The magnetic field of an electromagnetic wave | vedclass

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Question

(A) The magnitude of the electric field is given by $E_0 = c B_0$,where $c = 3 	imes 10^8 	ext{ m/s}$.Given $B_0 = 1.6 	imes 10^{-6} 	imes \sqrt{2

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$\vec E = 4.8 	imes 10^2 \cos(2 	imes 10^7 z + 6 	imes 10^{15} t) (-\hat i + 2\hat j) 	ext{ V/m}$

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(A) The magnitude of the electric field is given by $E_0 = c B_0$,where $c = 3 	imes 10^8 	ext{ m/s}$.Given $B_0 = 1.6 	imes 10^{-6} 	imes \sqrt{2^2 + 1^2} = 1.6 	imes 10^{-6} 	imes \sqrt{5}$.Thus,$E_0 = 3 	imes 10^8 	imes 1.6 	imes 10^{-6} 	imes \sqrt{5} = 4.8 	imes 10^2 \sqrt{5}$.The wave propagates in the direction of $-\hat k$ (since the phase is $kz + \omega t$).The direction of propagation is given by the direction of $\vec E 	imes \vec B$.Since $\vec E \cdot \vec B = 0$,the vector $\vec E$ must be perpendicular to $(2\hat i + \hat j)$.Testing option $(A)$: $\vec E \propto (-\hat i + 2\hat j)$.Dot product: $(-\hat i + 2\hat j) \cdot (2\hat i + \hat j) = -2 + 2 = 0$. This satisfies the perpendicular condition.Cross product: $(-\hat i + 2\hat j) 	imes (2\hat i + \hat j) = -\hat k - 4\hat k = -5\hat k$. This matches the direction of propagation $-\hat k$.

The magnetic field of an electromagnetic wave is given by $\vec B = 1.6 \times 10^{-6} \cos(2 \times 10^7 z + 6 \times 10^{15} t) (2\hat i + \hat j) \text{ Wb/m}^2$. The associated electric field will be

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