An electromagnetic wave of frequency 100 text | vedclass

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(A) The conducting current density is given by $j_c = \sigma E$. For an electromagnetic wave, $E = E_0 \sin(\omega t - kx)$, so the maximum conducting

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$1800$

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(A) The conducting current density is given by $j_c = \sigma E$. For an electromagnetic wave, $E = E_0 \sin(\omega t - kx)$, so the maximum conducting current density is $(j_c)_{\max} = \sigma E_0$. The displacement current density is given by $j_d = \epsilon_0 \frac{\partial E}{\partial t}$. Substituting $E$, we get $j_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$, so the maximum displacement current density is $(j_d)_{\max} = \epsilon_0 E_0 \omega$. The ratio is $\frac{(j_c)_{\max}}{(j_d)_{\max}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$. Given $\sigma = 10 	ext{ mho/m}$, $f = 100 	imes 10^6 	ext{ Hz}$, and $\omega = 2 \pi f = 2 \pi 	imes 10^8 	ext{ rad/s}$. Also, $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9$, so $\frac{1}{\epsilon_0} = 4 \pi 	imes 9 	imes 10^9 = 36 \pi 	imes 10^9$. Ratio $= \frac{10 	imes 36 \pi 	imes 10^9}{2 \pi 	imes 10^8} = \frac{360 \pi 	imes 10^9}{2 \pi 	imes 10^8} = 180 	imes 10 = 1800$.

List-$I$ (Relation)	List-$II$ (Law)
$A$. $\oint \overrightarrow{E} \cdot d\vec{l} = -\frac{d}{dt} \oint \overrightarrow{B} \cdot d\vec{a}$	$I$. Ampere's circuital law
$B$. $\oint \vec{B} \cdot d\vec{l} = \mu_0(I + \epsilon_0 \frac{d\phi_E}{dt})$	$II$. Faraday's laws of electromagnetic induction
$C$. $\oint \overrightarrow{E} \cdot d\vec{a} = \frac{1}{\epsilon_0} \int \rho dv$	$III$. Ampere-Maxwell law
$D$. $\oint \overrightarrow{B} \cdot d\vec{l} = \mu_0 I$	$IV$. Gauss's law of electrostatics

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