JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2021
- Hard
- Numerical
The electric field in a plane electromagnetic wave is given by \(\overrightarrow{{E}}=200 \cos \left[\left(\frac{0.5 \times 10^{3}}{{m}}\right) {x}-\left(1.5 \times 10^{11} \frac{{rad}}{{s}} \times {t}\right)\right] \frac{{V}}{{m}} \hat{{j}}\) If this wave falls normally on a perfectly reflecting surface having an area of \(100 \;{cm}^{2}\). If the radiation pressure exerted by the \(E.M.\) wave on the surface during a \(10\, minute\) exposure is \(\frac{{x}}{10^{9}} \frac{{N}}{{m}^{2}}\). Find the value of \({x}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(354\)
Step-by-step Solution
Detailed explanation
\({E}_{0}=200\)
\({I}=\frac{1}{2} \varepsilon_{0} {E}_{0}^{2} \cdot {C}\)
Radiation pressure
\(P=\frac{2 I}{C}\)
\(=\left(\frac{2}{C}\right)\left(\frac{1}{2} \varepsilon_{0} E_{0}^{2} C\right)\)
\(=\varepsilon_{0} E_{0}^{2}\)
\(=8.85 \times 10^{-12} \times 200^{2}\)
\(=8.85 \times 10^{-8} \times 4\)
\(=\frac{354}{10^{9}}\)
Ans. \(354.0\)
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