Question

The most intense beam of light that can propagate through dry air must have an electric field whose maximum amplitude is no greater than the breakdown value for air: \(E_{\max }^{\text {air }}=3.0 \cdot 10^{6} \mathrm{~V} / \mathrm{m}\), assuming that this value is unaffected by the frequency of the wave. a) Calculate the maximum amplitude the magnetic field of this wave can have. b) Calculate the intensity of this wave. c) What happens to a wave more intense than this?

Short answer

Answer: The maximum magnetic field amplitude of the given wave is \(1.0\cdot10^{-2}\,T\). The intensity of this wave is calculated as \(\frac{1}{2}\mu_0 c B_{\max}^2 = 4.5\cdot 10^2\,\mathrm{W/m^2}\). When a wave has a higher intensity than this limit, the air molecules experience a dielectric breakdown, resulting in a discharge or spark which prevents the wave from propagating further.

Step-by-step solution

a) Calculate the maximum amplitude of the magnetic field of this wave.

The amplitude of the electric field and the magnetic field in an electromagnetic wave are related by the equation: \(E_{\max} = B_{\max}\cdot c\) where \(E_{\max}\) is the maximum electric field amplitude, \(B_{\max}\) is the maximum magnetic field amplitude and \(c\) is the speed of light in vacuum (approximately \(3.0\cdot10^8 \,\mathrm{m/s}\)). We are given the value of \(E_{\max}^{\text{air}} = 3.0\cdot10^6\,\mathrm{V/m}\). Solving for \(B_{\max}\), we get: \(B_{\max} = \frac{E_{\max}^{\text{air}}}{c} = \frac{3.0\cdot10^6\,\mathrm{V/m}}{3.0\cdot10^8\,\mathrm{m/s}}\)

b) Calculate the intensity of this wave.

The intensity of an electromagnetic wave is given by: \(I = \frac{1}{2} \mu_0 c B_{\max}^2\) where \(\mu_0\) is the permeability of free space, with a value approximately equal to \(4\pi \cdot 10^{-7}\,\mathrm{T\cdot m/A}\). Using the value of \(B_{\max}\) calculated in the previous step, we can now calculate the intensity \(I\): \(I = \frac{1}{2} \left(4\pi\cdot 10^{-7}\,\mathrm{T\cdot m/A}\right) \left(3.0\cdot 10^8\,\mathrm{m/s}\right) B_{\max}^2\)

c) What happens to a wave more intense than this?

A wave more intense than the given limit has a higher electric field amplitude than the breakdown value for air. When the electric field amplitude exceeds this value, the electric field becomes so strong that the air molecules experience a dielectric breakdown. This results in a discharge or a spark, preventing the propagation of the wave.

Key concepts

Electric Field Intensity

Electric field intensity, often denoted as \(E\), represents the strength of an electric field at a particular point. It is defined as the force experienced by a unit positive charge placed at that point. When discussing electromagnetic wave propagation, the electric field intensity refers to the amplitude of the electric field component of the wave.

The intensity of an electromagnetic wave's electric field is crucial as it determines the energy transferred by the wave through space. In the context of our specific exercise, the electric field's maximum intensity that air can withstand without breaking down is \( 3.0 \cdot 10^{6} \mathrm{\, V/m} \). This threshold is known as the breakdown electric field intensity for air, beyond which the air can become ionized, leading to dielectric breakdown, a critical concept we'll explore later.

Magnetic Field Amplitude

The magnetic field component of an electromagnetic wave, denoted as \(B\), has an amplitude that correlates with the electric field intensity. This relationship is governed by Maxwell's equations, which in simple terms tell us that both electric and magnetic fields support and propagate each other in an electromagnetic wave.

In a theoretical vacuum, the maximum magnetic field amplitude can be calculated using the formula \(E_{max} = B_{max} \cdot c\), where \(c\) is the speed of light. For our dry air scenario, we use this relationship to determine the maximum value for the magnetic field's amplitude that would correspond to the given electric field intensity, ensuring that the electromagnetic wave does not exceed the breakdown threshold for air.

Electromagnetic Wave Intensity

Intensity, in the context of electromagnetic waves, refers to the power transferred per unit area and is represented typically by the symbol \( I \). For waves traveling in free space, the intensity can be computed using the formula \(I = \frac{1}{2} \mu_0 c B_{max}^2\), which showcases how it relies on the magnetic field amplitude squared. It measures how much energy is propagated in the form of the wave and is a key parameter when determining how much energy an electromagnetic wave can deliver.

Higher intensities mean more energy can be transmitted, but as we found in the problem, there's a limit to this intensity for propagation through materials like air to avoid dielectric breakdown.

Dielectric Breakdown

Dielectric breakdown is a phenomenon that occurs when the electric field within a material becomes strong enough to cause a portion of the material to become electrically conductive. This typically results in a discharge of electricity, such as lightning in the case of air or a spark in electrical insulators.

In our example, we consider what would happen to an electromagnetic wave that exceeds the maximum electric field intensity air can withstand (\(3.0 \cdot 10^{6} \mathrm{\, V/m}\)). When the threshold is crossed, the air molecules can no longer remain insulating, causing them to ionize and thus leading to dielectric breakdown. This creates a conductive path, and the electromagnetic wave's energy is dissipated as a discharge rather than propagating further.