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

### Related Formula I = frac12 epsilon_0 E_0^2 c$$I = \frac{1}{2} \epsilon_0 E_0^2 c$$ Where E_0$E_0$ is the amplitude of the electric field (200text V/m$200\text{ V/m}$) and c = 3 times 10^8text m/s$c = 3 \times 10^8\text{ m/s}$. ### Core Logic Substitute the constants into the equation: I = frac12 times (8.85 times 10^-12) times (200)^2 times (3 times 10^8)$$I = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (200)^2 \times (3 \times 10^8)$$ ### Step 1: Compute value I = frac12 times 8.85 times 10^-12 times 4 times 10^4 times 3 times 10^8$$I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8$$ I = 2 times 8.85 times 3 times 10^0$$I = 2 \times 8.85 \times 3 \times 10^0$$ I = 53.1text W/m^2$$I = 53.1\text{ W/m}^2$$ ### Pattern Recognition Isolate powers of ten first (10^-12 times 10^4 times 10^8 = 10^0$10^{-12} \times 10^4 \times 10^8 = 10^0$) to streamline intermediate tracking accuracy on calculation variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Core Logic Let's review the fundamental Maxwell's equations: 1. **Ampere - Maxwell Law** relates the magnetic path integral to conduction current and displacement current: oint vecB cdot dvecl = mu_0 i_c + mu_0 varepsilon_0 fracdphi_Edt implies textA - IV$$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d\phi_E}{dt} \implies \text{A - IV}$$ 2. **Faraday's Law of Induction** states that changing magnetic flux induces an electromotive force (EMF): oint vecE cdot dvecl = -fracdphi_Bdt implies textB - III$$\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt} \implies \text{B - III}$$ 3. **Gauss's Law for Electricity** relates net electric flux to enclosed charge: oint vecE cdot dvecA = fracQvarepsilon_0 implies textC - I$$\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0} \implies \text{C - I}$$ 4. **Gauss's Law for Magnetism** states that magnetic monopoles do not exist: oint vecB cdot dvecA = 0 implies textD - II$$\oint \vec{B} \cdot d\vec{A} = 0 \implies \text{D - II}$$ ### Step 1: Match Evaluation The match configurations are: * A rightarrow$\rightarrow$ IV * B rightarrow$\rightarrow$ III * C rightarrow$\rightarrow$ I * D rightarrow$\rightarrow$ II This perfectly corresponds to Option (3). ### Pattern Recognition Understand the integral geometries: Path integrals (line integrals oint cdot dvecl$\oint \cdot d\vec{l}$) correspond to circulating fields (induction laws like Ampere/Faraday). Surface integrals (flux integrals oint cdot dvecA$\oint \cdot d\vec{A}$) correspond to bounded charge states (Gauss laws). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatics Class 12 Physics: Magnetism and Matter

### Related Formula p = fracUc$$p = \frac{U}{c}$$ ### Core Logic For an electromagnetic wave incident on a surface that is completely absorbed, the total momentum transferred is equal to the total energy transferred divided by the speed of light in vacuum (c$c$). ### Step 1: Calculate Momentum Given total energy E = 6.48 times 10^5 mathrm~J$E = 6.48 \times 10^5 \mathrm{~J}$. Speed of light c = 3 times 10^8 mathrm~m/s$c = 3 \times 10^8 \mathrm{~m/s}$. p = fracEc = frac6.48 times 10^53 times 10^8$$p = \frac{E}{c} = \frac{6.48 \times 10^5}{3 \times 10^8}$$ p = 2.16 times 10^-3 mathrm~kg~m/s$$p = 2.16 \times 10^{-3} \mathrm{~kg~m/s}$$ ### Pattern Recognition Always check for "complete absorption" versus "perfect reflection". For complete absorption, p = E/c$p = E/c$. For perfect reflection, p = 2E/c$p = 2E/c$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Core Logic Match each law with its corresponding mathematical expression (Maxwell's equations). (A) Gauss's law of magnetostatics: The net magnetic flux through any closed surface is zero. oint vecB cdot mathrmdveca = 0$\oint \vec{B} \cdot \mathrm{d}\vec{a} = 0$ (Matches II). (B) Faraday's law of electromagnetic induction: The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux. oint vecE cdot mathrmdvecl = -fracmathrmdmathrmdt int vecB cdot mathrmdveca$\oint \vec{E} \cdot \mathrm{d}\vec{l} = -\frac{\mathrm{d}}{\mathrm{d}t} \int \vec{B} \cdot \mathrm{d}\vec{a}$ (Matches III). (C) Ampere's law: The line integral of the magnetic field around a closed loop is proportional to the electric current passing through the loop. oint vecB cdot mathrmdvecl = mu_0 I$\oint \vec{B} \cdot \mathrm{d}\vec{l} = \mu_0 I$ (Matches IV). (D) Gauss's law of electrostatics: The electric flux through any closed surface is proportional to the enclosed electric charge. oint vecE cdot mathrmdveca = frac1varepsilon_0 int rho mathrmdV$\oint \vec{E} \cdot \mathrm{d}\vec{a} = \frac{1}{\varepsilon_0} \int \rho \mathrm{d}V$ (Matches I). ### Step 1: Final Match A rightarrow$\rightarrow$ II B rightarrow$\rightarrow$ III C rightarrow$\rightarrow$ IV D rightarrow$\rightarrow$ I This matches option (4). ### Pattern Recognition These are the fundamental Maxwell equations in integral form. Memorizing their direct mappings guarantees quick marks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electromagnetic Induction

### Related Formula B_0 = fracE_0C$$B_0 = \frac{E_0}{C}$$ hatC = hatE times hatB$$\hat{C} = \hat{E} \times \hat{B}$$ ### Core Logic In an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by E_0 = c B_0$E_0 = c B_0$. Thus, B_0 = E_0 / C$B_0 = E_0 / C$. The direction of wave propagation is given by the cross product of the electric field and magnetic field vectors: hatC = hatE times hatB$\hat{C} = \hat{E} \times \hat{B}$. ### Step 1: Determine Wave Direction and Magnetic Field Direction Given the phase term (omega t - kz)$(\omega t - kz)$, the wave propagates in the +z$+z$ direction, so hatC = hatk$\hat{C} = \hat{k}$. The electric field oscillates in the +x$+x$ direction, so hatE = hati$\hat{E} = \hat{i}$. We know: hatk = hati times hatB$$\hat{k} = \hat{i} \times \hat{B}$$ Since hati times hatj = hatk$\hat{i} \times \hat{j} = \hat{k}$, the magnetic field must oscillate in the +y$+y$ direction (hatj$\hat{j}$). ### Step 2: Construct Final Vector The full magnetic field vector shares the same phase and applies the above amplitude and direction: vecB = fracE_0C cos(omega t - kz) hatj$$\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$$ ### Pattern Recognition Phase remains identical. Amplitude scales by 1/c$1/c$. Direction satisfies the right-hand triad (vecE, vecB, vecv)$(\vec{E}, \vec{B}, \vec{v})$ where vecv = vecE times vecB$\vec{v} = \vec{E} \times \vec{B}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves