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 $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · EM Waves

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 1 0^{- 9} T

.
(Given : Permittivity of vacuum

ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

, speed of light in vacuum

c = 3 \times 1 0^{8} ms^{- 1}

)

Answer: 43

Solution

Intensity of an EM wave is related to electric field amplitude by:

I = \frac{1}{2} ϵ_{0} E_{0}^{2} c

E_{0} = \frac{2 I}{ϵ _{0} c} = \frac{2 \times 0.22}{8.85 \times 1 0 ^{- 12} \times 3 \times 1 0 ^{8}} = 12.87

V/m

Magnetic field amplitude:

B_{0} = \frac{E _{0}}{c} = \frac{12.87}{3 \times 1 0 ^{8}} = 4.29 \times 1 0^{- 8}

= 43 \times 1 0^{- 9}

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer

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