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An electromagnetic wave travelling in the X – direction has a frequency of $2 \times 10^{14} H z$ and electric field amplitude of 18 V/m .From the option given below, which one describes the magnetic field for this wave ?

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

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

$\vec{B} (x, t) = (9 \times 10^{- 8} T) \hat{i} \sin 2 π (1.5 \times 10^{- 8} x - 3 \times 10^{14} t)$

$\vec{B} (x, t) = (6 \times 10^{- 8} T) \hat{k} \sin 2 π (0.6 \times 10^{- 6} x - 2 \times 10^{14} t)$

answer is D.

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Detailed Solution

$f = 2 \times 10^{14} H z, C = \frac{E_{0}}{B_{0}}$

$E_{0} = 18 V / m, B_{0} = \frac{E_{0}}{C}$

$= \frac{18}{3 \times 10^{8}} = 6 \times 10^{- 8} T$

$C = f λ$

$λ = \frac{c}{f} \Rightarrow \frac{1}{λ} = \frac{f}{c} = \frac{2 \times 10^{14}}{3 \times 10^{8}}$

$\frac{2}{3} \times 10^{6} = 0.6 \times 10^{6}$

$B = B_{0} \overset{Λ}{k} \sin 2 π (\frac{x}{λ} - f t)$

$= 6 \times 10^{- 8} \overset{\land}{k} \sin 2 π (0.6 \times 10^{6} - 2 \times 10^{14} t)$

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