Problem 11 - University Physics with Modern Physics

Question

An electromagnetic wave has an electric field given by \(\vec{E} (y, t)\) = (3.10 \(\times\) 10\(^5\) V/m) \(\hat{k}\) cos [ky - (12.65 \(\times\) 10\(^{12}\) rad/s)t]. (a) In which direction is the wave traveling? (b) What is the wavelength of the wave? (c) Write the vector equation for \(\vec{B} (y, t)\).

Step-by-Step Solution

Answer

The wave travels in the positive y-direction, with a wavelength of approximately 1.49\( \times \)10\( ^{-4} \) m. The magnetic field equation is \( \vec{B}(y,t) = (1.033 \times 10^{-3} \, \text{T}) \, \hat{i} \cos (ky - 12.65 \times 10^{12} \, t) \).

Step 1: Determine the Direction of Travel

The direction of the wave is determined by the term inside the cosine function. The electric field is given by \( \vec{E} = (3.10 \times 10^5 \, \text{V/m}) \, \hat{k} \cos (ky - (12.65 \times 10^{12} \, \text{rad/s})t) \). Since the expression is in the form \( ky - \omega t \), the wave travels in the positive \( y \)-direction.

Step 2: Calculate the Wavelength

The term \( \omega \) in the equation \( ky - \omega t \) is the angular frequency, and it is given as \( 12.65 \times 10^{12} \, \text{rad/s} \). The wave number \( k \) is related to the wavelength \( \lambda \) by the equation \( k = \frac{2\pi}{\lambda} \). Solve for \( \lambda \) using the equation \( \lambda = \frac{2\pi c}{\omega} \), where \( c \), the speed of light, is \( 3 \times 10^8 \, \text{m/s} \). Thus, \( \lambda = \frac{2\pi \cdot 3 \times 10^8}{12.65 \times 10^{12}} \approx 1.49 \times 10^{-4} \, \text{m} \).

Step 3: Determine the Magnetic Field Equation

The magnetic field \( \vec{B} \) is related to the electric field \( \vec{E} \) by the equation \( \vec{B} = \frac{\vec{E}}{c} \). Therefore, \( \vec{B}(y,t) = \frac{3.10 \times 10^5}{3 \times 10^8} \, \hat{i} \cos (ky - (12.65 \times 10^{12} \, \text{rad/s})t) \). Simplifying gives \( \vec{B}(y,t) = (1.033 \times 10^{-3} \, \text{T}) \, \hat{i} \cos (ky - 12.65 \times 10^{12} \, t) \).

Key Concepts

Electric Field Equation

Electromagnetic waves consist of a combination of electric and magnetic fields that propagate through space. The electric field in an electromagnetic wave is described by a specific equation. In this problem, the electric field equation is \[\vec{E}(y, t) = (3.10 \times 10^5 \text{ V/m}) \hat{k} \cos [ky - (12.65 \times 10^{12} \text{ rad/s})t]\]which tells us several things about the wave:

• The amplitude of the electric field is \(3.10 \times 10^5\) V/m, meaning this is the maximum strength of the field.
• The direction of the electric field is given by \(\hat{k}\), which corresponds to the direction of the unit vector along the z-axis.
• The part of the equation \(\cos [ky - (12.65 \times 10^{12} \text{ rad/s})t]\) represents how the field changes over time and space. It's a cosine wave, indicating oscillatory behavior.

Understanding the electric field equation is crucial for predicting how an electromagnetic wave interacts with materials it encounters.

Magnetic Field Equation

The magnetic field of an electromagnetic wave is just as important as its electric counterpart. The two fields are perpendicular and oscillate together. The magnetic field equation derived from the electric field is \[\vec{B}(y,t) = (1.033 \times 10^{-3} \text{ T}) \hat{i} \cos (ky - 12.65 \times 10^{12} \text{ rad/s} t)\]Here's what each part of the magnetic field equation signifies:

• The term \(1.033 \times 10^{-3}\) T is the amplitude of the magnetic field. The much smaller size compared to the electric field is because magnetic fields in electromagnetic waves are inherently weaker.
• The direction of the magnetic field is given by \(\hat{i}\), indicating it is aligned along the x-axis, perpendicular to the electric field described by \(\hat{k}\).
• The cosine function matches the phase and frequency with the electric field, maintaining their synchronized oscillations.

Recognizing these relationships and equations helps to understand the structure and behavior of electromagnetic waves as they travel through space.

Wave Direction

An electromagnetic wave's direction of travel can be determined from its field equations. In this scenario, the key term is found inside the cosine expression, \(ky - \omega t\), where \(k\) is the wave number and \(\omega\) is the angular frequency. This standard form implies that the wave travels in the positive direction of the axis that is not part of the dot product.
For the given electric field \[\vec{E}(y, t) = (3.10 \times 10^5 \text{ V/m}) \hat{k} \cos [ky - (12.65 \times 10^{12} \text{ rad/s})t]\]the wave motion happens along the y-axis. The minus sign in front of \(\omega t\) indicates the wave progresses in the positive y-direction.

• The harmonic expression \(ky - \omega t\) indicates that as time \(t\) increases, the phase of the wave continues to advance if it's moving in the positive y-direction.

Understanding direction is important in physics to know where energy is going and how it will interact with physical boundaries or different media.

Wavelength Calculation

The wavelength of an electromagnetic wave is related to its wave number \(k\) and angular frequency \(\omega\). First, relation between wave number and wavelength is given by\[k = \frac{2\pi}{\lambda}\]where \(\lambda\) is the wavelength. Rewriting, we find\[\lambda = \frac{2\pi}{k}\]Since the wave equation in this exercise offers \(\omega = 12.65 \times 10^{12}\) rad/s, we relate it to the wave's speed is by using the speed of light \(c = 3 \times 10^8 \text{ m/s}\) and understand\[\lambda = \frac{2\pi c}{\omega}\]Substituting the values, we have\[\lambda = \frac{2\pi \cdot 3 \times 10^8}{12.65 \times 10^{12}} \approx 1.49 \times 10^{-4} \text{ m}\]This calculation directly shows how the wavelength is derived from known constants and the angular frequency. Wavelengths tell us a lot about wave properties, such as energy and potential interactions with objects of similar sizes.