Question 52: EM waves

Question

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) :

Accepted Answer

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:
$$\overrightarrow{\mathrm{B}} = \frac{\overrightarrow{\mathrm{E}} 	imes \hat{\mathrm{k}}}{\mathrm{c}}$$
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:
$$\overrightarrow{\mathrm{B}} = \frac{(\hat{\mathrm{i}} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})) 	imes \hat{\mathrm{z}}}{\mathrm{c}}$$
The cross product $$\hat{\mathrm{i}} 	imes \hat{\mathrm{z}}$$ 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:
$$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$$

Answer

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

Answer

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

Answer

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

Answer

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

EM waves

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) :

Choose the correct answer: