A long, straight wire carries a current `i`. The magnetizing field intensity `H` is measured at a point `P` close to the wire. A long, cylindrical iron rod is brought close to the wire so that the point `P` is at the centre of the rod. The value of `H` at `P` will

Options

A. increase many times
B. decrease many times
C. remain almost constant
D. become zero

Answer

To solve the problem, we need to analyze how the introduction of a cylindrical iron rod near a long straight wire carrying a current affects the magnetic field intensity \( H \) at a point \( P \) located at the center of the rod.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Intensity \( H \)**:
- The magnetic field intensity \( H \) around a long straight wire carrying a current \( i \) is given by the formula:
\[ H = \frac{i}{2\pi r} \]
where \( r \) is the distance from the wire to the point where \( H \) is being measured.

2. **Position of Point \( P \)**:
- In this case, point \( P \) is located close to the wire, and we consider it to be at the center of the cylindrical iron rod that is being introduced.

3. **Effect of the Iron Rod**:
- The iron rod is a ferromagnetic material, which means it can become magnetized in the presence of a magnetic field. However, the magnetic field intensity \( H \) is defined independently of the medium.
- The magnetic field intensity \( H \) is a property of the current \( i \) flowing through the wire and does not change with the introduction of the iron rod.

4. **Conclusion**:
- Since \( H \) is independent of the medium surrounding the wire, the value of \( H \) at point \( P \) will remain constant even after the iron rod is brought close to the wire.
- Therefore, the correct answer is that the value of \( H \) at point \( P \) will **remain almost constant**.

### Final Answer:
The value of \( H \) at point \( P \) will remain almost constant.