A magnetic field vector in an electromagnetic | vedclass

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Question

(A) The relationship between the electric field $\vec{E}$ and the magnetic field $\vec{B}$ in an electromagnetic wave is given by $\vec{E} = c(\vec{B}

Vedclass · Accepted Answer

$\vec{E} = -v\lambda B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{k}$

Vedclass · Answer

(A) The relationship between the electric field $\vec{E}$ and the magnetic field $\vec{B}$ in an electromagnetic wave is given by $\vec{E} = c(\vec{B} 	imes \hat{n})$,where $\hat{n}$ is the direction of wave propagation.Here,the wave propagates in the $+x$ direction,so $\hat{n} = \hat{i}$.The magnetic field is given as $\vec{B} = B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{j}$.Using the relation $\vec{E} = c(\vec{B} 	imes \hat{i})$,we substitute $\vec{B}$:$\vec{E} = c B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) (\hat{j} 	imes \hat{i})$.Since $\hat{j} 	imes \hat{i} = -\hat{k}$ and the speed of light $c = v\lambda$ (where $v$ is frequency and $\lambda$ is wavelength),we get:$\vec{E} = -v\lambda B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{k}$.

$A$ magnetic field vector in an electromagnetic wave is represented by $\vec{B} = B_0 \sin(2\pi vt - \frac{2\pi x}{\lambda}) \hat{j}$. Its associated electric field vector is . . . . . . .

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