The equation of the electric field of an electromagnetic wave propagating through free space is given by : $E=\sqrt{377} \sin \left(6.27 \times 10^{3} t-2.09 \times 10^{-5} x\right)$ N/C The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right)$ W/m$^{2}$. The value of $\alpha$ is ______. (Take $\sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}=377$ in SI units)

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · EM Waves

The equation of the electric field of an electromagnetic wave propagating through free space is given by :

E = 377 sin (6.27 \times 1 0^{3} t - 2.09 \times 1 0^{- 5} x)

N/C
The average power of the electromagnetic wave is

(\frac{1}{α})

W/m

^{2}

. The value of

α

is ______.
(Take

\frac{μ _{0}}{ε _{0}} = 377

in SI units)

Answer: 2

The wave speed

\frac{ω}{k} = \frac{6.27 \times 1 0 ^{3}}{2.09 \times 1 0 ^{- 5}} = 3 \times 1 0^{8}

m/s confirms propagation in free space. The average intensity is

I = \frac{1}{2} ε_{0} c E_{0}^{2} = \frac{1}{2} \frac{ε _{0}}{μ _{0}} E_{0}^{2} = \frac{E _{0}^{2}}{2 \times 377}

. With

E_{0} = 377

N/C,

E_{0}^{2} = 377

, so

I = \frac{377}{2 \times 377} = \frac{1}{2}

W/m

^{2}

. Comparing with

\frac{1}{α}

gives

α = 2

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Integer answer