A light wave traveling in air along the x-dir | vedclass

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(A) The given equation of the electric field is $E_{y} = 540 \sin \pi 	imes 10^{4}(x - ct) 	ext{ Vm}^{-1}$.Comparing this with the standard wave equ

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$18$

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(A) The given equation of the electric field is $E_{y} = 540 \sin \pi 	imes 10^{4}(x - ct) 	ext{ Vm}^{-1}$.Comparing this with the standard wave equation $E = E_{0} \sin(kx - \omega t)$,we identify the peak value of the electric field as $E_{0} = 540 	ext{ Vm}^{-1}$.The relationship between the peak electric field $E_{0}$ and the peak magnetic field $B_{0}$ is given by $B_{0} = \frac{E_{0}}{c}$.Substituting the given values,$B_{0} = \frac{540}{3 	imes 10^{8}} 	ext{ T}$.$B_{0} = 180 	imes 10^{-8} 	ext{ T} = 18 	imes 10^{-7} 	ext{ T}$.Thus,the peak value of the magnetic field is $18 	imes 10^{-7} 	ext{ T}$.

$A$ light wave traveling in air along the $x$-direction is given by $E_{y} = 540 \sin \pi \times 10^{4}(x - ct) \text{ Vm}^{-1}$. Then,the peak value of the magnetic field of the wave will be $\dots \times 10^{-7} \text{ T}$ (Given $c = 3 \times 10^{8} \text{ ms}^{-1}$)

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