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Step-by-Step Solution
Step 1: Understand the Physical Situation
A long, straight wire of radius $a$ carries a steady current $I$ that is uniformly distributed across its cross-section. We need to determine the ratio of the magnetic field at a distance $r_1 = \frac{a}{2}$ (inside the wire) to the magnetic field at $r_2 = 2a$ (outside the wire).
Step 2: Magnetic Field Inside the Wire ($r_1 = \frac{a}{2}$)
If the current is uniformly distributed over the cross-sectional area $\pi a^2$, the current density $J$ is
$$
J = \frac{I}{\pi a^2}.
$$
The portion of the current enclosed by an imaginary circular Amperian loop of radius $r_1$ (where $r_1 \le a$) is
$$
I_{\text{enclosed}} = J \times \pi r_1^2 = \frac{I}{\pi a^2} \times \pi r_1^2 = I \frac{r_1^2}{a^2}.
$$
By Ampère’s circuital law,
$$
\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}.
$$
For a circular path of radius $r_1$, the magnetic field $B_1$ is uniform along the path, and the path length is $2\pi r_1$. Thus,
$$
B_1 (2\pi r_1) = \mu_0 I_{\text{enclosed}} = \mu_0 \left( I \frac{r_1^2}{a^2} \right).
$$
Hence,
$$
B_1 = \frac{\mu_0 I \, r_1^2}{2 \pi r_1 \, a^2} = \frac{\mu_0 I \, r_1}{2 \pi a^2}.
$$
Step 3: Magnetic Field Outside the Wire ($r_2 = 2a$)
For a point outside the wire (i.e., at $r > a$), the entire current $I$ is enclosed. Using Ampère’s law again for a circular path of radius $r_2$,
$$
B_2 \,(2\pi r_2) = \mu_0 I.
$$
Substituting $r_2 = 2a$,
$$
B_2 = \frac{\mu_0 I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a}.
$$
Step 4: Calculate the Required Ratio
The ratio of $B_1$ to $B_2$ is
$$
\frac{B_1}{B_2}
= \frac{\frac{\mu_0 I \, r_1}{2 \pi a^2}}{\frac{\mu_0 I}{4\pi a}}
= \left(\frac{\mu_0 I \, r_1}{2 \pi a^2} \right) \Big/ \left(\frac{\mu_0 I}{4\pi a}\right).
$$
Simplify:
$$
\frac{B_1}{B_2}
= \frac{\mu_0 I \, r_1}{2 \pi a^2} \times \frac{4\pi a}{\mu_0 I}
= 2 \,\frac{r_1}{a}.
$$
Since $r_1 = \frac{a}{2}$,
$$
2 \times \frac{\frac{a}{2}}{a} = 2 \times \frac{1}{2} = 1.
$$
Step 5: Conclusion
The ratio of the magnetic field at $r_1 = \frac{a}{2}$ to that at $r_2 = 2a$ is 1.
