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  • The intensity of the light from a bulb incident on a surface is . The

The intensity of the light from a bulb incident on a surface is 0.22 \mathrm{~W} / \mathrm{m}^{2}. The amplitude of the magnetic field in this light-wave is ___________\times 10^{-9} \mathrm{~T}.
(Given : Permittivity of vacuum \epsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1}-\mathrm{m}^{-2}, speed of light in vacuum \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1})

Option: 1

43


Option: 2

4


Option: 3

66


Option: 4

78


Answers (1)

best_answer

\mathrm{\begin{array}{r} \mathrm{I=\frac{1}{2} \; \varepsilon_{0} \; E_{0}^{2} \times C} \\ \end{array}}

     \mathrm{E_{0}=C B_{0}}

\mathrm{I=\frac{1}{2} \times \varepsilon_{0} \times C^{2} \times B_{0}^{2} \times C}

    \mathrm{=\frac{1}{2}\; \varepsilon_{0} \; C^{3} B_{0}^{2}}

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

\mathrm{B_{0}=43\times10^{-9}T}

Hence, correct answer is 43.

Posted by

manish painkra

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