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:

$$\overrightarrow{\mathrm{B}} = \frac{\overrightarrow{\mathrm{E}} \times \hat{\mathrm{k}}}{\mathrm{c}}$$

where:

• $$\overrightarrow{\mathrm{E}}$$ is the electric field.
• $$\hat{\mathrm{k}}$$ is the unit vector in the direction of propagation of the wave.
• $$\mathrm{c}$$ is the speed of light in a vacuum.

Given the electric field:

$$\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$$

The wave is propagating in the $$z$$-direction, so $$\hat{\mathrm{k}} = \hat{z}$$. The unit vector $$\hat{\mathrm{i}}$$ represents the $$x$$-direction.

The magnetic field induction is given by:

$$\overrightarrow{\mathrm{B}} = \frac{(\hat{\mathrm{i}} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})) \times \hat{\mathrm{z}}}{\mathrm{c}}$$

The cross product $$\hat{\mathrm{i}} \times \hat{\mathrm{z}}$$ yields $$\hat{\mathrm{j}}$$ (the unit vector in the $$y$$-direction):

$$\overrightarrow{\mathrm{B}} = \hat{\mathrm{j}} \frac{40 \cos \omega(\mathrm{t}-z / \mathrm{c})}{\mathrm{c}}$$

Therefore, the magnetic field induction is:

$$\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z/\mathrm{c})$$

The correct answer is:

Option A:

$$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$$