A long conducting wire of radius 1 cm carrying current of 1 A placed with its axis coinciding with X-axis. Find the magnetic field at point having coordinates `(0.5 cm, 0.5 cm )` .

Answer

To find the magnetic field at the point (0.5 cm, 0.5 cm) due to a long conducting wire carrying a current of 1 A, we can use Ampere's Circuital Law. Let's go through the solution step by step.

### Step 1: Understand the Geometry
The wire is a long straight conductor with its axis along the x-axis. The radius of the wire is 1 cm. The point where we need to find the magnetic field has coordinates (0.5 cm, 0.5 cm), which means it is located in the xy-plane.

**Hint:** Visualize the wire and the point in a coordinate system to understand their positions relative to each other.

### Step 2: Determine the Distance from the Wire
The point (0.5 cm, 0.5 cm) is at a distance from the wire. The distance \( r \) from the center of the wire to the point can be calculated using the Pythagorean theorem:
\[ r = \sqrt{(0.5)^2 + (0.5)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} \text{ cm} = 0.707 \text{ cm} \]

**Hint:** Use the Pythagorean theorem to find the distance from the wire to the point.

### Step 3: Apply Ampere's Circuital Law
According to Ampere's Circuital Law:
\[ \oint \mathbf{B} \cdot d\mathbf{L} = \mu_0 I_{\text{enclosed}} \]
For a circular path around the wire, the magnetic field \( B \) is constant along the path, so we can simplify the left side:
\[ B \cdot (2\pi r) = \mu_0 I_{\text{enclosed}} \]

**Hint:** Remember that \( I_{\text{enclosed}} \) is the current enclosed by the path. For a point outside the wire, the enclosed current is the total current flowing through the wire.

### Step 4: Calculate the Enclosed Current
Since the point is outside the wire, the enclosed current \( I_{\text{enclosed}} \) is simply the total current flowing through the wire, which is given as 1 A.

**Hint:** Identify that the current flowing through the wire is the same as the enclosed current when the point is outside the wire.

### Step 5: Substitute Values into the Equation
Now we can substitute the values into the equation:
\[ B \cdot (2\pi \cdot 0.707 \times 10^{-2}) = \mu_0 \cdot 1 \]
Where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

**Hint:** Make sure to convert all units to SI units for consistency.

### Step 6: Solve for B
Rearranging the equation gives:
\[ B = \frac{\mu_0 \cdot 1}{2\pi \cdot 0.707 \times 10^{-2}} \]
Substituting the value of \( \mu_0 \):
\[ B = \frac{4\pi \times 10^{-7}}{2\pi \cdot 0.707 \times 10^{-2}} = \frac{2 \times 10^{-7}}{0.707 \times 10^{-2}} = \frac{2 \times 10^{-5}}{0.707} \approx 2.83 \times 10^{-5} \, \text{T} \]

**Hint:** When simplifying, ensure to keep track of the powers of ten and the constants.

### Final Answer
The magnetic field at the point (0.5 cm, 0.5 cm) is approximately \( 2.83 \times 10^{-5} \, \text{T} \).