A beam contains `2 xx 10^(8)` doubly charged positive ions per cubic centimeter, all of which are moving with a speed of `10^(5) m//s` . The current density is

Options

A. `6.4A//m`
B. `3.2 A//m`
C. `1.6 A//m`
D. None of these

Answer

To find the current density \( J \) of the beam containing doubly charged positive ions, we can follow these steps:

### Step 1: Identify the parameters
- Number density of ions \( n = 2 \times 10^8 \) ions/cm³
- Speed of ions \( v_d = 10^5 \) m/s
- Charge of each doubly charged ion \( q = 2e \), where \( e = 1.6 \times 10^{-19} \) C

### Step 2: Convert the number density to SI units
The number density in SI units (m³) can be converted from cm³:
\[ n = 2 \times 10^8 \, \text{ions/cm}^3 = 2 \times 10^8 \, \text{ions} \times (10^{-2} \, \text{m})^3 = 2 \times 10^8 \times 10^{-6} \, \text{ions/m}^3 = 2 \times 10^2 \, \text{ions/m}^3 = 200 \, \text{ions/m}^3 \]

### Step 3: Calculate the total charge per unit volume
Since the ions are doubly charged, the charge \( q \) of each ion is:
\[ q = 2e = 2 \times (1.6 \times 10^{-19}) \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]

### Step 4: Use the formula for current density
The current density \( J \) can be calculated using the formula:
\[ J = n \cdot q \cdot v_d \]
Substituting the values:
\[ J = (2 \times 10^8 \, \text{ions/m}^3) \cdot (3.2 \times 10^{-19} \, \text{C}) \cdot (10^5 \, \text{m/s}) \]

### Step 5: Perform the calculation
Calculating \( J \):
\[ J = 2 \times 10^8 \cdot 3.2 \times 10^{-19} \cdot 10^5 \]
\[ = 2 \times 3.2 \times 10^{8 - 19 + 5} = 6.4 \times 10^{-6} \, \text{A/m}^2 \]

### Step 6: Convert to appropriate units
Since \( 1 \, \text{A/m}^2 = 1 \, \text{Ampere per meter square} \), we can express the current density as:
\[ J = 6.4 \, \text{A/m}^2 \]

### Final Answer
The current density \( J \) is \( 6.4 \, \text{A/m}^2 \).

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