (50)^2$$
$$U_{\text{avg}} = \frac{1}{2} \times 8.85 \times 10^{-12} \times 2500$$
$$U_{\text{avg}} = 1.10625 \times 10^{-8} \mathrm{\, J/m^3}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q35 (2026)
The electric field a plane electromagnetic wave is given by :
$$E_y = 69 \sin [ 0.6 \times 10^3 x - 1.8 \times 10^{11} t ]\text{ V/m}.$$
The expression for magnetic field associated with this electromagnetic wave is ____ T.
- $\mathrm{B_z} = 2.3\times 10^{-7}\sin [0.6\times 10^{3}\mathrm{x} - 1.8\times 10^{11}\mathrm{t}]$
- $\mathrm{B_z} = 2.3\times 10^{-7}\sin [0.6\times 10^{3}\mathrm{x} + 1.8\times 10^{11}\mathrm{t}]$
- $\mathrm{B_y} = 69\sin [0.6\times 10^3\mathrm{x} + 1.8\times 10^{11}\mathrm{t}]$
- $\mathrm{B_y} = 2.3\times 10^{-7}\sin [0.6\times 10^{3}\mathrm{x} - 1.8\times 10^{11}\mathrm{t}]$
### Related Formula
$$B_0 = \frac{E_0}{c}$$
$$\hat{c} = \hat{E} \times \hat{B}$$
### Core Logic
The phase of the wave is $(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$. This indicates the wave propagates in the $+x$ direction, so $\hat{c} = \hat{i}$.
The electric field oscillates along the $y$-axis, so $\hat{E} = \hat{j}$.
From $\hat{B} = \hat{c} \times \hat{E}$, we have $\hat{B} = \hat{i} \times \hat{j} = \hat{k}$.
So, the magnetic field is along the $z$-axis ($B_z$).
### Step 1: Calculate Amplitude of B
Wave speed $v = c = \frac{\omega}{k} = \frac{1.8 \times 10^{11}}{0.6 \times 10^3} = 3 \times 10^8\text{ m/s}$.
The amplitude of the magnetic field is:
$$B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 23 \times 10^{-8} = 2.3 \times 10^{-7}\text{ T}$$
The phase remains exactly the same as the electric field:
$$B_z = 2.3 \times 10^{-7} \sin(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$$
### Pattern Recognition
$B_0 = E_0/c$ gives the magnitude. The vector identity $\hat{B} = \hat{v} \times \hat{E}$ gives the direction. Phase part never changes sign or terms between E and B equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q46 (2026)
An electromagnetic wave of frequency $100 \text{ MHz}$ propagates through a medium of conductivity, $\sigma = 10 \text{ mho/m}$. The ratio of maximum conducting current density to maximum displacement current density is ________.
$[\text{Take } \frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9} \text{ N}\cdot\text{m}^{2}/\text{C}^{2}]$
### Related Formula
$J_c = \sigma E$
$$J_d = \epsilon_0 \frac{\partial E}{\partial t}$$
### Core Logic
Let the electric field of the wave be $E = E_0 \sin(\omega t - kx)$.
The conduction current density is:
$$J_c = \sigma E_0 \sin(\omega t - kx)$$
The maximum conduction current density is:
$$(J_c)_{\text{max}} = \sigma E_0 \quad \text{--- (i)}$$
The displacement current density is:
$$J_d = \frac{1}{A} \left( \epsilon_0 \frac{\partial (EA)}{\partial t} \right) = \epsilon_0 \frac{\partial E}{\partial t}$$
$$J_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$$
The maximum displacement current density is:
$$(J_d)_{\text{max}} = \epsilon_0 E_0 \omega \quad \text{--- (ii)}$$
### Step 1: Taking the Ratio
Dividing (i) by (ii):
$$\text{Ratio} = \frac{(J_c)_{\text{max}}}{(J_d)_{\text{max}}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$$
### Step 2: Substitution and Calculation
We know $f = 100 \text{ MHz} = 10^8 \text{ Hz}$, so $\omega = 2\pi f = 2\pi \times 10^8 \text{ rad/s}$.
$\sigma = 10 \text{ mho/m}$.
Also, $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \implies \frac{1}{\epsilon_0} = 4\pi \times 9 \times 10^9$.
$$\text{Ratio} = \frac{10 \times 4\pi \times 9 \times 10^9}{2\pi \times 10^8}$$
$$\text{Ratio} = \frac{360\pi \times 10^9}{2\pi \times 10^8} = \frac{3600}{2} = 1800$$
### Step 3: Final Conclusion
The required ratio is 1800.
### Pattern Recognition
The ratio of conduction to displacement current density in any medium is universally $\sigma / (\omega \epsilon_0)$. This dictates whether a medium behaves as a good conductor or a dielectric at a given frequency.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q47 (2026)
The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by,
$E_{y}=20\sin(3\times 10^{6}x-4.5\times 10^{14}t)$ V/m
(where x, t and other values have S.I. units). The dielectric constant of the medium is \_\_\_\_. (speed of light in free space is $3 \times 10^{8} \mathrm{~m/s}$)
### Related Formula
$$v = \frac{\omega}{k}, \quad n = \frac{c}{v} = \sqrt{\mu_r \epsilon_r}$$
### Core Logic
Wave velocity in medium:
$$v = \frac{\omega}{k} = \frac{4.5 \times 10^{14}}{3 \times 10^6} = 1.5 \times 10^8 \text{ m/s}$$
Refractive index:
$$n = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$$
For non-magnetic medium ($\mu_r = 1$):
$$n = \sqrt{\epsilon_r} \implies 2 = \sqrt{\epsilon_r} \implies \epsilon_r = 4$$
### Pattern Recognition
Sees: EM wave equation in dielectric medium.
Shortcut: Extract phase velocity from wave equation coefficients, find refractive index and dielectric constant.
Check: Numerical answer is 4. ✓
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q43 (2026)
A laser beam has intensity of $4.0 \times 10^{14}$ W/m$^2$. The amplitude of magnetic field associated with beam is ____ T. (Take $\epsilon_{0} = 8.85 \times 10^{-12}$ C$^2$/Nm$^2$ and c = 3 × 10$^8$ m/s)
- $2.0$
- $18.3$
- $5.5$
- $1.83$
### Related Formula
$$I = \frac{1}{2} \epsilon_0 E_0^2 c$$
$$B_0 = \frac{E_0}{c} = \frac{1}{c} \sqrt{\frac{2I}{\epsilon_0 c}}$$
### Core Logic
Expressing electric field amplitude $E_0$ in terms of intensity $I$:
$$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$$
Relating magnetic field amplitude $B_0$ to $E_0$ via $B_0 = \frac{E_0}{c}$:
$$B_0 = \frac{1}{c}\sqrt{\frac{2I}{\epsilon_0 c}} = \frac{1}{3 \times 10^8} \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}}$$
Simplifying terms under the radical:
$$B_0 = \frac{1}{3 \times 10^8} \sqrt{\frac{8 \times 10^{14}}{2.655 \times 10^{-3}}} = \frac{1}{3 \times 10^8} \sqrt{3.013 \times 10^{17}} = \frac{10}{3} \sqrt{\frac{8}{8.85 \times 3}} \approx 1.83 \mathrm{~T}$$
### Step 1: Final Conclusion
The amplitude of the magnetic field is $1.83 \mathrm{~T}$.
### Pattern Recognition
EM Wave Intensity: $I = \frac{1}{2} c \frac{B_0^2}{\mu_0} = \frac{1}{2} \epsilon_0 E_0^2 c$.
Direct sub: $B_0 = \sqrt{\frac{2 \mu_0 I}{c}}$ or $B_0 = \frac{1}{c} \sqrt{\frac{2I}{\epsilon_0 c}} \approx 1.83\mathrm{~T}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
