91Ó°ÊÓ

1. The average rate of protons per second through the surface is $1500\frac{\text{protons}}{\text{s}â‹\dots {\text{m}}^{\text{2}}}$.
2. Radius of the Earth, $\mathrm{R}=6.37×{10}^6\text{m}$.

Using the concept of current flow through a given area, we can get the net current intercepted by the spherical surface.

Formulae:

The net current flowing through a given area,

$\mathrm{i}=\mathrm{rate}\mathrm{of}\mathrm{protons}\mathrm{per}\mathrm{area}×\mathrm{area}×\mathrm{electronic}\mathrm{charge}\mathrm{of}\mathrm{the}\mathrm{particle}$ (i)

The net charge considering the number of charges present,$\mathrm{q}=\mathrm{ne}$ (ii)

The total surface area of the sphere,$\mathrm{A}=4{\mathrm{Ï€°ù}}^2$ . (iii)

The proton flux is given as 1500 protons per square meter per second, where each proton provides a charge of $\mathrm{q}=+\mathrm{e}$.

Now, we can get the surface area of the Earth using equation (iii) as

$\begin{array}{c}\mathrm{A}=4\mathrm{π}{(6.37×{10}^6\text{m})}^2\\ =5.1×{10}^{14}{\text{m}}^{\text{2}}\end{array}$

Thus, substituting this area and the equation (ii) in equation (i), the net current intercepting the Earth’s surface is

$\begin{array}{c}\mathrm{i}=(5.1×{10}^{14}{\text{m}}^{\text{2}})\left(1500\frac{\text{protons}}{\text{s}â‹\dots {\text{m}}^{\text{2}}}\right)(1.6×{10}^{−19}\text{C/proton})\\ =0.122\text{A}\end{array}$

Hence, the value of the current is$0.122\text{A}$ .