Problem 6 - University Physics with Modern Physics

Question

An electromagnetic wave of wavelength \(435 \mathrm{~nm}\) is traveling in vacuum in the \(-z\) -direction. The electric field has amplitude \(2.70 \times 10^{-3} \mathrm{~V} / \mathrm{m}\) and is parallel to the \(x\) -axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for \(\overrightarrow{\boldsymbol{E}}(z, t)\) and \(\overrightarrow{\boldsymbol{B}}(z, t)\).

Step-by-Step Solution

Answer

(a) Frequency is approximately \(6.90 \times 10^{14} \text{Hz}\). (b) Magnetic-field amplitude is \(9.00 \times 10^{-12} \text{T}\). (c) Vector equations: \(\overrightarrow{E}(z, t) = 2.70 \times 10^{-3} \cos(1.45\times10^7 z + 4.33\times10^{15} t) \hat{i}\), \(\overrightarrow{B}(z, t) = 9.00 \times 10^{-12} \cos(1.45\times10^7 z + 4.33\times10^{15} t) \hat{j}\).

Step 1: Convert Wavelength to Frequency

The frequency of an electromagnetic wave, \( f \), is related to its wavelength, \( \lambda \), by the speed of light in vacuum. The relationship is given by:\[f = \frac{c}{\lambda}\]where \( c = 3 \times 10^8 \ \text{m/s} \). Given \( \lambda = 435 \ \text{nm} \), first convert this to meters: \( \lambda = 435 \times 10^{-9} \ \text{m} \).Now plug into the formula:\[f = \frac{3 \times 10^8}{435 \times 10^{-9}} \approx 6.90 \times 10^{14} \ \text{Hz}\]

Step 2: Find Magnetic-Field Amplitude

The amplitude of the magnetic field, \( B_0 \), can be found using the relation between the electric field amplitude, \( E_0 \), and \( B_0 \):\[E_0 = c \cdot B_0\]Rearrange the formula to solve for \( B_0 \):\[B_0 = \frac{E_0}{c} = \frac{2.70 \times 10^{-3}}{3 \times 10^8} = 9.00 \times 10^{-12} \ \text{T}\]

Step 3: Write Electric Field Equation

For a wave traveling in the \(-z\) direction with the electric field in the \(x\)-axis, we write:\[\overrightarrow{E}(z, t) = E_0 \cos(kz + \omega t) \hat{i}\]where \( E_0 = 2.70 \times 10^{-3} \ \text{V/m} \).Since \( k = \frac{2\pi}{\lambda} \approx 1.45 \times 10^7 \ \text{m}^{-1} \), and the angular frequency \( \omega = 2\pi f \approx 4.33 \times 10^{15} \ \text{rad/s} \):\[\overrightarrow{E}(z, t) = (2.70 \times 10^{-3}) \cos(1.45 \times 10^7 z + 4.33 \times 10^{15} t) \hat{i}\]

Step 4: Write Magnetic Field Equation

For the magnetic field of a wave traveling in the \(-z\) direction, perpendicular to both \( \overrightarrow{E} \) and the direction of travel, we write:\[\overrightarrow{B}(z, t) = B_0 \cos(kz + \omega t) \hat{j}\]where \( B_0 = 9.00 \times 10^{-12} \ \text{T} \).Substituting the values from Step 2 and Step 3:\[\overrightarrow{B}(z, t) = (9.00 \times 10^{-12}) \cos(1.45 \times 10^7 z + 4.33 \times 10^{15} t) \hat{j}\]

Key Concepts

Wavelength to Frequency Conversion

Electromagnetic waves move at the speed of light in a vacuum. The speed of light, denoted as \( c \), is a well-known constant value of \( 3 \times 10^8 \ ext{m/s} \). This speed is crucial when you're converting between wavelength and frequency, two essential properties of waves. The formula connecting wavelength \( \lambda \) and frequency \( f \) is:

• \( f = \frac{c}{\lambda} \)

In this equation, if you know the wavelength of the wave, you can easily find its frequency by dividing the speed of light by the wavelength. Remember, the units are important here. Wavelength often comes in nanometers (nm) for these equations, so be sure to convert to meters (m) by multiplying by \( 10^{-9} \).
In our example, with a wavelength of 435 nm, the conversion goes as follows:

• \( \lambda = 435 \times 10^{-9} \ ext{m} \)
• \( f = \frac{3 \times 10^8}{435 \times 10^{-9}} \approx 6.90 \times 10^{14} \ ext{Hz} \)

This relationship helps us understand that as the wavelength decreases, the frequency increases—a fundamental concept in wave physics.

Electric Field and Magnetic Field Relationship

Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation. The relationship between the electric field \( E \) and the magnetic field \( B \) is direct and can be given by the formula:

• \( E = c \cdot B \)

Here, \( c \) is again the speed of light. The electric field's amplitude \( E_0 \) can be used to find the magnetic field's amplitude \( B_0 \) using:

• \( B_0 = \frac{E_0}{c} \)

In practice, if you know the amplitude of the electric field, finding the magnetic field's amplitude is straightforward. For our scenario, with an electric field amplitude of \( 2.70 \times 10^{-3} \ ext{V/m} \):

• \( B_0 = \frac{2.70 \times 10^{-3}}{3 \times 10^8} = 9.00 \times 10^{-12} \ ext{T} \)

This illustrates how intimately tied these two fields are in electromagnetic waves. They sustain each other as the wave propagates through space.

Wave Equation Derivation

Understanding wave equations is essential for describing electromagnetic wave behavior. The vector wave equations for the electric field \( \overrightarrow{E} \) and the magnetic field \( \overrightarrow{B} \) involve amplitude, wave number \( k \), and angular frequency \( \omega \).For an electric field propagating in the \(-z\)-direction and oscillating along the \(x\)-axis, the equation is:

• \( \overrightarrow{E}(z, t) = E_0 \cos(kz + \omega t) \hat{i} \)

Here:

• \( E_0 = 2.70 \times 10^{-3} \ ext{V/m} \)
• \( k = \frac{2\pi}{\lambda} \approx 1.45 \times 10^7 \ ext{m}^{-1} \)
• \( \omega = 2\pi f \approx 4.33 \times 10^{15} \ ext{rad/s} \)

The corresponding magnetic field, perpendicular and propagating in the same direction, can be expressed as:

• \( \overrightarrow{B}(z, t) = B_0 \cos(kz + \omega t) \hat{j} \)
• \( B_0 = 9.00 \times 10^{-12} \ ext{T} \)

Wave equations show the spatial and temporal dependence of the fields, depicting how they vary over time and space. These relationships govern how electromagnetic waves travel through a vacuum, maintaining the interconnected nature of their fields.