Question

At a point high in the Earth's atmosphere, He2+ ions in a concentration of 2.8 * 1012/m3 are moving due north
at a speed of 2.0 x 106 m/s. Also, a 7.0 * 1011/mº concentration of Oz ions is moving due south at a speed
of 7.2 x 106 m/s. Determine the magnitude and direction of the net current passing through unit area (A/m^2)​

Answer 1

Given that,

Concentration of He²⁺ ions $c= 2.8\times10^{12}\ m^3$

Speed $v=2.0\times10^{6}\ m/s$

Concentration of O⁻² ions $c= 7.0\times10^{11}\ m^3$

Speed $v=7.2\times10^{6}\ m/s$

The total charge passing per unit volume due to$He^{2+}$ is

$Q=2\times2.8\times10^{12}\times1.6\times10^{-19}$

$Q=8.96\times10^{-7}\ C$

We need to calculate the current density for $He^{2+}$

Using formula of current density

$J=\dfrac{charge}{volume}\times v$

Put the value into the formula

$J=8.96\times10^{-7}\times2.0\times10^{6}$

$J=1.792\ A/m^2$

Here, The charge of  He²⁺ ions is positive so the direction of current is in north.

We need to calculate the current density for $O^{-2}$

Using formula of current density

$J=\dfrac{charge}{volume}\times v$

Put the value into the formula

$J=7.0\times10^{11}\times1.6\times10^{-19}\times7.2\times10^{6}$

$J=0.8064\ A/m^2$

Here, The charge of O⁻² ions is also direction of current is in north because the current flows in the opposite direction then the direction of flow of negative value

The magnitude of the net current passing through unit area is

$I=1.792+0.8064$

$I=2.5984\ A/m^2$

Hence, The magnitude and direction of the net current passing through unit area is 2.5984 A/m² in north direction.