Electromagnetic Waves

JEE Main level, numerical answer, medium

17 of 165

Question 17 of 165

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, σ=10mho/m\sigma=10 \mathrm{mho} / \mathrm{m}. The ratio of maximum conduction current density to maximum displacement current density is ____\_\_\_\_.
[Take 14πϵo=9×109Nm2/C2]\left[\text{Take }\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]

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Answer: 1800

For an electromagnetic wave in a conducting medium, the maximum conduction current density is Jc,max=σE0J_{c,max} = \sigma E_0 and maximum displacement current density is Jd,max=ϵ0ωE0J_{d,max} = \epsilon_0 \omega E_0.
The ratio is: Jc,maxJd,max=σϵ0ω\frac{J_{c,max}}{J_{d,max}} = \frac{\sigma}{\epsilon_0 \omega}.
With σ=10\sigma = 10 mho/m, f=100f = 100 MHz
So ω=2π×108\omega = 2\pi \times 10^8 rad/s, and ϵ0=10−936π\epsilon_0 = \frac{10^{-9}}{36\pi} F/m: Jc,maxJd,max=10×36π10−9×2π×108=10×362×10−1=1800\frac{J_{c,max}}{J_{d,max}} = \frac{10 \times 36\pi}{10^{-9} \times 2\pi \times 10^8} = \frac{10 \times 36}{2 \times 10^{-1}} = 1800.

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