Near Earth's surface the air contains both negative and positive ions, due to radioactivity in the soil and cosmic rays from space. As a simplified model, assume there are 600.0 singly charged positive ions per \(\mathrm{cm}^{3}\) and 500.0 singly charged negative ions per \(\mathrm{cm}^{3} ;\) ignore the presence of multiply charged ions. The electric field is $100.0 \mathrm{V} / \mathrm{m},$ directed downward. (a) In which direction do the positive ions move? The negative ions? (b) What is the direction of the current due to these ions? (c) The measured resistivity of the air in the region is $4.0 \times 10^{13} \Omega \cdot \mathrm{m} .$ Calculate the drift speed of the ions, assuming it to be the same for positive and negative ions. [Hint: Consider a vertical tube of air of length \(L\) and cross-sectional area \(A\) How is the potential difference across the tube related to the electric field strength?] (d) If these conditions existed over the entire surface of the Earth, what is the total current due to the movement of ions in the air?

Positive ions will move in the same direction as the electric field, which is directed downward. Negative ions will move opposite to the direction of the electric field, which means they will move upward.

The direction of electric current is defined as the direction in which positive charges move. Since the positive ions are moving downward, the direction of the current due to these ions is also downward.

To calculate the drift speed, we first need to find the potential difference across a vertical tube of air of length L and cross-sectional area A, given the electric field strength E. The potential difference V can be found using the equation: V = E * L Now, we can use Ohm's Law to relate the current I, potential difference V, and resistance R: I = V / R Since we have resistivity ρ instead of resistance, we use the equation relating resistance to resistivity: R = (ρ * L) / A Substituting this into the previous equation: I = (V * A) / (ρ * L) Now, we can find the drift speed v_d of the ions using the equation for current, charge density, and drift speed: I = n * q * A * v_d Where n is the number of ions per unit volume, q is the charge on each ion, and v_d is the drift speed. Combining this with the previously derived equation and solving for v_d: v_d = (V * n * q) / (ρ * L) Now, plug in the given values for electric field E, resistivity ρ, and ion concentrations: total_n = 600 + 500 = 1100 ions/cm³ = 1.1 x \(10^{22}\) ions/m³ E = 100 V/m ρ = 4.0 x \(10^{13}\) \(\Omega\) m q = 1.6 x \(10^{-19}\) C (charge of positive and negative ions) Now calculating the drift speed: v_d = (E * total_n * q) / ρ v_d = (100 * 1.1 * \(10^{22}\) * 1.6 * \(10^{-19}\)) / (4.0 * \(10^{13}\)) v_d ≈ 4.4 x \(10^{-3}\) m/s The drift speed of the ions, assuming it to be the same for positive and negative ions, is approximately 4.4 x \(10^{-3}\) m/s.

To find the total current over the Earth's surface, we will first find the current over a unit area of the Earth's surface: I_A = total_n * q * v_d I_A = (1.1 * \(10^{22}\) * 1.6 * \(10^{-19}\) * 4.4 * \(10^{-3}\)) A/m² I_A ≈ 9.7 x \(10^{-2} \mathrm{A}/\mathrm{m}²\) Now, we can find the total current by multiplying the current per unit area by the Earth's surface area: I_total = I_A * A_Earth A_Earth = 4 * π * R_Earth² R_Earth ≈ 6.37 x \(10^{6}\) m A_Earth ≈ 5.1 x \(10^{14} \mathrm{m}^2\) Now calculate the total current: I_total = 9.7 x \(10^{-2} \mathrm{A}/\mathrm{m}²\) * 5.1 x \(10^{14} \mathrm{m}^2\) I_total ≈ 4.94 x \(10^{13}\) A If these conditions existed over the entire Earth's surface, the total current due to the movement of ions in the air is approximately 4.94 x \(10^{13}\) A.