An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $σ = 10 m h o / m$ . The ratio of maximum conduction current density to maximum displacement current density is ______.

$[Take \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} / C^{2}]$ [2026]

Question

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $σ = 10 m h o / m$ . The ratio of maximum conduction current density to maximum displacement current density is ______.

$[Take \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} / C^{2}]$ [2026]

Smart Achievers · Accepted Answer

(1800)

$A$

$j_{c} = σ E$

$[Take \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} / C^{2}]$

$j_{c} = σ E_{0} \sin (ω t - k x)$

$\Rightarrow {(j_{c})}_{\max} = σ E_{0} . . . (i)$

$J_{d} = \frac{i_{d}}{A} = \frac{1}{A} \times ε_{0} \frac{A d E}{d t}$

$= ε_{0} \times E_{0} ω \cos (ω t - k x)$

${(j_{d})}_{\max} = ε_{0} E_{0} ω . . . (ii)$

(i)/(ii)

$\frac{{(j_{c})}_{\max}}{{(j_{d})}_{\max}} = \frac{σ E_{0}}{ε_{0} ω E_{0}} \Rightarrow \frac{σ}{ε_{0} ω}$

$\Rightarrow \frac{10 \times 4 π \times 9 \times 10^{9}}{2 π \times 100 \times 10^{6}}$

$\Rightarrow 1800$

Solution

Q.
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $σ = 10 m h o / m$ . The ratio of maximum conduction current density to maximum displacement current density is ______.

$[Take \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} / C^{2}]$ [2026]

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Solution

Q. An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity,σ=10 mho/m. The ratio of maximum conduction current density to maximum displacement current density is ______. [Take 14πε0=9×109 N m2/C2] [2026]

Ans. (1800) A jc=σE E⇒E0sin(ωt-kx) jc=σE0sin(ωt-kx) ⇒(jc)max=σE0 ...(i) Jd=idA=1A×ε0AdEdt =ε0×E0ωcos(ωt-kx) (jd)max=ε0E0ω ...(ii) (i)/(ii) (jc)max(jd)max=σE0ε0ωE0⇒σε0ω ⇒10×4π×9×1092π×100×106 ⇒1800

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Q.
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $σ = 10 m h o / m$ . The ratio of maximum conduction current density to maximum displacement current density is ______.

$[Take \frac{1}{4 π ε_{0}} = 9 \times 10^{9} {N m}^{2} / C^{2}]$ [2026]