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,0,0)` and Q has the coordinates `(2cm, 2cm ,0)`. The integral has the magnitude (in SI units)`

Options

A. zero
B. `8pixx10^(-7)`
C. `2pixx10^(-7)`
D. `pixx10^(-7)`

Answer

To solve the problem, we need to compute the integral \( \oint \vec{B} \cdot d\vec{l} \) along the straight line from point P to point Q, where the wire carrying a current of 2A lies along the z-axis.

### Step 1: Understanding the Magnetic Field Due to a Long Straight Wire
The magnetic field \( \vec{B} \) due to a long straight wire carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( I \) is the current (2A in this case),
- \( r \) is the perpendicular distance from the wire to the point where the magnetic field is being calculated.

### Step 2: Determine the Path of Integration
The points P and Q are given as:
- \( P(2 \, \text{cm}, 0, 0) \)
- \( Q(2 \, \text{cm}, 2 \, \text{cm}, 0) \)

The path from P to Q is along the y-axis at a constant x-coordinate of 2 cm. The distance from the wire (located on the z-axis) to the path is:
\[ r = 2 \, \text{cm} = 0.02 \, \text{m} \]

### Step 3: Calculate the Magnetic Field at Points P and Q
Using the formula for the magnetic field:
\[ B = \frac{\mu_0 \cdot 2}{2 \pi \cdot 0.02} \]
Substituting \( \mu_0 = 4\pi \times 10^{-7} \):
\[ B = \frac{(4\pi \times 10^{-7}) \cdot 2}{2 \pi \cdot 0.02} = \frac{4 \times 10^{-7} \cdot 2}{0.04} = \frac{8 \times 10^{-7}}{0.04} = 2 \times 10^{-5} \, \text{T} \]

### Step 4: Set Up the Integral
The integral \( \oint \vec{B} \cdot d\vec{l} \) can be evaluated along the straight line from P to Q. Since the magnetic field is perpendicular to the path of integration (which is along the y-axis), the angle between \( \vec{B} \) and \( d\vec{l} \) is 90 degrees. Thus:
\[ \vec{B} \cdot d\vec{l} = B \cdot dl \cdot \cos(90^\circ) = 0 \]

### Step 5: Conclusion
Since the magnetic field is perpendicular to the path of integration, the integral evaluates to zero:
\[ \oint \vec{B} \cdot d\vec{l} = 0 \]

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