Question

Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is$\mathbf{8}\mathbf{.}\mathbf{70}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{3}}$ , and their speed is 470 km/s. (a) Find the current density of these protons. (b) If Earth’s magnetic field did not deflect the protons, what total current would Earth receive?

Short answer

a) Current density of protons is$6.54\times {10}^{-7}A/m^2$ .

b) Total current the earth would receive, if the earth’s magnetic field did not deflect the protons is$8.34\times {10}^{7}A$ .

Step-by-step solution

The given data

a) Proton concentration,$n=8.70c{m}^{-3}$

b) Speed of protons,$v=470km/s$

Understanding the concept of the current density

The term "current density" refers to the quantity of electric current moving across a certain cross-section. We have to use the relation between the current density and speed to calculate the current density of the protons. Using the relation between the current and the current density, we can find the total current the earth would receive by substituting the surface area of the earth.

Formulae:

The current density relation to the drift velocity,$J=ne{v}_d$ ...(i)

where, v is the speed of protons and n is the concentration of protons and e is the charge on a proton.

The current density for the current flowing through an area, $J=\frac{i}{A}$ ...(ii)

(a) Calculation of the current density of protons

Using the given data in equation (i), the magnitude of the current density of the protons can be given as follows:

$$\begin{gathered} J=\left(8.70\times {10}^6{m}^{-3}\right)\left(1.6\times {10}^{-19}C\right)\left(470\times {10}^{3}m/s\right) \\ =6.54\times {10}^{-7}A/m^2 \end{gathered}$$

Hence, the value of the current density is$6.54\times {10}^{-7}A/m^2$ .

(b) Calculation of the total current the earth would receive

The total surface area of the Earth is given by,

$A=4{\mathrm{\pi R}}_{\mathrm{E}}^2$

But, for the solar beam, the encounter is with a target of circular area${\mathrm{\pi R}}_{\mathrm{E}}^2$ . Thus, the rate of the charge transport can be given the current received by the Earth by using the given data in equation (ii) as follows: ($R_E=6.37\times {10}^{6}m$ is the radius of the earth.)

$$\begin{gathered} i=J\times {\mathrm{\pi R}}_{\mathrm{E}}^2 \\ =\left(6.54\times {10}^{-7}\mathrm{A}/{\mathrm{m}}^2\right)\times \mathrm{\pi }\times {\left(6.37\times {10}^6\mathrm{m}\right)}^2 \\ =8.336\times {10}^7\mathrm{A} \\ \approx 8.34\times {10}^7\mathrm{A} \end{gathered}$$

Hence, the value of the current received by the Earth is$8.34\times {10}^7\mathrm{A}$ .