Current density `vec(j )` at an area `vec( A) = ( 2 hat(i ) + 3 hat(j) ) m m^(2) ` is `vec(j) = ( 3 hat(j) + 4 hat(k)) xx 10^(3) A //m^(2)` . Current through the area is

Options

A. 9 mA
B. Zero
C. 18 mA
D. 12 mA

Answer

To find the current through the area given the current density and the area vector, we can use the formula:

\[ I = \vec{J} \cdot \vec{A} \]

where:
- \( I \) is the current,
- \( \vec{J} \) is the current density,
- \( \vec{A} \) is the area vector.

### Step 1: Convert the area vector to SI units
The area vector is given as:

\[ \vec{A} = (2 \hat{i} + 3 \hat{j}) \, \text{mm}^2 \]

To convert this to square meters, we use the conversion factor \( 1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2 \):

\[ \vec{A} = (2 \hat{i} + 3 \hat{j}) \times 10^{-6} \, \text{m}^2 \]

### Step 2: Write the current density vector
The current density vector is given as:

\[ \vec{J} = (0 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times 10^{3} \, \text{A/m}^2 \]

### Step 3: Calculate the dot product \( \vec{J} \cdot \vec{A} \)
Now we can calculate the dot product:

\[ I = \vec{J} \cdot \vec{A} = (0 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times 10^{3} \cdot (2 \hat{i} + 3 \hat{j}) \times 10^{-6} \]

Calculating the dot product:

\[ I = (0 \cdot 2) + (3 \cdot 3) + (4 \cdot 0) \times 10^{3} \times 10^{-6} \]

This simplifies to:

\[ I = 0 + 9 + 0 \times 10^{-3} \]

Thus,

\[ I = 9 \times 10^{-3} \, \text{A} \]

### Step 4: Convert to milliamperes
Since \( 1 \, \text{A} = 1000 \, \text{mA} \), we can convert the current to milliamperes:

\[ I = 9 \, \text{mA} \]

### Final Answer
The current through the area is:

\[ I = 9 \, \text{mA} \]