Electromagnetic Waves Previous Year Questions — JEE Main Physics
31 past-year Electromagnetic Waves questions from JEE Main (Physics).
Q18 (2025)
If $\mu_0$ and $\varepsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of $\left(\frac{1}{\mu_0\varepsilon_0}\right)$ is:
- $\mathrm{L} / \mathrm{T}^{2}$
- $\mathrm{L}^2 / \mathrm{T}^2$
- $\mathrm{T}^2 /\mathrm{L}$
- $\mathrm{T}^2 /\mathrm{L}^2$
### Related Formula
Maxwell's relation for the speed of light ($c$) in vacuum:
$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$
### Core Logic
Square both sides of the speed of light formula:
$$c^2 = \frac{1}{\mu_0 \varepsilon_0}$$
Since $c$ represents the speed of light (velocity), its dimensional formula is:
$$[c] = [L T^{-1}]$$
Therefore, the dimensions of $c^2$ are:
$$[c^2] = [L T^{-1}]^2 = [L^2 T^{-2}]$$
### Step 1: Match options
Express the dimensions in terms of the given variable ratios:
$$[c^2] = \frac{L^2}{T^2}$$
This perfectly matches Option (2).
### Pattern Recognition
Sees: Dimensions of vacuum permittivity-permeability inverse product.
Trap: Trying to find the separate dimensions of both $\mu_0$ and $\varepsilon_0$, then conducting manual division. This is extremely slow and prone to algebraic error.
Shortcut: Directly identify the term as $c^2$. The velocity squared has units of $\mathrm{m}^2\mathrm{s}^{-2}$, giving dimensions of $\mathrm{L}^2\mathrm{T}^{-2}$ immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Class 11 Physics: Units and Measurements
Q5 (2025)
If $\in_0$ denotes the permittivity of free space and $\Phi_{\mathrm{E}}$ is the flux of the electric field through the area bounded by the closed surface, then dimension of $\left(\in_0\frac{\mathrm{d}\phi_{\mathrm{E}}}{\mathrm{d}t}\right)$ are that of:
- Electric field
- Electric potential
- Electric charge
- Electric current
### Related Formula
According to the Maxwell-Ampere law, displacement current $I_d$ is defined as:
$$I_d = \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t}$$
### Core Logic
Since $I_d$ represents a physical current, its dimensions must match those of standard conduction electric current ($[A]$ or $[I]$).
### Step 1: Dimensional Analysis
Let us verify using fundamental dimensions:
- Permittivity of free space $[\epsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
- Electric flux $[\Phi_E] = [M L^3 T^{-3} A^{-1}]$
- Time $[t] = [T]$
$$\left[ \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t} \right] = [M^{-1} L^{-3} T^4 A^2] \times \frac{[M L^3 T^{-3} A^{-1}]}{[T]} = [A]$$
This confirms the quantity is dimensionally equivalent to Electric current.
### Pattern Recognition
Sees: $\epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t}$.
Shortcut: Instantly identify this as Maxwell's expression for displacement current. Displacement current has the exact same dimensions as regular conduction current.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q8 (2025)
A plane electromagnetic wave propagates along the $+x$ direction in free space. The components of the electric field, $\vec{E}$ and magnetic field, $\vec{B}$ vectors associated with the wave in Cartesian frame are:
- $E_y, B_x$
- $E_y, B_z$
- $E_x, B_y$
- $E_z, B_y$
### Related Formula
$$\hat{c} = \hat{E} \times \hat{B}$$
where,
$\hat{c}$ = unit vector in the direction of wave propagation
$\hat{E}$ = unit vector of the electric field
$\hat{B}$ = unit vector of the magnetic field
### Core Logic
Given that the wave propagates along the $+x$ direction:
$$\hat{c} = \hat{i}$$
Let us test the combination $E_y$ and $B_z$, which implies $\hat{E} = \hat{j}$ and $\hat{B} = \hat{k}$:
$$\hat{E} \times \hat{B} = \hat{j} \times \hat{k} = \hat{i}$$
{{SOL_IMG1}}
This cross product yields exactly the direction of propagation $+x$. Therefore, $E_y$ and $B_z$ are appropriate components representing the cross-orthogonal vectors of the EM wave.
### Pattern Recognition
Always remember the cyclic relation $\hat{i} \rightarrow \hat{j} \rightarrow \hat{k}$. Since propagation is $\hat{i}$, our cross product must be $\hat{j} \times \hat{k}$. Thus electric field along $y$ matches magnetic field along $z$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Q22 (2025)
A parallel plate capacitor consisting of two circular plates of radius $10\mathrm{~cm}$ is being charged by a constant current of $0.15\mathrm{~A}$. If the rate of change of potential difference between the plates is $7 \times 10^{8}\mathrm{~V/s}$ then the integer value of the distance between the parallel plates is ______ $\mu\mathrm{m}$.
$$\left(\text{Take } \epsilon_{0} = 9 \times 10^{-12}\frac{\mathrm{F}}{\mathrm{m}}, \pi = \frac{22}{7}\right)$$$$$$
### Related Formula
$$I_d = I_c = C \frac{dV}{dt}$$
$$C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 \pi r^2}{d}$$
### Core Logic
Combining the current expression with capacitance relations yields:
$$I = \left(\frac{\epsilon_0 \pi r^2}{d}\right) \frac{dV}{dt}$$
Isolating plate separation distance $d$:
$$d = \frac{\epsilon_0 \pi r^2}{I} \cdot \left(\frac{dV}{dt}\right)$$
Substitute the parameters specified by the problem layout:
- $r = 10\mathrm{~cm} = 0.1\mathrm{~m}$
- $\epsilon_0 = 9 \times 10^{-12}$
- $\pi = 22/7$
- $I = 0.15\mathrm{~A}$
- $\frac{dV}{dt} = 7 \times 10^{8}\mathrm{~V/s}$
$$d = \frac{(9 \times 10^{-12}) \times \left(\frac{22}{7}\right) \times (0.1)^2}{0.15} \times (7 \times 10^8)$$
$$d = \frac{9 \times 10^{-12} \times 22 \times 0.01 \times 10^8}{0.15} = \frac{198 \times 10^{-4}}{0.15}$$
$$d = 1320 \times 10^{-6}\mathrm{~m} = 1320\mu\mathrm{m}$$
Thus, the integer value for the distance is 1320.
### Pattern Recognition
The charging current matches displacement current identically. Treat the system as a standard linear differential capacitor setup using $I = C \frac{dV}{dt}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
Class 12 Physics: Electrostatic Potential and Capacitance
Q16 (2025)
Due to presence of an em-wave whose electric component is given by $\mathrm{E} = 100\sin (\omega t - kx)\mathrm{NC}^{-1}$ , a cylinder of length $200~\mathrm{cm}$ holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds the same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as
- $25\sin (\omega t - kx)\mathrm{NC}^{-1}$
- $200\sin (\omega t - kx)\mathrm{NC}^{-1}$
- $400\sin (\omega t - kx)\mathrm{NC}^{-1}$
- $50\sin (\omega t - kx)\mathrm{NC}^{-1}$
### Related Formula
$$\text{Energy Density} = \frac{1}{2} \epsilon_0 \mathrm{E}^2$$
$$\text{Total Energy} = \text{Energy Density} \times \text{Volume}$$
### Core Logic
Since both cylinders hold equal amounts of electromagnetic energy:
$$\left(\text{Energy}\right)_1 = \left(\text{Energy}\right)_2$$
$$\frac{1}{2} \epsilon_0 \mathrm{E}_1^2 \cdot c \pi \mathrm{R}_1^2 \times \mathrm{L}_1 = \frac{1}{2} \epsilon_0 \mathrm{E}_2^2 \cdot c \pi \mathrm{R}_2^2 \times \mathrm{L}_2$$
Since the lengths are identical ($\mathrm{L}_1 = \mathrm{L}_2$), this simplifies to:
$$\mathrm{E}_1^2 \mathrm{R}_1^2 = \mathrm{E}_2^2 \mathrm{R}_2^2 \implies \mathrm{E}_1 \mathrm{R}_1 = \mathrm{E}_2 \mathrm{R}_2$$
Given the second cylinder has half the diameter (and radius) of the first ($\mathrm{R}_2 = \frac{\mathrm{R}_1}{2}$):
$$100 \times \mathrm{R}_1 = \mathrm{E}_2 \times \frac{\mathrm{R}_1}{2}$$
$$\mathrm{E}_2 = 200 \mathrm{N/C}$$
### Step 1: Final Equation Match
The wave equation adjusts its amplitude factor to $200\sin (\omega t - kx)\mathrm{NC}^{-1}$, which matches option (2).
### Pattern Recognition
When energy is constant and volume scales down inversely by a factor of 4 (due to $\mathrm{R}^2$), the electric field strength must increase by a factor of $\sqrt{4} = 2$ to maintain balance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
