A very long straight conducting wire, lying along the z-axis, carries a current of 2A. The integral `ointvecB.dvecl` is computed along the straight line PQ, where P has the coordinates `(2cm,2cm,0)` and Q has the coordinates `(2cm, 2cm ,0)`. The integral has the magnitude (in SI units)`

Options

A. `pi/2xx10^-7`
B. `8pixx10^-7`
C. `2pixx10^-7`
D. `pixx10^-7`

Explanation

To solve the problem of calculating the integral \( \oint \vec{B} \cdot d\vec{l} \) along the straight line PQ, we can follow these steps:

### Step 1: Understand the Setup
We have a long straight conducting wire along the z-axis carrying a current \( I = 2 \, \text{A} \). The points P and Q are both located at the coordinates \( (2 \, \text{cm}, 2 \, \text{cm}, 0) \), which means they are the same point. Therefore, the line segment PQ has no length.

**Hint:** Recognize that if P and Q are the same point, the integral over a closed path will be zero.

### Step 2: Calculate the Magnetic Field \( \vec{B} \)
The magnetic field \( \vec{B} \) around a long straight wire carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( r \) is the distance from the wire to the point where we are calculating the magnetic field, and \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

**Hint:** Identify the distance \( r \) from the wire to the points P and Q.

### Step 3: Determine the Distance \( r \)
The coordinates of points P and Q are both \( (2 \, \text{cm}, 2 \, \text{cm}, 0) \). The distance from the wire (which lies along the z-axis) to the point (2 cm, 2 cm) can be calculated using the Pythagorean theorem:
\[ r = \sqrt{(2 \, \text{cm})^2 + (2 \, \text{cm})^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \, \text{cm} = 0.02\sqrt{2} \, \text{m} \]

**Hint:** Convert the distance from centimeters to meters for consistency in SI units.

### Step 4: Substitute into the Magnetic Field Formula
Now substituting \( I = 2 \, \text{A} \) and \( r = 0.02\sqrt{2} \, \text{m} \) into the magnetic field formula:
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 2}{2\pi \cdot (0.02\sqrt{2})} \]
Simplifying this gives:
\[ B = \frac{4 \times 10^{-7} \cdot 2}{0.04\sqrt{2}} = \frac{8 \times 10^{-7}}{0.04\sqrt{2}} = \frac{8 \times 10^{-7}}{0.04 \cdot 1.414} \approx \frac{8 \times 10^{-7}}{0.05656} \approx 1.415 \times 10^{-5} \, \text{T} \]

**Hint:** Ensure you simplify the fractions correctly.

### Step 5: Calculate \( \oint \vec{B} \cdot d\vec{l} \)
Since the integral is computed along a path where P and Q are the same point, the length of the path \( d\vec{l} \) is zero. Therefore:
\[ \oint \vec{B} \cdot d\vec{l} = 0 \]

**Hint:** Remember that the integral of a vector field over a path of zero length is zero.

### Final Answer
The magnitude of the integral \( \oint \vec{B} \cdot d\vec{l} \) is:
\[ \boxed{0} \]