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JEE Mains · Physics · STD 12 - Electromagnetic waves
  • JEE Mains 2026
  • Hard
  • Numerical

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, \(\sigma=10 mho / m\). The ratio of maximum conducting current density to maximum displacement current density is _________ .
\(\left[\right.\) Take \(\left.\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 Nm ^2 / C ^2\right]\)

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Answer & Solution

Correct Answer

(A) 1800

Step-by-step Solution

Detailed explanation

\(j _{ c }=\sigma E\)
\(E \Rightarrow E _0 \sin (\omega t - kx )\)
\(j _{ c }=\sigma E _0 \sin (\omega t - kx )\)
\(\Rightarrow\left( j _{ c }\right)_{\max }=\sigma E _0\)....(i)
\(J_d \Rightarrow \frac{i_d}{A}=\frac{1}{A} \times \varepsilon_0 \frac{A d E}{d t}\)
\(\Rightarrow \varepsilon_0 \times E _0 \omega \cos (\omega t- kx )\)
\(\left( j _{ d }\right)_{\max } \Rightarrow \varepsilon_0 E _0 \omega\)....(ii)
(i)/(ii)
\(\frac{\left( j _{ c }\right)_{\max }}{\left( j _{ d }\right)_{\max }}=\frac{\sigma E _0}{\varepsilon_0 \omega E _0} \Rightarrow \frac{\sigma}{\varepsilon_0 \omega}\)
\(\Rightarrow \frac{10 \times 4 \pi \times 9 \times 10^9}{2 \pi \times 100 \times 10^6}\)
\(\Rightarrow 1800\)
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