In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 10¹⁰ ~Hz$5 \times 10^{10} \mathrm{~Hz}$ and an amplitude of 50 ~Vm⁻¹$50 \mathrm{~Vm}^{-1}$. The total average energy density of the electromagnetic field of the wave is : [Use ε₀ = 8.85 × 10⁻¹² C² / Nm²$\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{Nm}^2$ ]

Related Formula

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

Core Logic

Square both sides of the speed of light formula:

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

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

Step 1: Match options

Express the dimensions in terms of the given variable ratios:

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

Related Formula

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

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]$[\epsilon_0] = [M^{-1} L^{-3} T^4 A^2]$
• Electric flux [ΦE] = [M L³ T⁻³ A⁻¹]$[\Phi_E] = [M L^3 T^{-3} A^{-1}]$
• Time [t] = [T]$[t] = [T]$

This confirms the quantity is dimensionally equivalent to Electric current.

Pattern Recognition

Sees: ε₀ dΦEdt$\epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t}$. Shortcut: Instantly identify this as Maxwell's expression for displacement current. Displacement current has the exact same dimensions as regular conduction current.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Related Formula

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:

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

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$\hat{i} \rightarrow \hat{j} \rightarrow \hat{k}$. Since propagation is i$\hat{i}$, our cross product must be j × 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

Related Formula

Core Logic

Combining the current expression with capacitance relations yields:

Isolating plate separation distance d$d$:

Substitute the parameters specified by the problem layout:

• r = 10~cm = 0.1~m$r = 10\mathrm{~cm} = 0.1\mathrm{~m}$
• ε₀ = 9 × 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 × 10⁸~V/s$\frac{dV}{dt} = 7 \times 10^{8}\mathrm{~V/s}$

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

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

Related Formula

Core Logic

Since both cylinders hold equal amounts of electromagnetic energy:

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

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

Step 1: Final Equation Match

The wave equation adjusts its amplitude factor to 200 (ω t - kx)NC⁻¹$200\sin (\omega t - kx)\mathrm{NC}^{-1}$, which matches option (2).

Pattern Recognition

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

Chapter Mix

Class 12 Physics: Electromagnetic Waves