An electromagnetic wave of frequency
100 text MHz$100 \text{ MHz}$ propagates through a medium of conductivity,
sigma = 10 text mho/m$\sigma = 10 \text{ mho/m}$. The ratio of
maximum conducting current density to maximum displacement current density is ________.
[textTake frac14piepsilon_0 = 9 times 10^9 text Ncdottextm^2/textC^2]$[\text{Take } \frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9} \text{ N}\cdot\text{m}^{2}/\text{C}^{2}]$Solution
### Related Formula
J_c = sigma E$J_c = \sigma E$
J_d = epsilon_0 fracpartial Epartial t$$J_d = \epsilon_0 \frac{\partial E}{\partial t}$$
### Core Logic
Let the electric field of the wave be E = E_0 sin(omega t - kx)$E = E_0 \sin(\omega t - kx)$.
The conduction current density is:
J_c = sigma E_0 sin(omega t - kx)$$J_c = \sigma E_0 \sin(\omega t - kx)$$
The maximum conduction current density is:
(J_c)_textmax = sigma E_0 quad text--- (i)$$(J_c)_{\text{max}} = \sigma E_0 \quad \text{--- (i)}$$
The displacement current density is:
J_d = frac1A left( epsilon_0 fracpartial (EA)partial t right) = epsilon_0 fracpartial Epartial t$$J_d = \frac{1}{A} \left( \epsilon_0 \frac{\partial (EA)}{\partial t} \right) = \epsilon_0 \frac{\partial E}{\partial t}$$
J_d = epsilon_0 E_0 omega cos(omega t - kx)$$J_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$$
The maximum displacement current density is:
(J_d)_textmax = epsilon_0 E_0 omega quad text--- (ii)$$(J_d)_{\text{max}} = \epsilon_0 E_0 \omega \quad \text{--- (ii)}$$
### Step 1: Taking the Ratio
Dividing (i) by (ii):
textRatio = frac(J_c)_textmax(J_d)_textmax = fracsigma E_0epsilon_0 omega E_0 = fracsigmaepsilon_0 omega$$\text{Ratio} = \frac{(J_c)_{\text{max}}}{(J_d)_{\text{max}}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$$
### Step 2: Substitution and Calculation
We know f = 100 text MHz = 10^8 text Hz$f = 100 \text{ MHz} = 10^8 \text{ Hz}$, so omega = 2pi f = 2pi times 10^8 text rad/s$\omega = 2\pi f = 2\pi \times 10^8 \text{ rad/s}$.
sigma = 10 text mho/m$\sigma = 10 \text{ mho/m}$.
Also, frac14piepsilon_0 = 9 times 10^9 implies frac1epsilon_0 = 4pi times 9 times 10^9$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \implies \frac{1}{\epsilon_0} = 4\pi \times 9 \times 10^9$.
textRatio = frac10 times 4pi times 9 times 10^92pi times 10^8$$\text{Ratio} = \frac{10 \times 4\pi \times 9 \times 10^9}{2\pi \times 10^8}$$
textRatio = frac360pi times 10^92pi times 10^8 = frac36002 = 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 sigma / (omega epsilon_0)$\sigma / (\omega \epsilon_0)$. This dictates whether a medium behaves as a good conductor or a dielectric at a given frequency.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
