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(D) The amplitude of the magnetic field is given by $B_0 = \frac{E_0}{c}$.Given $E_0 = 27 \, Vm^{-1}$ and $c = 3 	imes 10^8 \, ms^{-1}$,we have $B_0

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The amplitude of the magnetic field is given by $B_0 = \frac{E_0}{c}$.Given $E_0 = 27 \, Vm^{-1}$ and $c = 3 	imes 10^8 \, ms^{-1}$,we have $B_0 = \frac{27}{3 	imes 10^8} = 9 	imes 10^{-8} \, T$.Since the wave travels in the $x-$ direction,the magnetic field must oscillate in the $y-z$ plane. Thus,the direction is $\hat{j}$ or $\hat{k}$.The wave number $k$ is given by $k = \frac{\omega}{c} = \frac{2\pi f}{c} = \frac{2\pi 	imes 2 	imes 10^{14}}{3 	imes 10^8} = \frac{4\pi}{3} 	imes 10^6 \approx 4.19 	imes 10^6 \, m^{-1}$.Specifically,$k = \frac{2\pi}{\lambda} = \frac{2\pi f}{c} = 2\pi 	imes \frac{2 	imes 10^{14}}{3 	imes 10^8} = 2\pi 	imes (0.666 	imes 10^6) \approx 2\pi 	imes (1.5 	imes 10^{-6})^{-1}$ is not quite right; let's calculate $k = \frac{2\pi f}{c} = \frac{2\pi 	imes 2 	imes 10^{14}}{3 	imes 10^8} = \frac{4\pi}{3} 	imes 10^6 \approx 4.18 	imes 10^6 \, m^{-1}$.Comparing with the options,the term inside the sine function is $2\pi (\frac{x}{\lambda} - ft)$. Here $\frac{1}{\lambda} = \frac{f}{c} = \frac{2 	imes 10^{14}}{3 	imes 10^8} = 0.666 	imes 10^6 \approx 1.5 	imes 10^{-6} \, m^{-1}$.Thus,the correct form is $\vec{B}(x, t) = (9 	imes 10^{-8} \, T) \hat{k} \sin [2\pi (1.5 	imes 10^{-6} \, x - 2 	imes 10^{14} \, t)]$.

An electromagnetic wave travelling in the $x-$ direction has a frequency of $2 \times 10^{14} \, Hz$ and an 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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