Step 1: Understand the problem
We are given a long straight wire of radius \( a \) carrying a steady current \( I \), uniformly distributed across its cross-section.
We need to find the ratio of magnetic field at two points:
- Point 1: at a distance \( r = \frac{a}{2} \) (inside the wire)
- Point 2: at a distance \( r = 2a \) (outside the wire)
Step 2: Magnetic field inside the wire (r < a)
For \( r < a \), the magnetic field is given by Ampère’s Law:
\( B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2} \)
Substituting \( r = \frac{a}{2} \):
\( B_{a/2} = \frac{\mu_0 I (a/2)}{2\pi a^2} = \frac{\mu_0 I}{4\pi a} \)
Step 3: Magnetic field outside the wire (r > a)
For \( r > a \), the wire behaves like a thin current-carrying conductor. The magnetic field is:
\( B_{\text{outside}} = \frac{\mu_0 I}{2\pi r} \)
Substituting \( r = 2a \):
\( B_{2a} = \frac{\mu_0 I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a} \)
Step 4: Take the ratio
\( \frac{B_{a/2}}{B_{2a}} = \frac{\mu_0 I / 4\pi a}{\mu_0 I / 4\pi a} = 1 \)
So, the ratio is:
\( B_{a/2} : B_{2a} = 1 : 1 \)
Final Answer: \( 1:1 \)