A long straight wire of radius R carries a steady current which is uniformly distributed through the cross-section of the wire. The magnitudes of the magnetic field at distance `r_(1)` and `r_(2)` from the centre of the wire are `B_(1)` and `B_(2)` respectively. Then `

Options

A. `(B_(1))/(B_(2)) = (r_(1))/(r_(2)), if r_(1) lt R` and `r_(2) lt R`
B. `(B_(1))/(B_(2)) = (r_(1))/(r_(2)), if r_(1) gt R` and `r_(2) gt R`
C. `(B_(1))/(B_(2)) = (r_(2))/(r_(1)), if r_(1) lt R` and `r_(2) lt R`
D. none of these

Answer

To solve the problem, we will use Ampere's Circuital Law to find the magnetic field inside and outside a long straight wire carrying a steady current.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a long straight wire of radius \( R \) carrying a steady current \( I \) that is uniformly distributed across its cross-section. We need to find the magnetic fields \( B_1 \) and \( B_2 \) at distances \( r_1 \) and \( r_2 \) from the center of the wire, respectively.

2. **Applying Ampere's Circuital Law**:
Ampere's Circuital Law states that:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}} \]
where \( I_{\text{enclosed}} \) is the current enclosed by the loop.

3. **Case 1: When \( r_1 < R \)** (inside the wire):
- For a circular path of radius \( r_1 \) inside the wire, the enclosed current \( I_{\text{enclosed}} \) can be calculated using the current density \( J \):
\[ J = \frac{I}{\pi R^2} \]
- The area of the circle of radius \( r_1 \) is \( \pi r_1^2 \), so:
\[ I_{\text{enclosed}} = J \cdot \text{Area} = \frac{I}{\pi R^2} \cdot \pi r_1^2 = I \frac{r_1^2}{R^2} \]
- Now, applying Ampere's Law:
\[ B_1 \cdot 2\pi r_1 = \mu_0 I_{\text{enclosed}} = \mu_0 \left( I \frac{r_1^2}{R^2} \right) \]
- Therefore, we can solve for \( B_1 \):
\[ B_1 = \frac{\mu_0 I r_1}{2\pi R^2} \]

4. **Case 2: When \( r_2 > R \)** (outside the wire):
- For a circular path of radius \( r_2 \) outside the wire, the entire current \( I \) is enclosed:
\[ B_2 \cdot 2\pi r_2 = \mu_0 I \]
- Solving for \( B_2 \):
\[ B_2 = \frac{\mu_0 I}{2\pi r_2} \]

5. **Finding the Ratio \( \frac{B_1}{B_2} \)**:
- We can find the ratio of the magnetic fields \( B_1 \) and \( B_2 \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I r_1}{2\pi R^2}}{\frac{\mu_0 I}{2\pi r_2}} = \frac{r_1}{R^2} \cdot r_2 \]

6. **Conclusion**:
- For \( r_1 < R \) and \( r_2 > R \), the relationship between \( B_1 \) and \( B_2 \) can be summarized as:
\[ B_1 \propto r_1 \quad \text{and} \quad B_2 \propto \frac{1}{r_2} \]
- Thus, if \( r_1 \) and \( r_2 \) are both less than \( R \), then:
\[ \frac{B_1}{B_2} = \frac{r_1}{r_2} \]
- If both \( r_1 \) and \( r_2 \) are greater than \( R \), then:
\[ \frac{B_1}{B_2} = \frac{r_2}{r_1} \]