The figure shows a long striaght wire of a circular cross-section (radius a) carrying steady current I.The current I is unifromly distributed across this distance `a//2` and 2a from axis is

Options

A. `2 :1`
B. `1 :2`
C. `4: 1`
D. `1 :1`

Answer

To solve the problem, we need to find the ratio of the magnetic fields at two different distances from the axis of a long straight wire carrying a steady current. The two distances we will consider are \( \frac{a}{2} \) and \( 2a \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The wire has a circular cross-section with radius \( a \).
- The current \( I \) is uniformly distributed across the wire.
- We need to find the magnetic field at distances \( \frac{a}{2} \) and \( 2a \) from the axis of the wire.

2. **Apply Ampere's Circuital Law**:
- According to Ampere's Circuital Law, the line integral of the magnetic field \( B \) around a closed loop is equal to \( \mu_0 \) times the total current \( I \) enclosed by the loop:
\[ \oint B \cdot dl = \mu_0 I_{\text{enc}} \]

3. **Calculate the Magnetic Field at \( r = \frac{a}{2} \)**:
- For a circular loop of radius \( r = \frac{a}{2} \):
\[ B \cdot 2\pi \left(\frac{a}{2}\right) = \mu_0 I_{\text{enc}} \]
- The current enclosed \( I_{\text{enc}} \) can be calculated based on the area ratio:
\[ I_{\text{enc}} = I \cdot \frac{\text{Area of circle with radius } \frac{a}{2}}{\text{Area of circle with radius } a} = I \cdot \frac{\pi \left(\frac{a}{2}\right)^2}{\pi a^2} = I \cdot \frac{1}{4} \]
- Substituting this into the equation:
\[ B \cdot 2\pi \left(\frac{a}{2}\right) = \mu_0 \cdot \frac{I}{4} \]
- Simplifying gives:
\[ B \cdot \pi a = \mu_0 \cdot \frac{I}{4} \implies B = \frac{\mu_0 I}{4\pi a} \]

4. **Calculate the Magnetic Field at \( r = 2a \)**:
- For a circular loop of radius \( r = 2a \):
\[ B' \cdot 2\pi (2a) = \mu_0 I \]
- Here, the entire current \( I \) is enclosed:
\[ B' \cdot 4\pi a = \mu_0 I \implies B' = \frac{\mu_0 I}{4\pi a} \]

5. **Find the Ratio of the Magnetic Fields**:
- Now, we can find the ratio of the magnetic fields at the two distances:
\[ \frac{B}{B'} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1 \]

### Conclusion:
The ratio of the magnetic fields at distances \( \frac{a}{2} \) and \( 2a \) from the axis of the wire is \( 1:1 \).