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JEE Main 2021 (26 Aug, Shift 1) · Momentum & Radiation Pressure

Momentum & Radiation Pressure question from JEE Main 2021 (26 Aug, Shift 1)

Q8JEE Main 2021 · 26 AUG · SHIFT 1Medium

Momentum & Radiation Pressure

The electric field in a plane electromagnetic wave is given by

E⃗=200cos⁡[(0.5×103m)x−(1.5×1011rads×t)]Vmj^\vec{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 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is x109Nm2\frac{x}{10^{9}} \frac{N}{m^2}. Find the value of x.

Numerical answer type. Work it out, then open the answer.

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Answer: 354

  1. E0=200 V/mE_0 = 200\ \text{V/m}; I=12ε0cE02=12×8.85×10−12×3×108×4×104≈53.1 W/m2I = \tfrac{1}{2}\varepsilon_0 c E_0^2 = \tfrac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times 4 \times 10^4 \approx 53.1\ \text{W/m}^2
  2. Perfect reflector: Prad=2Ic=106.23×108P_{rad} = \dfrac{2I}{c} = \dfrac{106.2}{3 \times 10^8} (area and time do not matter)
  3. Prad≈3.54×10−7=354109 N/m2P_{rad} \approx 3.54 \times 10^{-7} = \dfrac{354}{10^9}\ \text{N/m}^2
  4. So x=354x = 354.
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