4 The electric field of a traveling electromagnetic wave in vacuum is given by $$ \vec{E}=10(3 \hat{x}+4 \hat{z}) \sin \left(k y+2 \pi imes 10^{9} t ight) $$ statvolt $$ / \mathrm{cm} $$ ). 4.1 All given and requested numerical quantities are in CGS units. Describe fully the wave (giving the direction of its propagation, its frequency $$ f $$, wavelength $$ \lambda $$, the amplitudes of the electric and magnetic fields $$ E_{o} $$ and $$ B_{o} $$ respectively?

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4 The electric field of a traveling electromagnetic wave in vacuum is given by $$ \vec{E}=10(3 \hat{x}+4 \hat{z}) \sin \left(k y+2 \pi 	imes 10^{9} tight) $$ statvolt $$ / \mathrm{cm} $$ ). 4.1 All given and requested numerical quantities are in CGS units. Describe fully the wave (giving the direction of its propagation, its frequency $$ f $$, wavelength $$ \lambda $$, the amplitudes of the electric and magnetic fields $$ E_{o} $$ and $$ B_{o} $$ respectively?

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To analyze the given electric field of the electromagnetic wave, we start with the expression:

$$
\vec{E} = 10(3 \hat{x} + 4 \hat{z}) \sin \left(k y + 2 \pi \times 10^{9} t\right) \text{ statvolt/cm}
$$

### 1. Direction of Propagation

The argument of the sine function is $$ k y + 2 \pi \times 10^{9} t $$. This indicates that the wave is propagating in the $$ y $$-direction. The general form of a traveling wave is $$ \sin(kx - \omega t) $$, and in this case, we can see that the wave is traveling in the positive $$ y $$-direction since it can be rewritten as $$ \sin(k y - (-2 \pi \times 10^{9}) t) $$.

### 2. Wave Number $$ k $$

The wave number $$ k $$ is related to the wavelength $$ \lambda $$ by the equation:

$$
k = \frac{2\pi}{\lambda}
$$

From the sine function, we can identify that $$ k $$ is the coefficient of $$ y $$. However, we need to find $$ k $$ explicitly. The term $$ 2\pi \times 10^{9} $$ corresponds to the angular frequency $$ \omega $$:

$$
\omega = 2\pi \times 10^{9} \text{ rad/s}
$$

### 3. Frequency $$ f $$

The frequency $$ f $$ is related to the angular frequency $$ \omega $$ by:

$$
f = \frac{\omega}{2\pi} = 10^{9} \text{ Hz}
$$

### 4. Wavelength $$ \lambda $$

To find the wavelength $$ \lambda $$, we can use the relationship between $$ \omega $$ and $$ k $$:

$$
\lambda = \frac{v}{f}
$$

In vacuum, the speed of light $$ v $$ is approximately $$ 3 \times 10^{10} $$ cm/s. Thus,

$$
\lambda = \frac{3 \times 10^{10} \text{ cm/s}}{10^{9} \text{ Hz}} = 30 \text{ cm}
$$

### 5. Amplitude of the Electric Field $$ E_0 $$

The amplitude of the electric field $$ E_0 $$ can be determined from the expression for $$ \vec{E} $$:

$$
E_0 = 10 \text{ statvolt/cm}
$$

### 6. Amplitude of the Magnetic Field $$ B_0 $$

The amplitude of the magnetic field $$ B_0 $$ can be found using the relationship between the electric field and the magnetic field in an electromagnetic wave:

$$
B_0 = \frac{E_0}{c}
$$

where $$ c $$ is the speed of light in CGS units, approximately $$ 3 \times 10^{10} $$ cm/s. Thus,

$$
B_0 = \frac{10 \text{ statvolt/cm}}{3 \times 10^{10} \text{ cm/s}} \approx 3.33 \times 10^{-10} \text{ gauss}
$$

### Summary of Results

- **Direction of Propagation**: Positive $$ y $$-direction
- **Frequency $$ f $$**: $$ 10^{9} \text{ Hz} $$
- **Wavelength $$ \lambda $$**: $$ 30 \text{ cm} $$
- **Amplitude of Electric Field $$ E_0 $$**: $$ 10 \text{ statvolt/cm} $$
- **Amplitude of Magnetic Field $$ B_0 $$**: $$ 3.33 \times 10^{-10} \text{ gauss} $$
The wave propagates in the positive $$ y $$-direction with a frequency of $$ 10^{9} \text{ Hz} $$, a wavelength of $$ 30 \text{ cm} $$, an electric field amplitude of $$ 10 \text{ statvolt/cm} $$, and a magnetic field amplitude of $$ 3.33 \times 10^{-10} \text{ gauss} $$.

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