Electromagnetic Waves Previous Year Questions — JEE Main Physics
31 past-year Electromagnetic Waves questions from JEE Main (Physics) — page 3.
Q1219 (2025)
The magnetic field of an E.M. wave is given by $\vec{B}=\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)30~\sin\left[\omega\left(t-\frac{z}{c}\right)\right]$ (S.I. Units). The corresponding electric field in S.I. units is:
- $\vec{\mathrm{E}} = \left(\frac{1}{2}\hat{\mathrm{i}} -\frac{\sqrt{3}}{2}\hat{\mathrm{j}}\right)30\mathrm{c}\sin \left[\omega \left(\mathrm{t} - \frac{\mathrm{z}}{\mathrm{c}}\right)\right]$
- \vec{E} = \left(\frac{3}{4}\hat{i} + \frac{1}{4}\hat{j}\right) 30c \cos \left[\omega \left(t - \frac{z}{c}\right)\right]
- $\vec{\mathrm{E}} = \left(\frac{1}{2}\hat{\mathrm{i}} +\frac{\sqrt{3}}{2}\hat{\mathrm{j}}\right)30\mathrm{c}\sin \left[\omega \left(\mathrm{t} + \frac{\mathrm{z}}{\mathrm{c}}\right)\right]$
- $\vec{\mathrm{E}} = \left(\frac{\sqrt{3}}{2}\hat{\mathrm{i}} -\frac{1}{2}\hat{\mathrm{j}}\right)30\mathrm{c}\sin \left[\omega \left(\mathrm{t} + \frac{\mathrm{z}}{\mathrm{c}}\right)\right]$
### Related Formula
For a plane electromagnetic wave propagating in a given direction:
1. Peak electric field amplitude relates to peak magnetic field amplitude via:
$$E_0 = B_0 \cdot c$$
2. The directional orientation unit vectors satisfy the cross product relation:
$$\hat{E} = \hat{B} \times \hat{c}$$
where $\hat{c}$ points along the wave propagation vector direction.
### Core Logic
Given the wave equation format, the phase term $\left(t - \frac{z}{c}\right)$ shows that propagation is along the positive z-axis :
$$\hat{c} = \hat{k}$$
The magnetic field direction unit vector is :
$$\hat{B} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$$
Compute the electric field direction vector using the cross product relation :
$$\hat{E} = \hat{B} \times \hat{k} = \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) \times \hat{k}$$
$$\hat{E} = \frac{\sqrt{3}}{2}(\hat{i} \times \hat{k}) + \frac{1}{2}(\hat{j} \times \hat{k})$$
Using unit vector properties ($\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$):
$$\hat{E} = -\frac{\sqrt{3}}{2}\hat{j} + \frac{1}{2}\hat{i} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j} \quad \text{}$$
With peak amplitude $E_0 = 30c$ , the resulting vector equation is:
$$\vec{E} = \left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right]$$
### Pattern Recognition
The vectors $\vec{E}$, $\vec{B}$, and the propagation direction are always mutually perpendicular. Since $\vec{E} \cdot \vec{B} = 0$, you can quickly double-check your answer by verifying that the \dot product of the final $\vec{E}$ and $\vec{B}$ direction options equals zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q1295 (2025)
Given below are two statements : one is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A) : Electromagnetic waves carry energy but not momentum.
Reason (R): Mass of a photon is zero.
In the light of the above statements, choose the most appropriate answer from the options given below:
- (A) is true but (R) is false.
- (A) is false but (R) is true.
- Both (A) and (R) are true but (R) is not the correct explanation of (A).
- Both (A) and (R) are true and (R) is the correct explanation of (A).
### Related Formula
$$p = \frac{E}{c}$$
### Core Logic
Assertion (A) is false because electromagnetic waves carry both energy and finite radiation momentum ($p = E/c$).
Reason (R) is correct because the rest mass of a photon equals zero.
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q38 (2024)
A parallel plate capacitor has a capacitance $C = 200\mathrm{~pF}$. It is connected to $230\mathrm{~V}$ ac supply with an angular frequency $300\mathrm{~rad/s}$. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are:
- $1.38\mathrm{~\mu A}\text{ and }1.38\mathrm{~\mu A}$
- $14.3\mathrm{~\mu A}\text{ and }143\mathrm{~\mu A}$
- $13.8\mathrm{~\mu A}\text{ and }138\mathrm{~\mu A}$
- $13.8\mathrm{~\mu A}\text{ and }13.8\mathrm{~\mu A}$
### Related Formula
Capacitive reactance:
$$X_C = \frac{1}{\omega C}$$
Conduction Current ($I_{\text{rms}}$):
$$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}} \cdot \omega C$$
Continuity relation:
$I_c = I_d$
### Core Logic
Given parameters:
$C = 200\mathrm{~pF} = 200 \times 10^{-12}\mathrm{~F}$, $V_{\text{rms}} = 230\mathrm{~V}$, $\omega = 300\mathrm{~rad/s}$.
Calculate current:
$$I = V_{\text{rms}} \cdot \omega C = 230 \times 300 \times 200 \times 10^{-12}$$
### Step 1: Simplify Numerical Calculation
$I = 230 \times 60000 \times 10^{-12} = 13.8 \times 10^{-6}\mathrm{~A} = 13.8\mathrm{~\mu A}$
By Maxwell's electromagnetic formulation, the displacement current $I_d$ within the dielectric space matches the exterior conduction current $I_c$ seamlessly in magnitude.
### Pattern Recognition
Factual invariant: Conduction current always equals displacement current inside a standard layout loop context ($I_c = I_d$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Class 12 Physics: Alternating Current
Q45 (2024)
A plane electromagnetic wave of frequency $35\text{ MHz}$ travels in free space along the $X$-direction. At a particular point (in space and time) $\vec{E} = 9.6\hat{j}\text{ V/m}$. The value of magnetic field at this point is:
- $3.2 \times 10^{-8}\hat{k}\text{ T}$
- $3.2 \times 10^{-8}\hat{i}\text{ T}$
- $9.6\hat{j}\text{ T}$
- $9.6 \times 10^{-8}\hat{k}\text{ T}$
### Related Formula
The relation between the amplitudes of electric field $E$ and magnetic field $B$ in an EM wave is:
$$c = \frac{E}{B}$$
where:
* $c = 3 \times 10^8\text{ m/s}$ is the speed of light.
The directions are related by the vector cross product:
$$\hat{E} \times \hat{B} = \hat{v}$$
where $\hat{v}$ is the direction of propagation of the EM wave.
### Core Logic
Given:
* Wave propagation direction, $\hat{v} = \hat{i}$ (along X-direction)
* Electric field vector direction, $\hat{E} = \hat{j}$
* Electric field magnitude, $E = 9.6\text{ V/m}$
### Step 1: Calculate Magnetic Field Magnitude
Using the magnitude relation:
$$B = \frac{E}{c} = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8}\text{ T}$$
### Step 2: Determine Magnetic Field Direction
Using the direction cross-product rule:
$$\hat{E} \times \hat{B} = \hat{v}$$
$$\hat{j} \times \hat{B} = \hat{i}$$
Since we know that $\hat{j} \times \hat{k} = \hat{i}$, the direction of the magnetic field must be $\hat{k}$:
$$\hat{B} = \hat{k}$$
Combining the magnitude and direction:
$$\vec{B} = 3.2 \times 10^{-8}\hat{k}\text{ T}$$
### Pattern Recognition
A propagation along $+x$ with electric field along $+y$ strictly mandates the magnetic field must point along $+z$ ($+y \times +z = +x$). This instantly eliminates Option 2 and Option 3, leaving only magnitude verification.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q46 (2024)
A plane electromagnetic wave propagating in x-direction is described by $E_{y} = (200\text{ Vm}^{-1})\sin[1.5 \times 10^{7}t - 0.05x]$. The intensity of the wave is (Use $\epsilon_{0} = 8.85 \times 10^{-12}\text{ C}^{2}\text{N}^{-1}\text{m}^{-2}$):
- $35.4\text{ Wm}^{-2}$
- $53.1\text{ Wm}^{-2}$
- $26.6\text{ Wm}^{-2}$
- $106.2\text{ Wm}^{-2}$
### Related Formula
$$I = \frac{1}{2} \epsilon_0 E_0^2 c$$
Where $E_0$ is the amplitude of the electric field ($200\text{ V/m}$) and $c = 3 \times 10^8\text{ m/s}$.
### Core Logic
Substitute the constants into the equation:
$$I = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (200)^2 \times (3 \times 10^8)$$
### Step 1: Compute value
$$I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8$$
$$I = 2 \times 8.85 \times 3 \times 10^0$$
$$I = 53.1\text{ W/m}^2$$
### Pattern Recognition
Isolate powers of ten first ($10^{-12} \times 10^4 \times 10^8 = 10^0$) to streamline intermediate tracking accuracy on calculation variables.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
