Problem 8 - University Physics with Modern Physics

Question

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the -\(z\)-direction. The electric field has amplitude 2.70 \(\times\) 10\(^{-3}\) V/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 \(\vec{E} (z, t)\) and \(\vec{B} (z, t)\).

Step-by-Step Solution

Answer

(a) Frequency: 6.90 × 10^14 Hz, (b) Magnetic field amplitude: 9.00 × 10^-12 T, (c) \( \vec{E}(z, t) = 2.70 \times 10^{-3} \cos(1.44 \times 10^7 z - 4.34 \times 10^{15} t) \hat{i} \); \( \vec{B}(z, t) = 9.00 \times 10^{-12} \cos(1.44 \times 10^7 z - 4.34 \times 10^{15} t) \hat{j} \).

Step 1: Convert Wavelength to Frequency

We start by calculating the frequency of the wave using the formula \( f = \frac{c}{\lambda} \), where \( c = 3 \times 10^8 \) m/s is the speed of light and \( \lambda = 435 \times 10^{-9} \) m is the wavelength.Calculating:\[ f = \frac{3 \times 10^8 \text{ m/s}}{435 \times 10^{-9} \text{ m}} = 6.90 \times 10^{14} \text{ Hz} \]

Step 2: Calculate Magnetic Field Amplitude

The amplitude of the magnetic field \( B_{m} \) can be found using the relation between the electric field amplitude \( E_{m} \) and the magnetic field amplitude: \( B_{m} = \frac{E_{m}}{c} \).Given \( E_m = 2.70 \times 10^{-3} \text{ V/m} \) and \( c = 3 \times 10^8 \text{ m/s} \):\[ B_{m} = \frac{2.70 \times 10^{-3} \text{ V/m}}{3 \times 10^8 \text{ m/s}} = 9.00 \times 10^{-12} \text{ T} \]

Step 3: Write the Electric Field Equation

The electric field \( \vec{E}(z, t) \) for a wave traveling in the -\( z \)-direction and polarized along the \( x \)-axis is given by:\[ \vec{E}(z, t) = E_{m} \cos(kz - \omega t) \hat{i} \]where \( k = \frac{2\pi}{\lambda} \) and \( \omega = 2\pi f \).Calculating\( k = \frac{2\pi}{435 \times 10^{-9}} \text{ m} = 1.44 \times 10^7 \text{ m}^{-1} \)and\( \omega = 2\pi \times 6.90 \times 10^{14} = 4.34 \times 10^{15} \text{ rad/s} \).Thus, the equation becomes:\[ \vec{E}(z, t) = 2.70 \times 10^{-3} \cos(1.44 \times 10^7 z - 4.34 \times 10^{15} t) \hat{i} \]

Step 4: Write the Magnetic Field Equation

The magnetic field \( \vec{B}(z, t) \) is perpendicular to both \( \vec{E}(z, t) \) and the direction of propagation. Therefore, it is along the \( y \)-axis:\[ \vec{B}(z, t) = B_{m} \cos(kz - \omega t) \hat{j} \]Using the previously calculated values, the equation becomes:\[ \vec{B}(z, t) = 9.00 \times 10^{-12} \cos(1.44 \times 10^7 z - 4.34 \times 10^{15} t) \hat{j} \]

Key Concepts

Wavelength

The wavelength of an electromagnetic wave is the distance between two consecutive peaks (or troughs) of the wave. It is usually denoted as \( \lambda \). For the provided exercise, the wavelength is specified as 435 nm (nanometers).
To convert this to meters, multiply by \( 10^{-9} \), giving \( \lambda = 435 \times 10^{-9} \) meters. This is crucial because all calculations involving the speed of light, which is in meters per second (m/s), require the wavelength to be in meters.
The relationship that comes into play with wavelength is how it dictates the energy and behavior of the electromagnetic wave. Shorter wavelengths correspond to higher energy and frequency, making them crucial in applications like visible light and X-rays.

Electric Field

The electric field of an electromagnetic wave is a vector field that represents the force that would be experienced by a positive charge placed in the field. In the exercise, the electric field amplitude is given as \( 2.70 \times 10^{-3} \text{ V/m} \).
The electric field oscillates perpendicular to the direction of wave propagation. Here, it is aligned along the \( x \)-axis. This means the wave propagates in the -\( z \)-direction while the electric field oscillates in the \( x \)-direction.
Understanding the electric field allows us to predict how the wave will interact with charged particles and matter, influencing technologies like antennas and capacitors in electronics.

Magnetic Field Amplitude

The magnetic field amplitude is a crucial component of electromagnetic waves but often less intuitive than the electric field. For the given wave, the relationship between electric field amplitude \( E_m \) and magnetic field amplitude \( B_m \) is given by:

• \( B_m = \frac{E_m}{c} \)

where \( c \) is the speed of light, \( 3 \times 10^8 \text{ m/s} \).
Substituting the provided values, \( B_m \) calculates to \( 9.00 \times 10^{-12} \text{ T} \) (teslas).
Like the electric field, the magnetic field is perpendicular to both the electric field and the direction of wave propagation. Here, it is oriented along the \( y \)-axis, perpendicular to both the \( x \)-aligned electric field and the -\( z \)-propagation direction, forming a right-hand coordinate system with the electric field and propagation direction.

Frequency

Frequency is the number of wave cycles that pass a point in one second and is a key descriptor of electromagnetic waves. In the given problem, frequency \( f \) can be found using:

• \( f = \frac{c}{\lambda} \)

This employs the speed of light \( c = 3 \times 10^8 \text{ m/s} \) and the wavelength to calculate frequency.
The resulting frequency \( 6.90 \times 10^{14} \text{ Hz} \) (hertz) corresponds to many cycles per second. Frequencies like these fall in the visible spectrum, crucial to understanding phenomena like color perception in light.
This also lays the foundation for technologies like radio, where varying frequencies correspond to different channels and types of communication.