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

<p>To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field $\overrightarrow{\mathrm{E}}$ and the magnetic field $\overrightarrow{\mathrm{B}}$ in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following relation holds:</p> <p>$$\overrightarrow{\mathrm{B}} = \frac{\overrightarrow{\mathrm{E}} \times \hat{\mathrm{k}}}{\mathrm{c}}$$</p> <p>where:</p> <ul> <li>$\overrightarrow{\mathrm{E}}$ is the electric field.</li> <li>$\hat{\mathrm{k}}$ is the unit vector in the direction of propagation of the wave.</li> <li>$\mathrm{c}$ is the speed of light in a vacuum.</li> </ul> <p>Given the electric field:</p> <p>$$\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$$</p> <p>The wave is propagating in the $z$-direction, so $\hat{\mathrm{k}} = \hat{z}$. The unit vector $\hat{\mathrm{i}}$ represents the $x$-direction.</p> <p>The magnetic field induction is given by:</p> <p>$$\overrightarrow{\mathrm{B}} = \frac{(\hat{\mathrm{i}} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})) \times \hat{\mathrm{z}}}{\mathrm{c}}$$</p> <p>The cross product $\hat{\mathrm{i}} \times \hat{\mathrm{z}}$ yields $\hat{\mathrm{j}}$ (the unit vector in the $y$-direction):</p> <p>$$\overrightarrow{\mathrm{B}} = \hat{\mathrm{j}} \frac{40 \cos \omega(\mathrm{t}-z / \mathrm{c})}{\mathrm{c}}$$</p> <p>Therefore, the magnetic field induction is:</p> <p>$$\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z/\mathrm{c})$$</p> <p>The correct answer is:</p> <p>Option A:</p> <p>$$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$$</p>