Question

Earth’s lower atmosphere contains negative and positive ions that are produced by radioactive elements in the soil and cosmic rays from space. In a certain region, the atmospheric electric field strength is 120V/mand the field is directed vertically down. This field causes singly charged positive ions, at a density of 620cm^-3 , to drift downward and singly charged negative ions, at a density of 550cm^-3 , to drift upward (Figure). The measured conductivity of the air in that region is$\mathbf{2}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}{\left(\Omega ·m\right)}^{\mathbf{-}\mathbf{1}}$. Calculate (a) the magnitude of the current density and (b) the ion drift speed, assumed to be the same for positive and negative ions.

Short answer

1. The magnitude of the current density is $3.24\times {10}^{-12}\mathrm{A}/{\mathrm{m}}^2$
2. The ion-drift speed is 1.73cm/s

Step-by-step solution

Identification of given data

1. Electric field strength,$E=120\mathrm{V}/\mathrm{m}$
2. Density of positive ions,$n_{+}=620{\mathrm{cm}}^{-3}$
3. Density of negative ions,$n_{-}=550{\mathrm{cm}}^{-3}$
4. Conductivity of air, $\sigma =2.70\times {10}^{-14}{\left(\mathrm{\Omega }.\mathrm{m}\right)}^{-1}$

Significance of current density and drift velocity

The current density is defined as the number of charges flowing through the conductor per unit time per unit area. The drift velocity is the average velocity achieved by the electrons.

Using the formulae for current density and drift velocity, we can find their required values.

Formulae:

The current density of the material, $J=\sigma E$ …(i)

Here, $J$is the current density,$\sigma$ is conductivity, and $E$is the electric field.

The ion-drift velocity of the conducting electrons, $v=\frac{J}{ne}$ …(ii)

Here, $v$is the drift velocity, $J$is the current density, $n$is number of electrons and e is electric charge on the electron.

(a) Determining the current density

The magnitude of the current density can be calculated using the given data in equation (i) as follows:

$$\begin{gathered} J=\left(2.70\times {10}^{-14}{\left(\mathrm{\Omega }·\mathrm{m}\right)}^{-1}\right)\left(120\mathrm{V}/\mathrm{m}\right) \\ =3.24\times {10}^{-12}\mathrm{A}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the current density is $3.24\times {10}^{-12}\mathrm{A}/{\mathrm{m}}^2$.

(b) Determining the ion-drift velocity

The value of ion-drift speed can be calculated using the given data in equation (ii) as follows:

$$\begin{gathered} v=\frac{\mathrm{J}}{\left[\left({\mathrm{n}}_{+}+{\mathrm{n}}_{-}\right)\mathrm{e}\right]} \\ =\frac{3.24\times {10}^{-12}\mathrm{A}/{\mathrm{m}}^2}{\left[\left(620+550\right){\mathrm{cm}}^{-3}\left(1.6\times {10}^{-19}\mathrm{C}\right)\right]} \\ =1.73\mathrm{cm}/\mathrm{s} \end{gathered}$$

Hence, the value of the speed is $1.73\mathrm{cm}/\mathrm{s}$.