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 ______________ $\times$ 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)
Answer (integer)
43
Solution
<p>$$I = {1 \over 2}{\varepsilon _0}E_0^2\,.\,c = {1 \over 2}{\varepsilon _0}{(c{B_0})^2}c$$</p>
<p>$\Rightarrow I = {1 \over 2}{\varepsilon _0}{c^3}B_0^2$</p>
<p>$$ \Rightarrow 0.22 = {1 \over 2}\left( {8.85 \times {{10}^{ - 12}}} \right){\left( {3 \times {{10}^8}} \right)^3}B_0^2$$</p>
<p>$\Rightarrow {B_0} \simeq 43 \times {10^{ - 9}}$ T</p>
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Properties of EM Waves
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