JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

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 ______.

$(Take \frac{μ _{0}}{ε _{0}} = 377 in SI units)$

Your answer

Solution

Step 1 — Verify the wave is in vacuum: $v = \frac{ω}{k} = \frac{6.27 \times 1 0 ^{3}}{2.09 \times 1 0 ^{- 5}} = 3 \times 1 0^{8} m/s$ . ✓

Step 2 — Time-averaged intensity:

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

With $E_{0}^{2} = 377$ and $μ_{0} / ε_{0} = 377$ :

$I = \frac{1}{2} \cdot \frac{377}{377} = \frac{1}{2} W/m^{2}$

$∴ α = 2$ .

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