The magnetic field of a plane electromagnetic | vedclass

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Question

(C) The relationship between the amplitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ is given by $E_0 = cB_0$, where $c$ is the spe

Vedclass · Accepted Answer

$6 	ext{ Vm}^{-1}$ along $z$-axis

Vedclass · Answer

(C) The relationship between the amplitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ is given by $E_0 = cB_0$, where $c$ is the speed of light in vacuum $(c = 3 	imes 10^8 	ext{ ms}^{-1})$.Given $B_0 = 2 	imes 10^{-8} 	ext{ T}$.Substituting the values: $E_0 = (3 	imes 10^8) 	imes (2 	imes 10^{-8}) = 6 	ext{ Vm}^{-1}$.The wave propagates in the negative $x$-direction (as indicated by the $+kx$ term). The magnetic field is along the $y$-axis $(\hat{j})$. Since the direction of propagation is $\vec{E} 	imes \vec{B}$, the electric field must be along the $z$-axis $(\hat{k})$ because $(-\hat{k}) 	imes \hat{j} = -\hat{i}$ (negative $x$-direction). Thus, the electric field is $6 	ext{ Vm}^{-1}$ along the $z$-axis.

The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{ B } = 2 \times 10^{-8} \sin (0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{ j } \text{ T}$. The amplitude of the electric field would be:

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The energy density associated with the electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$ of an electromagnetic wave in free space is given by ($\epsilon_0$ - permittivity of free space,$\mu_0$ - permeability of free space):

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