Question

up in the atmosphere of Earth, there are He²+ ions moving northwards and O₂‐ lons moving Southwards. The concentration and speed for He²+ lons are 2.8 × 10¹²m.s‐³and 2.0×10⁶m.s‐1 , while O₂- ions they are 7,0x10¹¹m.s‐³ and 7.2×10⁶m.s‐¹, Therefore, what will the direction and magnitude of the net current through unit area be ? Up moving Speed​

Answer 1

Given that,

Concentration of $\mathrm{He}^{2+}$ ions $c=2.8 \times 10^{12} \mathrm{~m}^{3}$

Speed $v=2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}$

Concentration of $\mathrm{O}^{-2}$ions $c=7.0 \times 10^{11} \mathrm{~m}^{3}$

Speed $v=7.2 \times 10^{6} \mathrm{~m} / \mathrm{s}$

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

$Q=2 \times 2.8 \times 10^{12} \times 1.6 \times 10^{-19}$

$Q=8.96 \times 10^{-7} C$

We need to calculate the current density f$or H e^{2+}$

Using the formula of current density

$J=\frac{\text { charge }}{\text { volume }} \times v$

Put the value into the formula

$J=8.96 \times 10^{-7} \times 2.0 \times 10^{6}$

$J=1.792 \mathrm{~A} / \mathrm{m}^{2}$

Here, The charge of$\mathrm{He}^{2+}$ ions is positive so the direction of current is in north.

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

Using the formula of current density

$J=\frac{\text { charge }}{\text { volume }} \times v$

Put the value into the formula

$J=7.0 \times 10^{11} \times 1.6 \times 10^{-19} \times 7.2 \times 10^{6}$

$J=0.8064 \mathrm{~A} / \mathrm{m}^{2}$

Here, The charge of $\mathrm{O}^{-2}$ ions is also the direction of current is in north because the current flows in the opposite direction than the direction of flow of negative value

The magnitude of the net current passing through unit area is

$I=1.792+0.8064$

$I=2.5984 \mathrm{~A} / \mathrm{m}^{2}$

Hence, The magnitude and direction of the net current passing through unit area is$2.5984 \mathrm{~A} / \mathrm{m}^{2} i$ n the north direction.

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