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 - 10-9 T-Given - Permittivity of vacuum -0-8-85 - 10-12 C2 N-1-m-2- speed of light in vacuum c-3 - 108 -ms-1

I=12ε_0 E_0^2· c=12ε_0(c B_0)^2 c ⇒ I=12ε_0 c^3 B_0^2 ⇒ 0.22=12(8.85 × 10^ 12)(3 × 10^8)^3 B_0^2 ⇒ B_0≃ 43 × 10^ 9 T nbsp; ...

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Byju's Answer

Standard XII

Physics

YDSE Problems II

The intensity...

Question

The intensity of the light from a bulb incident on a surface is $0.22 W / m^{2}$ . The amplitude of the magnetic field in this light-wave is $\times 10^{-9} \mathrm{~T} $. (G i v e n : P e r m i t t i v i t y o f v a c u u m$ \varepsilon_{0}=8.85 \times 10^{-12}C^{2} N^{-1}-m^{-2} $, s p e e d o f l i g h t i n v a c u u m$ ~(c)=3 \times 10^{8}\left.\mathrm{ms}^{-1}\right)$

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Solution

$I = \frac{1}{2} ε_{0} E_{0}^{2} \cdot c = \frac{1}{2} ε_{0} {(c B_{0})}_{0}^{2} c$

$\Rightarrow I = \frac{1}{2} ε_{0} c^{3} B_{0}^{2}$

$\Rightarrow 0.22 = \frac{1}{2} (8.85 \times 10^{- 12}) {(3 \times 10^{8})}^{3} B_{0}^{2}$

$\Rightarrow B_{0} ≃ 43 \times 10^{- 9} T$

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I=12ε0E20⋅c=12ε0(cB0)2c ⇒I=12ε0c3B20 ⇒0.22=12(8.85×10−12)(3×108)3B20 ⇒B0≃43×10−9T

$I = \frac{1}{2} ε_{0} E_{0}^{2} \cdot c = \frac{1}{2} ε_{0} {(c B_{0})}_{0}^{2} c$

$\Rightarrow I = \frac{1}{2} ε_{0} c^{3} B_{0}^{2}$

$\Rightarrow 0.22 = \frac{1}{2} (8.85 \times 10^{- 12}) {(3 \times 10^{8})}^{3} B_{0}^{2}$

$\Rightarrow B_{0} ≃ 43 \times 10^{- 9} T$