Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-1 , a cylinder of length 200~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

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). When energy is constant and volume scales down inversely by a factor of 4 (due to R2), the electric field strength must increase by a factor of sqrt(4) = 2 to maintain balance. Class 12 Physics: Electromagnetic Waves

Due to presence of an em-wave whose electric component is given by E = 100 (omega t - kx)NC-1 , a cylinder of length 200~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

Since both cylinders hold equal amounts of electromagnetic energy: Since the lengths are identical (L1 = L2), this simplifies to: Given the second cylinder has half the diameter (and radius) of the first (R2 = R12): The wave equation adjusts its amplitude factor to 200 (omega t - kx)NC-1, which matches option (2). When energy is constant and volume scales down inversely by a factor of 4 (due to R2), the electric field strength must increase by a factor of sqrt(4) = 2 to maintain balance. Class 12 Physics: Electromagnetic Waves

Due to presence of an em-wave whose electric component is given by E = 100 (omega t

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Questions	10	12	9	31

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Energy Density = (1)/(2) ε₀ E²

Total Energy = Energy Density \times Volume

Core Logic

Since both cylinders hold equal amounts of electromagnetic energy:

(Energy)₁ = (Energy)₂

(1)/(2) ε₀ E₁² \cdot c π R₁² \times L₁ = (1)/(2) ε₀ E₂² \cdot c π R₂² \times L₂

Since the lengths are identical ( $L₁ = L₂$ $\mathrm{L}_1 = \mathrm{L}_2$), this simplifies to:

E₁² R₁² = E₂² R₂² E₁ R₁ = E₂ R₂

Given the second cylinder has half the diameter (and radius) of the first ( $R₂ = R₁2$ $\mathrm{R}_2 = \frac{\mathrm{R}_1}{2}$):

100 \times R₁ = E₂ \times R₁2

E₂ = 200 N/C

Step 1: Final Equation Match

The wave equation adjusts its amplitude factor to $200 (ω t - kx)NC⁻¹$ , which matches option (2).

Pattern Recognition

When energy is constant and volume scales down inversely by a factor of 4 (due to $R²$ $\mathrm{R}^2$), the electric field strength must increase by a factor of $\sqrt(4) = 2$ $\sqrt{4} = 2$ to maintain balance.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

More Electromagnetic Waves Previous-Year Questions — Page 3

Q18 jee_main_2025_02_april_evening Speed of Electromagnetic Waves

μ₀

and

ε₀

are the permeability and permittivity of free space, respectively, then the dimension of

((1)/(μ₀ε₀))

is:

A. $L / T²$ $\mathrm{L} / \mathrm{T}^{2}$
B. $L² / T²$ $\mathrm{L}^2 / \mathrm{T}^2$
C. $T² /L$ $\mathrm{T}^2 /\mathrm{L}$
D. $T² /L²$ $\mathrm{T}^2 /\mathrm{L}^2$

Solution

Related Formula

Maxwell's relation for the speed of light ( $c$ $c$) in vacuum:

c = 1\sqrt(μ₀ ε₀)

Core Logic

Square both sides of the speed of light formula:

c² = (1)/(μ₀ ε₀)

Since $c$ $c$ represents the speed of light (velocity), its dimensional formula is:

[c] = [L T⁻¹]

Therefore, the dimensions of $c²$ $c^2$ are:

[c²] = [L T⁻¹]² = [L² T⁻²]

Step 1: Match options

Express the dimensions in terms of the given variable ratios:

[c²] = (L²)/(T²)

This perfectly matches Option (2).

Pattern Recognition

Sees: Dimensions of vacuum permittivity-permeability inverse product. Trap: Trying to find the separate dimensions of both $μ₀$ $\mu_0$ and $ε₀$ $\varepsilon_0$, then conducting manual division. This is extremely slow and prone to algebraic error. Shortcut: Directly identify the term as $c²$ $c^2$. The velocity squared has units of $m²s⁻²$ $\mathrm{m}^2\mathrm{s}^{-2}$, giving dimensions of $L²T⁻²$ $\mathrm{L}^2\mathrm{T}^{-2}$ immediately.

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 11 Physics: Units and Measurements

Q5 jee_main_2025_07_april_morning Displacement Current

in₀

denotes the permittivity of free space and

ΦE

is the flux of the electric field through the area bounded by the closed surface, then dimension of

(in₀ dφEdt)

are that of:

A. Electric field
B. Electric potential
C. Electric charge
D. Electric current

Solution

Related Formula

According to the Maxwell-Ampere law, displacement current $Id$ $I_d$ is defined as:

Id = ε₀ dΦEdt

Core Logic

Since $Id$ $I_d$ represents a physical current, its dimensions must match those of standard conduction electric current ( $[A]$ $[A]$ or $[I]$ $[I]$).

Step 1: Dimensional Analysis

Let us verify using fundamental dimensions:

Permittivity of free space $[ε₀] = [M⁻¹ L⁻³ T⁴ A²]$
Electric flux $[ΦE] = [M L³ T⁻³ A⁻¹]$ $[\Phi_E] = [M L^3 T^{-3} A^{-1}]$
Time $[t] = [T]$ $[t] = [T]$

[ ε₀ dΦEdt ] = [M⁻¹ L⁻³ T⁴ A²] \times [M L³ T⁻³ A⁻¹][T] = [A]

This confirms the quantity is dimensionally equivalent to Electric current.

Pattern Recognition

Sees: $ε₀ dΦEdt$ . Shortcut: Instantly identify this as Maxwell's expression for displacement current. Displacement current has the exact same dimensions as regular conduction current.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q8 jee_main_2025_29_jan_evening Propagation of Electromagnetic Waves

A plane electromagnetic wave propagates along the

+x

direction in free space. The components of the electric field,

E

and magnetic field,

B

vectors associated with the wave in Cartesian frame are:

A. $Ey, Bₓ$ $E_y, B_x$
B. $Ey, Bz$ $E_y, B_z$
C. $Eₓ, By$ $E_x, B_y$
D. $Ez, By$ $E_z, B_y$

Solution

Related Formula

c = E \times B

where, $c$ $\hat{c}$ = unit vector in the direction of wave propagation $E$ $\hat{E}$ = unit vector of the electric field $B$ $\hat{B}$ = unit vector of the magnetic field

Core Logic

Given that the wave propagates along the $+x$ $+x$ direction:

c = i

Let us test the combination $Ey$ $E_y$ and $Bz$ $B_z$, which implies $E = j$ $\hat{E} = \hat{j}$ and $B = k$ $\hat{B} = \hat{k}$:

E \times B = j \times k = i

Propagation of Electromagnetic Waves diagram for Q8 - JEE Main 2025 Evening

This cross product yields exactly the direction of propagation $+x$ $+x$. Therefore, $Ey$ $E_y$ and $Bz$ $B_z$ are appropriate components representing the cross-orthogonal vectors of the EM wave.

Pattern Recognition

Always remember the cyclic relation $i arrow j arrow k$ . Since propagation is $i$ $\hat{i}$, our cross product must be $j \times k$ $\hat{j} \times \hat{k}$. Thus electric field along $y$ $y$ matches magnetic field along $z$ $z$.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q22 jee_main_2025_29_jan_evening Displacement Current

A parallel plate capacitor consisting of two circular plates of radius

10~cm

is being charged by a constant current of

0.15~A

. If the rate of change of potential difference between the plates is

7 \times 10⁸~V/s

then the integer value of the distance between the parallel plates is ______

(Take ε₀ = 9 \times 10⁻¹² Fm, π = (22)/(7))

$$$$

Numerical Answer. Answer: 1320 to 1320

Solution

Related Formula

Id = Ic = C (dV)/(dt)

C = (ε₀ A)/(d) = (ε₀ π r²)/(d)

Core Logic

Combining the current expression with capacitance relations yields:

I = ((ε₀ π r²)/(d)) (dV)/(dt)

Isolating plate separation distance $d$ $d$:

d = (ε₀ π r²)/(I) \cdot ((dV)/(dt))

Substitute the parameters specified by the problem layout:

$r = 10~cm = 0.1~m$ $r = 10\mathrm{~cm} = 0.1\mathrm{~m}$
$ε₀ = 9 \times 10⁻¹²$ $\epsilon_0 = 9 \times 10^{-12}$
$π = 22/7$ $\pi = 22/7$
$I = 0.15~A$ $I = 0.15\mathrm{~A}$
$(dV)/(dt) = 7 \times 10⁸~V/s$

d = (9 \times 10⁻¹²) \times ((22)/(7)) \times (0.1)²0.15 \times (7 \times 10⁸)

d = 9 \times 10⁻¹² \times 22 \times 0.01 \times 10⁸0.15 = 198 \times 10⁻⁴0.15

d = 1320 \times 10⁻⁶~m = 1320

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 $I = C (dV)/(dt)$ $I = C \frac{dV}{dt}$.

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatic Potential and Capacitance

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Equation Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Match options

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Dimensional Analysis

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

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