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Electromagnetic Waves question

2021 · 26 Feb · Shift 1 · Q70
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  5. /2021 · 26 Feb · Shift 1 · Q70

Electromagnetic Waves question

2021 · 26 Feb · Shift 1 · Q70

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x ×\times× 10 −-− 1 V/m. Value of x is ‾\underline{\hspace{2cm}}​. (Rounded off to the nearest integer) [Take ε0=8.85×10−12{\varepsilon _0} = 8.85 \times {10^{ - 12}}ε0​=8.85×10−12 C2N −-− 1 m −-− 2, c = 3×1083 \times {10^8}3×108 ms −-− 1]
Numerical answer
View written solutionFree

Correct answer: 137

  1. Useful power emitted as radiation

The bulb is rated at 1000 W1000\,\text{W}1000W, but its efficiency is only 1.25%1.25\%1.25%.

So the power radiated as electromagnetic waves is

P=1000×1.25100=12.5 WP = 1000 \times \frac{1.25}{100} = 12.5\,\text{W}P=1000×1001.25​=12.5W
  1. Intensity at distance r=2 mr=2\,\text{m}r=2m

Assuming isotropic emission, intensity at distance rrr is

I=P4πr2I = \frac{P}{4\pi r^2}I=4πr2P​

Thus,

I=12.54π(2)2=12.516πI = \frac{12.5}{4\pi (2)^2} = \frac{12.5}{16\pi}I=4π(2)212.5​=16π12.5​

Using π≈3.14\pi \approx 3.14π≈3.14,

I≈12.550.24≈0.249 W/m2I \approx \frac{12.5}{50.24} \approx 0.249\,\text{W/m}^2I≈50.2412.5​≈0.249W/m2
  1. Relation between intensity and peak electric field

For an electromagnetic wave,

I=12ε0cE02I = \frac{1}{2} \varepsilon_0 c E_0^2I=21​ε0​cE02​

where E0E_0E0​ is the peak electric field.

So,

E0=2Iε0cE_0 = \sqrt{\frac{2I}{\varepsilon_0 c}}E0​=ε0​c2I​​

Substitute values:

E0=2(0.249)(8.85×10−12)(3×108)E_0 = \sqrt{\frac{2(0.249)}{(8.85\times 10^{-12})(3\times 10^8)}}E0​=(8.85×10−12)(3×108)2(0.249)​​

First compute denominator:

ε0c=8.85×10−12×3×108=2.655×10−3\varepsilon_0 c = 8.85\times 10^{-12} \times 3\times 10^8 = 2.655\times 10^{-3}ε0​c=8.85×10−12×3×108=2.655×10−3

Then,

2Iε0c=0.4982.655×10−3≈187.6\frac{2I}{\varepsilon_0 c} = \frac{0.498}{2.655\times 10^{-3}} \approx 187.6ε0​c2I​=2.655×10−30.498​≈187.6

Hence,

E0≈187.6≈13.7 V/mE_0 \approx \sqrt{187.6} \approx 13.7\,\text{V/m}E0​≈187.6​≈13.7V/m
  1. Match with the given form

Given,

E0=x×10−1 V/mE_0 = x \times 10^{-1}\,\text{V/m}E0​=x×10−1V/m

So,

13.7=x×10−113.7 = x \times 10^{-1}13.7=x×10−1 x=137x = 137x=137
  1. Final answer

Rounded to nearest integer,

137\boxed{137}137​
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