Calculating Electric Field Amplitude of Electromagnetic Wave

An electromagnetic wave, or EM wave, is a form of energy that travels through space or through a medium in the form of waves. These waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of propagation.

In a vacuum, electromagnetic waves travel at a constant speed, which is the speed of light, denoted by \(c\). There is a fundamental relationship between the amplitudes of the electric field (\(E_0\)) and the magnetic field (\(B_0\)) of a harmonic electromagnetic wave in a vacuum. This relationship is given by the equation:

\( c = \frac{E_0}{B_0} \)

Where:

• \( E_0 \) is the amplitude of the electric field.
• \( B_0 \) is the amplitude of the magnetic field.
• \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

The question provides the amplitude of the magnetic field (\(B_0\)) and asks for the amplitude of the electric field (\(E_0\)). We can rearrange the formula to solve for \(E_0\):

\( E_0 = c \times B_0 \)

Applying the Formula to Find Electric Field Amplitude

We are given:

• Magnetic field amplitude, \( B_0 = 610 \, \text{nT} \).

First, we need to convert the magnetic field amplitude from nanotesla (nT) to Tesla (T). The conversion is \( 1 \, \text{nT} = 10^{-9} \, \text{T} \).

\( B_0 = 610 \, \text{nT} = 610 \times 10^{-9} \, \text{T} \)

Now we can substitute the values of \(c\) and \(B_0\) into the rearranged formula:

\( E_0 = (3 \times 10^8 \, \text{m/s}) \times (610 \times 10^{-9} \, \text{T}) \)

Let's perform the multiplication:

\( E_0 = 3 \times 610 \times 10^8 \times 10^{-9} \, \frac{\text{m}}{\text{s}} \cdot \text{T} \)

\( E_0 = 1830 \times 10^{8 - 9} \, \frac{\text{m}}{\text{s}} \cdot \text{T} \)

\( E_0 = 1830 \times 10^{-1} \, \frac{\text{m}}{\text{s}} \cdot \text{T} \)

\( E_0 = 183 \, \frac{\text{m}}{\text{s}} \cdot \text{T} \)

The units for electric field amplitude are typically Newtons per Coulomb (N/C) or Volts per meter (V/m). Let's check if the units match:

Recall that \( 1 \, \text{T} = 1 \, \frac{\text{N}}{\text{A} \cdot \text{m}} \) and \( 1 \, \text{A} = 1 \, \frac{\text{C}}{\text{s}} \). So, \( 1 \, \text{T} = 1 \, \frac{\text{N}}{(\text{C/s}) \cdot \text{m}} = 1 \, \frac{\text{N} \cdot \text{s}}{\text{C} \cdot \text{m}} \).

Now, let's look at the combined units from our calculation:

\( \frac{\text{m}}{\text{s}} \cdot \text{T} = \frac{\text{m}}{\text{s}} \cdot \frac{\text{N} \cdot \text{s}}{\text{C} \cdot \text{m}} \)

The 'm' and 's' units cancel out:

\( \frac{\cancel{\text{m}}}{\cancel{\text{s}}} \cdot \frac{\text{N} \cdot \cancel{\text{s}}}{\text{C} \cdot \cancel{\text{m}}} = \frac{\text{N}}{\text{C}} \)

The units match the standard unit for electric field. Therefore, the amplitude of the electric field is \( 183 \, \text{N/C} \).

Comparing this result with the given options, we find that it matches one of them.

Conclusion on Electric Field Amplitude

Based on the calculation using the relationship \( E_0 = c B_0 \) and the given magnetic field amplitude of \( 610 \, \text{nT} \), the amplitude of the electric field is \( 183 \, \text{NC}^{-1} \).

Revision Table: Electromagnetic Wave Properties

Additional Information: Electromagnetic Waves and Fields

Electromagnetic waves are a fascinating phenomenon in physics that unify electricity and magnetism. They are produced by accelerating electric charges. These waves don't require a medium to travel, which is why sunlight can reach us through the vacuum of space.

• Harmonic Waves: A harmonic electromagnetic wave is one where the electric and magnetic fields oscillate sinusoidally with time and position. This is the simplest type of EM wave to analyze and is a good model for understanding more complex light waves.
• Speed of Light \(c\): The speed of light in a vacuum is a fundamental constant of nature. Its value is exactly \( 299,792,458 \, \text{meters per second} \), but \( 3 \times 10^8 \, \text{m/s} \) is a commonly used approximation in calculations. This speed arises directly from the properties of the vacuum, specifically its permittivity (\( \epsilon_0 \)) and permeability (\( \mu_0 \)).
• Units: The electric field is typically measured in Newtons per Coulomb (N/C) or Volts per meter (V/m). The magnetic field is measured in Tesla (T). The relationship \( E_0 = c B_0 \) connects these units via the speed of light (m/s), confirming that \( \text{m/s} \times \text{T} \) is equivalent to N/C or V/m.
• Energy Density: The energy density of an electromagnetic wave is shared equally between the electric field and the magnetic field. The total energy density is proportional to \( E_0^2 \) or \( B_0^2 \).