## Related questions with answers

Suppose that we have $2.10 \times 10^\pi$ doubly charged positive ions per cubic centimeter, all moving north with a speed of $1.40 \times 10^5 \mathrm{~m} / \mathrm{s}$. (a) Calculate the current density, in magnitude and direction. (b) Can you calculate the total current in this ion beam? If not, what additional information is needed?

Solution

Verifieda)

Equation (29-7) gives us the correlation of current density and charge propagation speed for electrons. We have here positive ions as moving particles, so we have to change the sign of the charge and because one ion caries two elementary charges we have to multiply it by to. After that modification we have:

$\begin{align*} \overrightarrow{j} &= 2e n \overrightarrow{v} \\ j &= 2 \cdot 1.602 \cdot 10^{-19} \ \text{C} \cdot 2.1 \cdot 10^{14} \ \frac{1}{\text{m}^3} \cdot 1.4 \cdot 10^{5} \ \frac{\text{m}}{\text{s}} \end{align*}$

$\boxed{j \approx 9.42 \ \frac{\text{A}}{\text{m}^{2}} \ \ \text{In the direction of the moving ions.}}$

b)

$\boxed{\text{No, we need the know surface cross-section of the conductor.}}$

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