An electromagnetic wave travelling in the x-direction has frequency of 3×1014 Hz and electric field amplitude of 27 Vm–1. From the options given below, which one describes the magnetic field for this wave ?

c=E0B0 ; B0=E0C=273 x 108=9 x 10-8B¯=B0 sinkx-ωt =B0sin2πxλ-nt =B0sin2πnxc-nt ; c=E0B0 =9 x 10-8sin2π3 x 10143 x 108x-3 x 1014tB→=9 x 10-8 sin2π1 x 106x-3 x 1014t k^

An electromagnetic wave travelling in the x-direction has frequency of 3×1014 Hz and electric field amplitude of 27 Vm–1. From the options given below, which one describes the magnetic field for this wave ?

c=E0B0 ; B0=E0C=273 x 108=9 x 10-8B¯=B0 sinkx-ωt =B0sin2πxλ-nt =B0sin2πnxc-nt ; c=E0B0 =9 x 10-8sin2π3 x 10143 x 108x-3 x 1014tB→=9 x 10-8 sin2π1 x 106x-3 x 1014t k^

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An electromagnetic wave travelling in the x-direction has frequency of 3×10¹⁴Hz and electric field amplitude of 27 Vm^–1. From the options given below, which one describes the magnetic field for this wave ?

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$\vec{B} (x, t) = (9 x 10^{- 8} T) \hat{j} \sin [1.5 x 10^{- 8} x - 3 x 10^{14} t]$

$\vec{B} (x, t) = (9 x 10^{- 8} T) \hat{j} \sin [1.5 x 10^{- 6} x - 2 x 10^{14} t]$

$\vec{B} (x, t) = (9 x 10^{- 8} T) \hat{k} \sin 2 π [1.0 x 10^{6} x - 3 x 10^{14} t]$

$\vec{B} (x, t) = (3 x 10^{- 8} T) \hat{j} \sin [1.0 x 10^{6} x - 3 x 10^{14} t]$

answer is D.

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Detailed Solution

$\begin{array}{rcl} c & = & \frac{E_{0}}{B_{0}}; B_{0} = \frac{E_{0}}{C} = \frac{27}{3 x 10^{8}} = 9 x 10^{- 8} \\ \bar{B} & = & B_{0} \sin (kx - ωt) \\ = & B_{0} \sin 2 π (\frac{x}{λ} - nt) \\ = & B_{0} \sin 2 π (\frac{nx}{c} - nt); c = \frac{E_{0}}{B_{0}} \\ = & 9 x 10^{- 8} \sin 2 π (\frac{3 x 10^{14}}{3 x 10^{8}} x - 3 x 10^{14} t) \\ \vec{B} & = & 9 x 10^{- 8} \sin 2 π [1 x 10^{6} x - 3 x 10^{14} t] \hat{k} \end{array}$