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Question

Suppose that we have 2.10×10π2.10 \times 10^\pi doubly charged positive ions per cubic centimeter, all moving north with a speed of 1.40×105 m/s1.40 \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

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Answered 2 years ago
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a)

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:

j=2envj=21.6021019 C2.11014 1m31.4105 ms\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*}

j9.42 Am2  In the direction of the moving ions.\boxed{j \approx 9.42 \ \frac{\text{A}}{\text{m}^{2}} \ \ \text{In the direction of the moving ions.}}

b)

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

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