Related Formula
Jc = σ E$J_c = \sigma E$
Jd = ε₀ (∂ E)/(∂ t)$$J_d = \epsilon_0 \frac{\partial E}{\partial t}$$
Core Logic
Let the electric field of the wave be E = E₀ (ω t - kx)$E = E_0 \sin(\omega t - kx)$.
The conduction current density is:
Jc = σ E₀ (ω t - kx)$$J_c = \sigma E_0 \sin(\omega t - kx)$$
The maximum conduction current density is:
(Jc)max = σ E₀ --- (i)$$(J_c)_{\text{max}} = \sigma E_0 \quad \text{--- (i)}$$
The displacement current density is:
Jd = (1)/(A) ( ε₀ (∂ (EA))/(∂ t) ) = ε₀ (∂ E)/(∂ t)$$J_d = \frac{1}{A} \left( \epsilon_0 \frac{\partial (EA)}{\partial t} \right) = \epsilon_0 \frac{\partial E}{\partial t}$$
Jd = ε₀ E₀ ω (ω t - kx)$$J_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$$
The maximum displacement current density is:
(Jd)max = ε₀ E₀ ω --- (ii)$$(J_d)_{\text{max}} = \epsilon_0 E_0 \omega \quad \text{--- (ii)}$$
Step 1: Taking the Ratio
Dividing (i) by (ii):
Ratio = (Jc)max(Jd)max = (σ E₀)/(ε₀ ω E₀) = (σ)/(ε₀ ω)$$\text{Ratio} = \frac{(J_c)_{\text{max}}}{(J_d)_{\text{max}}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$$
Step 2: Substitution and Calculation
We know f = 100 MHz = 10⁸ Hz$f = 100 \text{ MHz} = 10^8 \text{ Hz}$, so ω = 2π f = 2π × 10⁸ rad/s$\omega = 2\pi f = 2\pi \times 10^8 \text{ rad/s}$.
σ = 10 mho/m$\sigma = 10 \text{ mho/m}$.
Also, (1)/(4πε₀) = 9 × 10⁹ (1)/(ε₀) = 4π × 9 × 10⁹$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \implies \frac{1}{\epsilon_0} = 4\pi \times 9 \times 10^9$.
Ratio = (10 × 4π × 9 × 10⁹)/(2π × 10⁸)$$\text{Ratio} = \frac{10 \times 4\pi \times 9 \times 10^9}{2\pi \times 10^8}$$
Ratio = (360π × 10⁹)/(2π × 10⁸) = (3600)/(2) = 1800$$\text{Ratio} = \frac{360\pi \times 10^9}{2\pi \times 10^8} = \frac{3600}{2} = 1800$$
Step 3: Final Conclusion
The required ratio is 1800.
Pattern Recognition
The ratio of conduction to displacement current density in any medium is universally σ / (ω ε₀)$\sigma / (\omega \epsilon_0)$. This dictates whether a medium behaves as a good conductor or a dielectric at a given frequency.
Chapter Mix
Class 12 Physics: Electromagnetic Waves
