A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is

Options

A. inversely proportional to r in the region `r lt a` and uniform throughout in the region `r gt a`
B. directly proportional to r in the region `r lt a` and inversely proportional to r in the region` r gt a`
C. zero in the region `r lt a` and inversely proportional to r in the region `r gt a`
D. uniform in the region ``r lt a and inversely proportional to distance r from the axis, in the region `r gt a`

Answer

To solve the problem of finding the magnetic field around a long straight wire of circular cross-section carrying a steady current \( I \), we will use Ampère's Law and the concept of magnetic fields in cylindrical symmetry.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a long straight wire with a circular cross-section of radius \( a \).
- The current \( I \) is uniformly distributed across this cross-section.

2. **Magnetic Field Inside the Wire (r < a)**:
- For points inside the wire (where the distance \( r \) from the center of the wire is less than the radius \( a \)), we can use Ampère's Law.
- The magnetic field \( B \) inside the wire can be given by the formula:
\[ B = \frac{\mu_0 I_{\text{enc}}}{2 \pi r} \]
- Here, \( I_{\text{enc}} \) is the current enclosed by the Amperian loop of radius \( r \). Since the current is uniformly distributed, we can find \( I_{\text{enc}} \) as:
\[ I_{\text{enc}} = I \cdot \frac{r^2}{a^2} \]
- Substituting this into the equation for \( B \):
\[ B = \frac{\mu_0 \left(I \cdot \frac{r^2}{a^2}\right)}{2 \pi r} = \frac{\mu_0 I r}{2 \pi a^2} \]
- This shows that the magnetic field \( B \) inside the wire is directly proportional to \( r \).

3. **Magnetic Field Outside the Wire (r > a)**:
- For points outside the wire (where the distance \( r \) is greater than the radius \( a \)), the entire current \( I \) flows through any Amperian loop we consider.
- The magnetic field \( B \) outside the wire can be expressed as:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
- This indicates that the magnetic field \( B \) outside the wire is inversely proportional to \( r \).

4. **Conclusion**:
- Thus, we summarize:
- For \( r < a \): \( B \propto r \) (directly proportional)
- For \( r > a \): \( B \propto \frac{1}{r} \) (inversely proportional)

### Final Answer:
- The magnetic field is directly proportional to \( r \) when \( r < a \) and inversely proportional to \( r \) when \( r > a \).