Physics · 2021
JEE Main · 26 August 2021, Shift 1 · Q26
The electric field in a plane electromagnetic wave is given by E =200 cos [((0.5 × 10^3)/m) x-(1.5 × 10^11 rad/s × t )] V/m j If this wave falls…
The electric field in a plane electromagnetic wave is given by
$$\overrightarrow{\mathrm{E}}=200 \cos \left[\left(\frac{0.5 \times 10^3}{\mathrm{~m}}\right) x-\left(1.5 \times 10^{11} \frac{\mathrm{rad}}{\mathrm{~s}} \times \mathrm{t}\right)\right] \frac{\mathrm{V}}{\mathrm{~m}} \hat{j}
$$
If this wave falls normally on a perfectly reflecting surface having an area of $\displaystyle 100 \mathrm{~cm}^2$. If the radiation pressure exerted by the E.M. wave on the surface during a $\displaystyle 10$ minute exposure is $\displaystyle \frac{x}{10^9} \frac{\mathrm{~N}}{\mathrm{~m}^2}$. Find the value of $\displaystyle x$.
Official answer
354
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.