A plane EM wave propagates in free space with electric field E(z,t)=E_0 cos(ω t - kz),x, where E_0 = 1.0×10^4 V/m and ω = 2π×10^9 rad/s. What is the magnitude of the time-averaged Poynting vector of this wave?

1.33×10^5 W/m^2 (approximately).

A plane EM wave propagates in free space with electric field $E (z, t) = E_{0} cos (ω t - k z), \overset{x}{^}$ , where $E_{0} = 1.0 \times 1 0^{4}$ V/m and $ω = 2 π \times 1 0^{9}$ rad/s. What is the magnitude of the time-averaged Poynting vector of this wave?

medium 2 marks Class 12 · Physics · Electromagnetic Waves

Answer

1.33 \times 1 0^{5}

W/m

^{2}

(approximately).

The correct answer is $1.33 \times 1 0^{5}$ W/m $^{2}$ (approximately). — Insert $E_{0} = 1.0 \times 1 0^{4}$ V/m: $I = \frac{( 1.0 \times 1 0 ^{4} ) ^{2}}{2 \times 377} \approx 1.33 \times 1 0^{5}$ W/m $^{2}$ .

Step-by-step solution

For a plane wave in free space the intrinsic impedance is $e t a_{0} = 377$ Ω.
Magnetic field amplitude is $H_{0} = E_{0} / e t a_{0}$ .
Time‑averaged intensity $I = \frac{1}{2} E_{0} H_{0} = \frac{E _{0}^{2}}{2 η _{0}}$ .
Insert $E_{0} = 1.0 \times 1 0^{4}$ V/m: $I = \frac{( 1.0 \times 1 0 ^{4} ) ^{2}}{2 \times 377} \approx 1.33 \times 1 0^{5}$ W/m $^{2}$ .

Common mistake: Forgetting the factor

1/2

for time averaging

Concept tested: Poynting vector and intensity of EM wave

Published 4 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

← All Electromagnetic Waves questions

A plane EM wave propagates in free space with electric field E(z,t)=E0​cos(ωt−kz),x^, where E0​=1.0×104 V/m and ω=2π×109 rad/s. What is the magnitude of the time-averaged Poynting vector of this wave?

Step-by-step solution

A plane EM wave propagates in free space with electric field $E (z, t) = E_{0} cos (ω t - k z), \overset{x}{^}$ , where $E_{0} = 1.0 \times 1 0^{4}$ V/m and $ω = 2 π \times 1 0^{9}$ rad/s. What is the magnitude of the time-averaged Poynting vector of this wave?