There exist magnetic field `vec(B)=yhat(i)-xhat(j)` in space. What is current (in amperes) crossing through the area enclosed by a circle `x^(2)+y^(2)=a^(2),z=0`. Here radius of circle `a=1`mm

Explanation

To solve the problem, we need to find the current crossing through the area enclosed by the circle defined by \( x^2 + y^2 = a^2 \) in the xy-plane, where \( a = 1 \) mm. The magnetic field is given as \( \vec{B} = y \hat{i} - x \hat{j} \).

### Step-by-Step Solution:

1. **Identify the Circle and Magnetic Field**:
- The circle is in the xy-plane with the equation \( x^2 + y^2 = a^2 \) where \( a = 1 \) mm or \( 1 \times 10^{-3} \) m.
- The magnetic field is given as \( \vec{B} = y \hat{i} - x \hat{j} \).

2. **Calculate the Magnitude of the Magnetic Field**:
- The magnitude of the magnetic field \( B \) can be calculated as:
\[ B = \sqrt{y^2 + (-x)^2} = \sqrt{y^2 + x^2} \]
- On the boundary of the circle, since \( x^2 + y^2 = a^2 \), we have:
\[ B = \sqrt{a^2} = a \]

3. **Apply Ampère's Law**:
- According to Ampère's law, the line integral of the magnetic field around a closed loop is equal to \( \mu_0 \) times the current enclosed:
\[ \oint \vec{B} \cdot d\vec{L} = \mu_0 I_{\text{enclosed}} \]
- For a circular path of radius \( a \), the line integral becomes:
\[ \oint \vec{B} \cdot d\vec{L} = B \cdot (2\pi a) \]
- Substituting \( B = a \):
\[ \oint \vec{B} \cdot d\vec{L} = a \cdot (2\pi a) = 2\pi a^2 \]

4. **Set Up the Equation**:
- Now we equate the two expressions:
\[ 2\pi a^2 = \mu_0 I_{\text{enclosed}} \]

5. **Substitute Values**:
- We know \( a = 1 \times 10^{-3} \) m and \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A.
- Substitute these into the equation:
\[ 2\pi (1 \times 10^{-3})^2 = (4\pi \times 10^{-7}) I_{\text{enclosed}} \]

6. **Solve for \( I_{\text{enclosed}} \)**:
- Simplifying the left side:
\[ 2\pi \times 10^{-6} = (4\pi \times 10^{-7}) I_{\text{enclosed}} \]
- Dividing both sides by \( 4\pi \times 10^{-7} \):
\[ I_{\text{enclosed}} = \frac{2\pi \times 10^{-6}}{4\pi \times 10^{-7}} = \frac{2 \times 10^{-6}}{4 \times 10^{-7}} = 5 \text{ A} \]

### Final Answer:
The current crossing through the area enclosed by the circle is \( I_{\text{enclosed}} = 5 \) A.