JEE Main 2026 Jan 21 Shift 2, Physics Q21: Displacement Current

JEE Main2026Jan 21, Shift 2Physics

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

$[Take \frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} Nm^{2} / C^{2}]$

Your answer

Solution

For a sinusoidal wave with $E = E_{0} sin (ω t - k x)$ :

Conduction current density: $j_{c} = σ E$ , so $(j_{c})_{ma x} = σ E_{0}$ .

Displacement current density: $j_{d} = ϵ_{0} \frac{\partial E}{\partial t} = ϵ_{0} E_{0} ω cos (ω t - k x)$ , so $(j_{d})_{ma x} = ϵ_{0} E_{0} ω$ .

Ratio: $\frac{( j _{c} ) _{ma x}}{( j _{d} ) _{ma x}} = \frac{σ}{ϵ _{0} ω} = \frac{σ \cdot 4 π \cdot ( 1/4 π ϵ _{0} )}{ω} = \frac{4 π \times 9 \times 1 0 ^{9} \times 10}{2 π \times 100 \times 1 0 ^{6}}$ .

$= \frac{10 \times 4 π \times 9 \times 1 0 ^{9}}{2 π \times 1 0 ^{8}} = \frac{2 \times 9 \times 1 0 ^{10}}{1 0 ^{8}} = 1800$ .

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