Question

The electric field near the Earth's surface has a magnitude of \(150 . \mathrm{N} / \mathrm{C}\) and the magnitude of the Earth's magnetic field near the surface is typically \(50.0 \mu \mathrm{T}\). Calculate and compare the energy densities associated with these two fields. Assume that the electric and magnetic properties of air are essentially those of a vacuum.

Short answer

Question: Calculate the energy densities associated with Earth's electric and magnetic fields near the surface, and compare them. Answer: Using the given magnitudes and constants, the energy densities are calculated as follows: Step 1: Constants - Electric field near Earth's surface, E = 150 N/C - Magnetic field near Earth's surface, B = 50.0 μT = 50.0 × 10^-6 T - Vacuum permittivity, ε₀ = 8.854 × 10^-12 F/m - Vacuum permeability, μ₀ = 4π × 10^-7 T⋅m/A Step 2: Calculate the electric field energy density u_E = (1/2) × (8.854 × 10^-12 F/m) × (150 N/C)^2 ≈ 9.452 × 10^-8 J/m³ Step 3: Calculate the magnetic field energy density u_B = (1/2) × ((50.0 × 10^-6 T)^2) / (4π × 10^-7 T⋅m/A) ≈ 1.989 × 10^-9 J/m³ Step 4: Compare the energy densities To compare the energy densities, find the ratio between them: u_E / u_B = (9.452 × 10^-8 J/m³) / (1.989 × 10^-9 J/m³) ≈ 47.5 The energy density due to Earth's electric field is approximately 47.5 times greater than the energy density due to its magnetic field near the surface.

Step-by-step solution

Constants

Begin by listing the magnitudes of known values, and constants, which are needed to solve the problem. - Electric field near Earth's surface, \(E = 150 \, \text{N/C}\) - Magnetic field near Earth's surface, \(B = 50.0 \, \mu \text{T} = 50.0 \times 10^{-6} \, \text{T}\) - Vacuum permittivity, \(\epsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\) - Vacuum permeability, \(\mu_0 = 4 \pi \times 10^{-7} \, \text{T}\cdot \text{m/A}\)

Calculate the electric field energy density

Using the formula \(u_E = \frac{1}{2} \epsilon_0 E^2\), we can plug in the values for \(\epsilon_0\) and \(E\) to find the energy density due to the electric field: $$ u_E = \frac{1}{2} \cdot (8.854 \times 10^{-12} \, \text{F/m}) \cdot (150 \, \text{N/C})^{2} $$ Calculate the value of \(u_E\).

Calculate the magnetic field energy density

Now, using the formula \(u_B = \frac{1}{2} \frac{B^2}{\mu_0}\), we plug in the values for \(\mu_0\) and \(B\) to calculate the energy density due to the magnetic field: $$ u_B = \frac{1}{2} \cdot \frac{(50.0 \times 10^{-6} \, \text{T})^{2}}{4 \pi \times 10^{-7} \, \text{T}\cdot \text{m/A}} $$ Calculate the value of \(u_B\).

Compare the energy densities

Finally, compare the calculated energy densities associated with Earth's electric and magnetic fields, \(u_E\) and \(u_B\). The comparison can be done by dividing the energy densities: \(\frac{u_E}{u_B}\). Find the ratio between the energy densities and discuss the comparison.

Key concepts

Earth's Electric Field

The Earth's atmosphere possesses an electric field, which is vital for phenomena such as lightning storms and the aurora. This field is generally directed downward, with its strength being measured at approximately 150 Newtons per Coulomb (N/C) near Earth's surface. The magnitude of an electric field describes how much force a charged particle would experience in that field. To understand this better, consider the field as the electric component that contributes to the energy density around us.

The energy density, specifically related to an electric field in a vacuum, is given by the equation \(u_E = \frac{1}{2} \epsilon_0 E^2\), where \(\