A long staright wire of radius `a` carries a steady current `I`. The curent is unifromly distributed over its cross-section. The ratio of the magnetic fields `B` and `B'` , at radial distances `(a)/(2)` and `2a` respectively from the axis of the wire is:

Options

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

Answer

To find the ratio of the magnetic fields \( B \) and \( B' \) at radial distances \( \frac{a}{2} \) and \( 2a \) respectively from the axis of a long straight wire carrying a steady current \( I \), we can use Ampere's Law.

### Step-by-Step Solution:

1. **Identify the distances**:
- The first distance is \( r_1 = \frac{a}{2} \).
- The second distance is \( r_2 = 2a \).

2. **Calculate the magnetic field \( B \) at \( r_1 = \frac{a}{2} \)**:
- According to Ampere's Law, the magnetic field \( B \) at a distance \( r \) from a long straight wire is given by:
\[ B = \frac{\mu_0 I_{\text{enc}}}{2 \pi r} \]
- For \( r_1 = \frac{a}{2} \), we need to find the current enclosed \( I_{\text{enc}} \) within a circle of radius \( \frac{a}{2} \).
- The area of the cross-section of the wire is \( \pi a^2 \) and the area of the circle of radius \( \frac{a}{2} \) is \( \pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{4} \).
- Since the current is uniformly distributed, the current enclosed is:
\[ I_{\text{enc}} = I \times \frac{\text{Area of circle}}{\text{Total area}} = I \times \frac{\frac{\pi a^2}{4}}{\pi a^2} = \frac{I}{4} \]
- Now substituting into the formula for \( B \):
\[ B = \frac{\mu_0 \left(\frac{I}{4}\right)}{2 \pi \left(\frac{a}{2}\right)} = \frac{\mu_0 I}{4 \pi} \cdot \frac{2}{a} = \frac{\mu_0 I}{2 \pi a} \]

3. **Calculate the magnetic field \( B' \) at \( r_2 = 2a \)**:
- For \( r_2 = 2a \), the entire current \( I \) is enclosed since \( 2a \) is outside the wire.
- Thus, the magnetic field \( B' \) is:
\[ B' = \frac{\mu_0 I}{2 \pi (2a)} = \frac{\mu_0 I}{4 \pi a} \]

4. **Find the ratio \( \frac{B}{B'} \)**:
- Now we can find the ratio of the two magnetic fields:
\[ \frac{B}{B'} = \frac{\frac{\mu_0 I}{2 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2 \]

### Final Result:
The ratio of the magnetic fields \( B \) and \( B' \) at distances \( \frac{a}{2} \) and \( 2a \) respectively is:
\[ \frac{B}{B'} = 2 \]