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Electromagnetic Waves question

2022 · 24 Jun · Shift 2 · Q61
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  5. /2022 · 24 Jun · Shift 2 · Q61

Electromagnetic Waves question

2022 · 24 Jun · Shift 2 · Q61

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.
  1. A
    1.19 ×\times× 10 −-− 8T
  2. B
    1.71 ×\times× 10 −-− 8T
  3. C
    0.84 ×\times× 10 −-− 8T
  4. D
    3.36 ×\times× 10 −-− 8T
View written solutionFree

Correct answer: B

  1. Useful radiated power

The bulb is rated at 200 W200\,\text{W}200W, but only 3.5%3.5\%3.5% of this is emitted as radiation.

P=0.035×200=7 WP = 0.035 \times 200 = 7\,\text{W}P=0.035×200=7W

  1. Intensity at distance r=4 mr=4\,\text{m}r=4m

Treating the bulb as a point source, the radiation spreads uniformly over a sphere of radius rrr.

I=P4πr2=74π(4)2=764πI = \frac{P}{4\pi r^2} = \frac{7}{4\pi (4)^2} = \frac{7}{64\pi}I=4πr2P​=4π(4)27​=64π7​

Numerically,

I≈7201.06≈3.48×10−2 W/m2I \approx \frac{7}{201.06} \approx 3.48 \times 10^{-2}\,\text{W/m}^2I≈201.067​≈3.48×10−2W/m2

  1. Relation between intensity and peak magnetic field

For an electromagnetic wave,

I=c2μ0B02I = \frac{c}{2\mu_0} B_0^2I=2μ0​c​B02​

So,

B0=2μ0IcB_0 = \sqrt{\frac{2\mu_0 I}{c}}B0​=c2μ0​I​​

Using:

  • μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7}\,\text{H/m}μ0​=4π×10−7H/m
  • c=3×108 m/sc = 3 \times 10^8\,\text{m/s}c=3×108m/s
  • I=3.48×10−2 W/m2I = 3.48 \times 10^{-2}\,\text{W/m}^2I=3.48×10−2W/m2

B0=2(4π×10−7)(3.48×10−2)3×108B_0 = \sqrt{\frac{2(4\pi \times 10^{-7})(3.48 \times 10^{-2})}{3 \times 10^8}}B0​=3×1082(4π×10−7)(3.48×10−2)​​

  1. Calculation

First evaluate the numerator:

2(4π×10−7)(3.48×10−2)≈8π×3.48×10−9≈87.5×10−9=8.75×10−82(4\pi \times 10^{-7})(3.48 \times 10^{-2}) \approx 8\pi \times 3.48 \times 10^{-9} \approx 87.5 \times 10^{-9} = 8.75 \times 10^{-8}2(4π×10−7)(3.48×10−2)≈8π×3.48×10−9≈87.5×10−9=8.75×10−8

Then,

8.75×10−83×108≈2.92×10−16\frac{8.75 \times 10^{-8}}{3 \times 10^8} \approx 2.92 \times 10^{-16}3×1088.75×10−8​≈2.92×10−16

Therefore,

B0=2.92×10−16≈1.71×10−8 TB_0 = \sqrt{2.92 \times 10^{-16}} \approx 1.71 \times 10^{-8}\,\text{T}B0​=2.92×10−16​≈1.71×10−8T

  1. Option check
  • A: 1.19×10−8 T1.19 \times 10^{-8}\,\text{T}1.19×10−8T
  • B: 1.71×10−8 T1.71 \times 10^{-8}\,\text{T}1.71×10−8T
  • C: 0.84×10−8 T0.84 \times 10^{-8}\,\text{T}0.84×10−8T
  • D: 3.36×10−8 T3.36 \times 10^{-8}\,\text{T}3.36×10−8T

Hence, the correct option is B.

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