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Step-by-Step Solution
Step 1: Understand the scenario
A long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distributed across its circular cross-section. We need to compare the magnetic field at two points:
Inside the wire: at a radial distance $r = \frac{a}{2}$ from the axis.
Outside the wire: at a radial distance $r = 2a$ from the axis.
Step 2: Magnetic field inside the wire ($r \leq a$)
When the point is inside the wire, the magnetic field $B$ at distance $r$ is given by the formula (derived using Ampere's law):
$B = \frac{\mu_0 \, I \, r}{2\pi a^2}$
Here, $a$ is the radius of the wire, and $I$ is the total current. The factor $\frac{r}{a^2}$ reflects that the current enclosed by the Amperian loop is proportional to the area enclosed (uniform current distribution).
Step 3: Magnetic field at $r = \frac{a}{2}$
Substituting $r = \frac{a}{2}$ into the inside formula:
$B = \frac{\mu_0 \, I \, \left(\frac{a}{2}\right)}{2\pi a^2} = \frac{\mu_0 I}{2\pi a^2} \cdot \frac{a}{2} = \frac{\mu_0 I}{4\pi a}$
Step 4: Magnetic field outside the wire ($r \geq a$)
When the point is outside the wire, the wire can be treated as though all the current $I$ passes through its cross-sectional radius $a$. Then, at a radial distance $r$ from the axis:
$B' = \frac{\mu_0 \, I}{2 \pi \, r}$
Step 5: Magnetic field at $r = 2a$
Substituting $r = 2a$ into the outside formula:
$B' = \frac{\mu_0 \, I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a}$
Step 6: Compute the ratio $\frac{B}{B'}$
From the expressions, we have:
$B \bigg|_{r=\frac{a}{2}} = \frac{\mu_0 I}{4\pi a}$, $B' \bigg|_{r=2a} = \frac{\mu_0 I}{4\pi a}$
Therefore,
$\displaystyle \frac{B}{B'} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1$
Step 7: Conclude the result
The ratio of the magnetic fields at $r = \frac{a}{2}$ and $r = 2a$ is $1 : 1$.
