A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r

Options

A. `Bpropr^2`
B. `Bpropr`
C. `Bprop1/(r^2)`
D. `Bprop1/r`

Explanation

To find the variation of the magnetic field \( B \) due to a long straight wire carrying a steady current \( I \) with distance \( r \) from its center (for \( r < R \)), we can use Ampère's law.

### Step-by-Step Solution:

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

2. **Use Ampère's Law**: According to Ampère's law, the magnetic field \( B \) around a long straight current-carrying conductor can be calculated using the formula:
\[ B \cdot 2\pi r = \mu_0 I_{\text{enc}} \]
where \( I_{\text{enc}} \) is the current enclosed by the Amperian loop of radius \( r \).

3. **Calculate the Enclosed Current**: Since the current is uniformly distributed, the current density \( J \) is given by:
\[ J = \frac{I}{\pi R^2} \]
The current enclosed by a circular loop of radius \( r \) (where \( r < R \)) is:
\[ I_{\text{enc}} = J \cdot A = J \cdot \pi r^2 = \frac{I}{\pi R^2} \cdot \pi r^2 = \frac{I r^2}{R^2} \]

4. **Substitute into Ampère's Law**: Now substituting \( I_{\text{enc}} \) back into Ampère's law:
\[ B \cdot 2\pi r = \mu_0 \frac{I r^2}{R^2} \]

5. **Solve for \( B \)**: Rearranging the equation to solve for \( B \):
\[ B = \frac{\mu_0 I r}{2\pi R^2} \]

6. **Conclusion**: The magnetic field \( B \) inside the wire (for \( r < R \)) varies linearly with the distance \( r \) from the center of the wire:
\[ B \propto r \]

### Final Answer:
The variation of the magnetic field \( B \) with distance \( r \) from the center of the wire (for \( r < R \)) is given by:
\[ B = \frac{\mu_0 I r}{2\pi R^2} \]