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 follow these steps:

### Step 1: Understand the given values
We have:
- Area vector \( \vec{A} = (2 \hat{i} + 3 \hat{j}) \, \text{mm}^2 \)
- Current density \( \vec{j} = (3 \hat{j} + 4 \hat{k}) \times 10^3 \, \text{A/m}^2 \)

### Step 2: Convert the area from mm² to m²
To convert the area from millimeters squared to meters squared, we use the conversion factor:
\[ 1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2 \]
Thus,
\[ \vec{A} = (2 \hat{i} + 3 \hat{j}) \, \text{mm}^2 = (2 \hat{i} + 3 \hat{j}) \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the current using the dot product
The current \( I \) through the area can be calculated using the formula:
\[ I = \vec{j} \cdot \vec{A} \]
Substituting the values:
\[ \vec{j} = (3 \hat{j} + 4 \hat{k}) \times 10^3 \, \text{A/m}^2 \]
\[ \vec{A} = (2 \hat{i} + 3 \hat{j}) \times 10^{-6} \, \text{m}^2 \]

Now, we compute the dot product:
\[ I = \left( (3 \hat{j} + 4 \hat{k}) \times 10^3 \right) \cdot \left( (2 \hat{i} + 3 \hat{j}) \times 10^{-6} \right) \]

### Step 4: Evaluate the dot product
The dot product involves multiplying the corresponding components:
- The \( \hat{i} \) component from \( \vec{A} \) does not contribute because there is no \( \hat{i} \) component in \( \vec{j} \).
- The \( \hat{j} \) components contribute: \( 3 \hat{j} \cdot 3 \hat{j} = 9 \)
- The \( \hat{k} \) component does not contribute because there is no \( \hat{k} \) component in \( \vec{A} \).

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

### Step 5: Convert to milliampere
To convert amperes to milliamperes:
\[ I = 9 \, \text{mA} \]

### Final Answer
The current through the area is:
\[ I = 9 \, \text{mA} \]

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