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Electromagnetic Waves question

2021 · 26 Aug · Shift 1 · Q64
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Electromagnetic Waves question

2021 · 26 Aug · Shift 1 · Q64

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
The electric field in a plane electromagnetic wave is given by E→=200cos⁡[(0.5×103m)x−(1.5×1011rads×t)]Vmj^\overrightarrow E = 200\cos \left[ {\left( {{{0.5 \times {{10}^3}} \over m}} \right)x - \left( {1.5 \times {{10}^{11}}{{rad} \over s} \times t} \right)} \right]{V \over m}\widehat jE=200cos[(m0.5×103​)x−(1.5×1011srad​×t)]mV​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{x \over {{{10}^9}}}{N \over {{m^2}}}109x​m2N​. Find the value of x .
Numerical answer
View written solutionFree

Correct answer: 354

  1. Given electric field of the plane wave
E⃗=200cos⁡[(0.5×103)x−(1.5×1011t)] j^  V/m\vec E = 200\cos\left[\left(0.5\times 10^3\right)x - \left(1.5\times 10^{11} t\right)\right]\,\hat j\; \text{V/m}E=200cos[(0.5×103)x−(1.5×1011t)]j^​V/m

So the amplitude of electric field is

E0=200  V/mE_0 = 200\;\text{V/m}E0​=200V/m
  1. Intensity of the electromagnetic wave

For a plane electromagnetic wave, average intensity is

I=12cε0E02I = \frac{1}{2}c\varepsilon_0 E_0^2I=21​cε0​E02​

Substitute values:

c=3×108  m/s,ε0=8.85×10−12  F/m,E0=200c = 3\times 10^8\;\text{m/s}, \qquad \varepsilon_0 = 8.85\times 10^{-12}\;\text{F/m}, \qquad E_0=200c=3×108m/s,ε0​=8.85×10−12F/m,E0​=200 I=12(3×108)(8.85×10−12)(200)2I = \frac{1}{2}(3\times 10^8)(8.85\times 10^{-12})(200)^2I=21​(3×108)(8.85×10−12)(200)2 (200)2=40000=4×104(200)^2 = 40000 = 4\times 10^4(200)2=40000=4×104

So,

I=12×3×8.85×4×108−12+4I = \frac{1}{2}\times 3\times 8.85\times 4\times 10^{8-12+4}I=21​×3×8.85×4×108−12+4 I=12×106.2=53.1  W/m2I = \frac{1}{2}\times 106.2 = 53.1\;\text{W/m}^2I=21​×106.2=53.1W/m2
  1. Radiation pressure on a perfectly reflecting surface

For normal incidence on a perfectly reflecting surface,

P=2IcP = \frac{2I}{c}P=c2I​

Therefore,

P=2×53.13×108P = \frac{2\times 53.1}{3\times 10^8}P=3×1082×53.1​ P=106.23×108=35.4×10−8=3.54×10−7  N/m2P = \frac{106.2}{3\times 10^8} = 35.4\times 10^{-8} = 3.54\times 10^{-7}\;\text{N/m}^2P=3×108106.2​=35.4×10−8=3.54×10−7N/m2
  1. Match with the required form

Given,

P=x109  N/m2P = \frac{x}{10^9}\;\text{N/m}^2P=109x​N/m2

So,

x109=3.54×10−7\frac{x}{10^9} = 3.54\times 10^{-7}109x​=3.54×10−7

Multiplying both sides by 10910^9109,

x=3.54×10−7×109=354x = 3.54\times 10^{-7}\times 10^9 = 354x=3.54×10−7×109=354
  1. Note about area and exposure time

The question mentions area 100 cm2100\,\text{cm}^2100cm2 and exposure time 101010 min, but for radiation pressure these are not needed. They would be needed for total force or impulse, not pressure.

Hence, the required integer is

354\boxed{354}354​
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