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Electromagnetic Waves question

2022 · 29 Jun · Shift 1 · Q60
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Electromagnetic Waves question

2022 · 29 Jun · Shift 1 · Q60

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
The intensity of the light from a bulb incident on a surface is 0.22 W/m2. The amplitude of the magnetic field in this light-wave is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 9 T. (Given : Permittivity of vacuum ∈\in∈ 0 = 8.85 ×\times× 10 −-− 12 C2 N −-− 1-m −-− 2, speed of light in vacuum c = 3 ×\times× 108 ms −-− 1)
Numerical answer
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Correct answer: 43

  1. For an electromagnetic wave, the average intensity is I=12cε0E02I=\frac{1}{2}c\varepsilon_0 E_0^2I=21​cε0​E02​ where E0E_0E0​ is the amplitude of electric field.

  2. Also, for an electromagnetic wave, E0=cB0E_0=cB_0E0​=cB0​ where B0B_0B0​ is the amplitude of magnetic field.

  3. Substitute E0=cB0E_0=cB_0E0​=cB0​ into the intensity formula: I=12cε0(cB0)2=12ε0c3B02I=\frac{1}{2}c\varepsilon_0(cB_0)^2=\frac{1}{2}\varepsilon_0 c^3 B_0^2I=21​cε0​(cB0​)2=21​ε0​c3B02​

So, B0=2Iε0c3B_0=\sqrt{\frac{2I}{\varepsilon_0 c^3}}B0​=ε0​c32I​​

  1. Put the given values: I=0.22 W/m2,ε0=8.85×10−12,c=3×108 m/sI=0.22\ \text{W/m}^2,\quad \varepsilon_0=8.85\times 10^{-12},\quad c=3\times 10^8\ \text{m/s}I=0.22 W/m2,ε0​=8.85×10−12,c=3×108 m/s

Then c3=(3×108)3=27×1024c^3=(3\times 10^8)^3=27\times 10^{24}c3=(3×108)3=27×1024

and ε0c3=(8.85×10−12)(27×1024)\varepsilon_0 c^3=(8.85\times 10^{-12})(27\times 10^{24})ε0​c3=(8.85×10−12)(27×1024) =238.95×1012=2.3895×1014=238.95\times 10^{12}=2.3895\times 10^{14}=238.95×1012=2.3895×1014

Also, 2I=0.442I=0.442I=0.44

Hence, B02=0.442.3895×1014≈1.841×10−15B_0^2=\frac{0.44}{2.3895\times 10^{14}}\approx 1.841\times 10^{-15}B02​=2.3895×10140.44​≈1.841×10−15

Therefore, B0=1.841×10−15≈4.29×10−8 TB_0=\sqrt{1.841\times 10^{-15}}\approx 4.29\times 10^{-8}\ \text{T}B0​=1.841×10−15​≈4.29×10−8 T

  1. Write in the required form: 4.29×10−8=42.9×10−9 T4.29\times 10^{-8}=42.9\times 10^{-9}\ \text{T}4.29×10−8=42.9×10−9 T

So the blank is 43\boxed{43}43​

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