A current of 2 amp. flows in a long, straight wire of radius `2 mm`. The intensity of magnetic field on the axis of the wire is

Options

A. `((mu_(0))/(pi))xx10^(3)` Tesla
B. `((mu_(0))/(2pi))xx10^(3)` Tesla
C. `((2mu_(0))/(pi))xx10^(3)` Tesla
D. Zero

Answer

To find the intensity of the magnetic field on the axis of a long straight wire carrying a current, we can use the Biot-Savart law. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a long straight wire with a radius of 2 mm carrying a current of 2 A. We need to find the magnetic field intensity at a point on the axis of the wire.

2. **Biot-Savart Law**:
The Biot-Savart law states that the magnetic field \( B \) at a point due to a small segment of current-carrying wire is given by:
\[ B = \frac{\mu_0}{4\pi} \int \frac{I \, dl \, \sin \theta}{r^2} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current,
- \( dl \) is the length of the current element,
- \( \theta \) is the angle between the current element and the line connecting the current element to the point where the field is being calculated,
- \( r \) is the distance from the current element to the point.

3. **Analyzing the Geometry**:
For a point on the axis of the wire, the angle \( \theta \) between the current element \( dl \) and the line connecting \( dl \) to the point on the axis is 0 degrees.

4. **Calculating \( \sin \theta \)**:
Since \( \theta = 0^\circ \), we have:
\[ \sin(0) = 0 \]

5. **Substituting into the Biot-Savart Law**:
Substituting \( \sin \theta = 0 \) into the Biot-Savart law gives:
\[ B = \frac{\mu_0}{4\pi} \int \frac{I \, dl \, \cdot 0}{r^2} = 0 \]

6. **Conclusion**:
Therefore, the intensity of the magnetic field on the axis of the wire is:
\[ B = 0 \, \text{T} \]

### Final Answer:
The intensity of the magnetic field on the axis of the wire is **0 T**.

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