Intensity of EM Waves: Videos & Practice Problems

All waves, including mechanical, sound, and electromagnetic waves, carry energy, which can be measured as intensity (I). Intensity is defined as power (P) per area (A), expressed as I^2 = PA. For electromagnetic waves, intensity relates to electric field (E) and magnetic field (B) strengths through equations like I = 1/2cEmax. Understanding these relationships is crucial for solving problems involving light sources and their intensity at various distances.

Intensity of EM Waves

Waves, whether mechanical or electromagnetic, are known to carry energy. A crucial concept in understanding wave behavior is intensity, which is defined as the energy per time per unit area, or equivalently, power per area. The formula for intensity (I) can be expressed as:

$$ I = \frac{P}{A} $$

where P is power and A is area. For electromagnetic waves, intensity can also be related to the magnitudes of the electric field (E) and magnetic field (B) through specific equations. Two key equations for intensity are:

$$ I = \frac{1}{2} c \epsilon_0 E_{\text{max}}^2 $$

and

$$ I = \frac{1}{2} \frac{c}{\mu_0} B_{\text{max}}^2 $$

In these equations, \(c\) is the speed of light, \(\epsilon_0\) is the permittivity of free space, and \(\mu_0\) is the permeability of free space. Both equations share a common factor of \(\frac{1}{2} c\) and involve the square of the maximum values of the electric and magnetic fields. The units of intensity are watts per square meter (W/m²).

When solving problems involving intensity, it is important to determine whether the source of the wave emits energy uniformly in all directions or directionally. For isotropic sources, such as a light bulb, the area can be calculated using the formula for the surface area of a sphere:

$$ A = 4 \pi r^2 $$

For directional sources, such as lasers, the area must be determined based on the specific geometry of the emission.

For example, consider an incandescent light bulb emitting 50 watts in all directions. To find the intensity at a distance of 5 meters, we first calculate the area:

$$ A = 4 \pi (5)^2 = 100 \pi \, \text{m}^2 $$

Then, using the intensity formula:

$$ I = \frac{50 \, \text{W}}{100 \pi} \approx 0.16 \, \text{W/m}^2 $$

Next, to find the maximum electric field (E_max), we rearrange the intensity equation:

$$ E_{\text{max}} = \sqrt{\frac{2I}{c \epsilon_0}} $$

Substituting the known values, we find:

$$ E_{\text{max}} = \sqrt{\frac{2 \times 0.16}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 10.98 \, \text{N/C} $$

For the root mean square (RMS) values of the electric and magnetic fields, the relationships are:

$$ E_{\text{RMS}} = \frac{E_{\text{max}}}{\sqrt{2}} $$

and

$$ B_{\text{RMS}} = \frac{B_{\text{max}}}{\sqrt{2}} $$

To find B_max, we use the relationship between E_max and B_max:

$$ B_{\text{max}} = \frac{E_{\text{max}}}{c} $$

Substituting the calculated E_max:

$$ B_{\text{max}} = \frac{10.98}{3 \times 10^8} \approx 3.66 \times 10^{-8} \, \text{T} $$

Finally, we can calculate B_RMS:

$$ B_{\text{RMS}} = \frac{3.66 \times 10^{-8}}{\sqrt{2}} \approx 2.59 \times 10^{-8} \, \text{T} $$

Understanding these relationships and calculations is essential for analyzing wave behavior in various contexts, from sound to light and beyond.

Example 1

In this problem, we begin by understanding the average intensity of sunlight at the Earth's surface, which is approximately 1400 watts per square meter (I = 1400 W/m²). The first part of the problem requires us to calculate the amplitudes of the electric field (E_max) and the magnetic field (B_max) associated with this intensity.

To find E_max, we use the relationship between intensity and the electric field, given by the equation:

I = \frac{1}{2} \epsilon_0 c E_{max}^2

Here, ε₀ is the permittivity of free space (8.85 × 10^-12 F/m), and c is the speed of light in a vacuum (approximately 3 × 10⁸ m/s). Rearranging the equation to solve for E_max gives:

E_{max} = \sqrt{\frac{2I}{\epsilon_0 c}}

Substituting the known values:

E_{max} = \sqrt{\frac{2 \times 1400}{8.85 \times 10^{-12} \times 3 \times 10^{8}}}

Calculating this yields E_max ≈ 1027 N/C.

Next, to find B_max, we can use the relationship between the electric and magnetic fields in electromagnetic waves:

E_{max} = c B_{max}

Rearranging gives:

B_{max} = \frac{E_{max}}{c}

Substituting the previously calculated E_max:

B_{max} = \frac{1027}{3 \times 10^{8}}

This results in B_max ≈ 3.42 × 10^-6 T.

In the second part of the problem, we need to determine the average power output of the Sun. The power (P) can be calculated using the intensity and the area over which it is distributed. Since the Sun radiates energy uniformly in all directions, we consider the area of a sphere:

A = 4 \pi r^2

where r is the distance from the Earth to the Sun, approximately 1.5 × 10¹¹ m. The power output can be expressed as:

P = I \times A = I \times 4 \pi r^2

Substituting the known values:

P = 1400 \times 4 \pi (1.5 \times 10^{11})^2

Calculating this gives P ≈ 3.96 × 10²⁶ W, which is also referred to as the luminosity of the Sun.

In summary, we have determined that the maximum electric field amplitude is approximately 1027 N/C, the maximum magnetic field amplitude is about 3.42 × 10^-6 T, and the average power output of the Sun is approximately 3.96 × 10²⁶ watts.

Example 2

In this problem, we analyze a radio tower emitting a signal with an average power output of 50,000 watts, which we denote as \( P = 50,000 \, \text{W} \). The tower radiates signals uniformly in a hemisphere above the Earth's surface, meaning the signals do not penetrate the ground. Our goal is to calculate the intensity of the signal detected by a satellite positioned 100 kilometers above the tower.

To find the intensity \( I \), we use the formula:

\( I = \frac{P}{A} \)

where \( A \) is the area over which the power is distributed. Since the tower emits signals in a hemisphere, the area is given by:

\( A = 2\pi r^2 \)

Here, \( r \) is the distance from the tower to the satellite, which is 100 kilometers or \( 1 \times 10^5 \, \text{m} \). Thus, the area becomes:

\( A = 2\pi (1 \times 10^5)^2 \)

Substituting this into the intensity formula, we calculate:

\( I = \frac{50,000}{2\pi (1 \times 10^5)^2} \approx 7.96 \times 10^{-7} \, \text{W/m}^2 \)

Next, we determine the power received by the satellite's circular radio antenna, which has a radius of 0.5 meters. The area of the antenna \( A_{\text{antenna}} \) is calculated using the formula for the area of a circle:

\( A_{\text{antenna}} = \pi r_{\text{antenna}}^2 = \pi (0.5)^2 \)

Now, we can find the power \( P_{\text{antenna}} \) received by the antenna using the intensity we calculated earlier:

\( P_{\text{antenna}} = I \times A_{\text{antenna}} \)

Substituting the values, we have:

\( P_{\text{antenna}} = (7.96 \times 10^{-7}) \times \left(\pi (0.5)^2\right) \approx 6.25 \times 10^{-7} \, \text{W} \)

In summary, we first calculated the intensity of the signal at the satellite's location, which was \( 7.96 \times 10^{-7} \, \text{W/m}^2 \). Then, we used this intensity to find the power received by the satellite's antenna, resulting in \( 6.25 \times 10^{-7} \, \text{W} \). This process illustrates the relationship between power, intensity, and area in the context of electromagnetic wave propagation.

Here’s what students ask on this topic:

What is the intensity of electromagnetic waves?

Intensity (I) of electromagnetic waves is defined as the power (P) per unit area (A) through which the wave propagates. Mathematically, it is expressed as:

$I=\frac{P}{A}$

For electromagnetic waves, intensity can also be related to the electric field (E) and magnetic field (B) strengths. The equations are:

$I=\frac{1}{2}c\epsilon$₀Emax2

and

$I=\frac{1}{2}\frac{c}{\mu }$₀Bmax2

where $c$ is the speed of light, $\epsilon$₀ is the permittivity of free space, and $\mu$₀ is the permeability of free space.

How do you calculate the intensity of light at a certain distance from a source?

To calculate the intensity of light at a certain distance from a source, you need to know the power (P) of the light source and the area (A) over which the light spreads. For a source emitting light uniformly in all directions, the area is the surface area of a sphere:

$A=4\pi r^2$

where $r$ is the distance from the source. The intensity (I) is then given by:

$I=\frac{P}{A}=\frac{P}{4\pi r^2}$

This formula assumes no energy losses and uniform emission in all directions.

What is the relationship between intensity and electric field in electromagnetic waves?

The intensity (I) of electromagnetic waves is related to the maximum electric field (E_max) by the equation:

$I=\frac{1}{2}c\epsilon$₀Emax2

where $c$ is the speed of light in a vacuum, and $\epsilon$₀ is the permittivity of free space. This equation shows that the intensity is proportional to the square of the maximum electric field.

How do you find the RMS value of the magnetic field in an electromagnetic wave?

The RMS (root mean square) value of the magnetic field (B_RMS) in an electromagnetic wave can be found using the maximum magnetic field (B_max) with the relationship:

$B$_RMS=B_max√2

To find B_max, you can use the relationship between the electric field and magnetic field in an electromagnetic wave:

$E$_max=cB_max

where $c$ is the speed of light. Rearrange to solve for B_max:

$B$_max=E_maxc

Then, use the first equation to find B_RMS.

What are the units of intensity for electromagnetic waves?

The units of intensity (I) for electromagnetic waves are watts per square meter (W/m²). This unit is derived from the definition of intensity as power (P) per unit area (A):

$I=\frac{P}{A}$

Since power is measured in watts (W) and area in square meters (m²), the resulting unit for intensity is W/m².