The magnetic field of a plane electromagnetic | vedclass

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Question

(D) Given the magnetic field $\overrightarrow{B} = B_0 \cos(kx + \omega t) \hat{j}$,where $B_0 = 3 	imes 10^{-8} 	ext{ T}$.The relationship between

Vedclass · Accepted Answer

$9 \cos (1.6 	imes 10^3 x + 48 	imes 10^{10} t) \hat{k} 	ext{ V/m}$

Vedclass · Answer

(D) Given the magnetic field $\overrightarrow{B} = B_0 \cos(kx + \omega t) \hat{j}$,where $B_0 = 3 	imes 10^{-8} 	ext{ T}$.The relationship between the amplitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ is $E_0 = c B_0$,where $c = 3 	imes 10^8 	ext{ m/s}$.$E_0 = (3 	imes 10^8 	ext{ m/s}) 	imes (3 	imes 10^{-8} 	ext{ T}) = 9 	ext{ V/m}$.The wave propagates in the negative $x$-direction (since the argument is $kx + \omega t$).The direction of propagation is given by the direction of $\overrightarrow{E} 	imes \overrightarrow{B}$.Since $\overrightarrow{B}$ is along $\hat{j}$ and the wave travels along $-\hat{i}$,we have $\hat{k} 	imes \hat{j} = -\hat{i}$.Thus,the electric field must be along the $\hat{k}$ direction.Therefore,$\overrightarrow{E} = 9 \cos (1.6 	imes 10^3 x + 48 	imes 10^{10} t) \hat{k} 	ext{ V/m}$.

The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{B} = 3 \times 10^{-8} \cos (1.6 \times 10^3 x + 48 \times 10^{10} t) \hat{j} \text{ T}$. The associated electric field is:

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