JEE Main · 2026 · 21 Jan 2026 · Shift 2 · EM Waves

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}]

Answer: 1800

Solution

The maximum conduction current density is

(j_{c})_{max} = σ E_{0}

. The maximum displacement current density is

(j_{d})_{max} = ϵ_{0} ω E_{0}

, where

ω = 2 π f

. The ratio is

\frac{( j _{c} ) _{max}}{( j _{d} ) _{max}} = \frac{σ}{ϵ _{0} ω} = \frac{σ \cdot 4 π \cdot 9 \times 1 0 ^{9}}{2 π f} = \frac{10 \times 4 π \times 9 \times 1 0 ^{9}}{2 π \times 100 \times 1 0 ^{6}} = \frac{10 \times 2 \times 9 \times 1 0 ^{9}}{100 \times 1 0 ^{6}} = \frac{180 \times 1 0 ^{9}}{100 \times 1 0 ^{6}} = 1800

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer