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 will use Ampere's Circuital Law, which states that the line integral of the magnetic field \( \vec{B} \) around a closed loop is equal to the permeability of free space \( \mu_0 \) times the total current \( I \) enclosed by that loop.

### Step-by-Step Solution:

1. **Identify the Magnetic Field**:
The magnetic field is given as:
\[ \vec{B} = y \hat{i} - x \hat{j} \]

2. **Determine the Area of Interest**:
We need to find the current crossing through the area enclosed by the circle defined by:
\[ x^2 + y^2 = a^2, \quad z = 0 \]
where the radius \( a = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \).

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

4. **Apply Ampere's Circuital Law**:
According to Ampere's Circuital Law:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \]
For a circular path of radius \( a \):
\[ \oint \vec{B} \cdot d\vec{l} = |\vec{B}| \cdot (2\pi a) \]
Substitute \( |\vec{B}| = a \):
\[ a \cdot (2\pi a) = \mu_0 I_{\text{enc}} \]

5. **Substitute Known Values**:
Substituting \( a = 1 \times 10^{-3} \text{ m} \) and \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \):
\[ (1 \times 10^{-3}) \cdot (2\pi \cdot 1 \times 10^{-3}) = (4\pi \times 10^{-7}) I_{\text{enc}} \]

6. **Simplify the Equation**:
The left side becomes:
\[ 2\pi \times 10^{-6} = 4\pi \times 10^{-7} I_{\text{enc}} \]
Dividing both sides by \( 4\pi \times 10^{-7} \):
\[ I_{\text{enc}} = \frac{2\pi \times 10^{-6}}{4\pi \times 10^{-7}} = \frac{2 \times 10^{-6}}{4 \times 10^{-7}} = 5 \text{ A} \]

### Final Answer:
The current crossing through the area enclosed by the circle is:
\[ I_{\text{enc}} = 5 \text{ A} \]