An electromagnetic wave is propagating in vac | vedclass

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Question

(B) We know that for an electromagnetic wave,the relation between the amplitudes of electric and magnetic fields is $E_0 = B_0 c$.Given $B_0 = 2 	ime

Vedclass · Accepted Answer

$\vec{E} = (6) \cos [\pi 	imes 10^{15}(t + \frac{y}{c})] \hat{i} 	ext{ V m}^{-1}$

Vedclass · Answer

(B) We know that for an electromagnetic wave,the relation between the amplitudes of electric and magnetic fields is $E_0 = B_0 c$.Given $B_0 = 2 	imes 10^{-8} 	ext{ T}$ and $c = 3 	imes 10^8 	ext{ m/s}$.$E_0 = (2 	imes 10^{-8}) 	imes (3 	imes 10^8) = 6 	ext{ V m}^{-1}$.The direction of propagation of the wave is given by the direction of the vector $\vec{E} 	imes \vec{B}$.Here,the wave propagates along $-\hat{j}$ and the magnetic field is along $\hat{k}$.Let the direction of $\vec{E}$ be $\hat{n}$. Then $\hat{n} 	imes \hat{k} = -\hat{j}$.Since $\hat{i} 	imes \hat{k} = -\hat{j}$,the electric field must be along the $\hat{i}$ direction.The phase of the wave remains the same,so the argument of the cosine function is $\pi 	imes 10^{15}(t + \frac{y}{c})$.Thus,$\vec{E} = (6) \cos [\pi 	imes 10^{15}(t + \frac{y}{c})] \hat{i} 	ext{ V m}^{-1}$.

An electromagnetic wave is propagating in vacuum along $-\hat{j}$ direction. The magnetic field of the wave is given by $\vec{B} = (2 \times 10^{-8}) \cos [\pi \times 10^{15}(t + \frac{y}{c})] \hat{k} \text{ T}$. The electric field $\vec{E}$ of this wave is $(c = \text{speed of light})$

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