JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 19)
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
43
Explanation
$$I = {1 \over 2}{\varepsilon _0}E_0^2\,.\,c = {1 \over 2}{\varepsilon _0}{(c{B_0})^2}c$$
$$ \Rightarrow I = {1 \over 2}{\varepsilon _0}{c^3}B_0^2$$
$$ \Rightarrow 0.22 = {1 \over 2}\left( {8.85 \times {{10}^{ - 12}}} \right){\left( {3 \times {{10}^8}} \right)^3}B_0^2$$
$$ \Rightarrow {B_0} \simeq 43 \times {10^{ - 9}}$$ T
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