The electric field in an electromagnetic wave is given by $\vec{E}=\overset{^}{i}40\cos \omega (t-z/c)N{C}^{-1}$. The magnetic field induction of this wave is (in SI unit) :

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Correct answer: c) $\vec{B}=\overset{^}{j}\frac{40}{c}\cos \omega (t-z/c)$. 1. Direction: The wave propagates along the $+z$-axis (from $t - z/c$). The electric field $\vec{E}$ is along the $+x$-axis ($\hat{i}$). The direction of propagation is given by $\vec{E} \times \vec{B}$. Since $\hat{i} \times \hat{j} = \hat{k}$, the magnetic field $\vec{B}$ must be along the $+y$-axis ($\hat{j}$). 2. Amplitude: The amplitude of the magnetic field is $B_{0}=\frac{E_{0}}{c}$. Given $E_{0}=40$, so $B_{0}=\frac{40}{c}$. Combining these, $\vec{B}=\hat{j}\frac{40}{c}\cos\omega(t-z/c)$.

The electric field in an electromagnetic wave is given by \(\vec{E}=\overset{^}{i}40\cos \omega (t-z/c)N{C}^{-1}\). The …

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