We use Ampere's Law to find the magnetic field B at a distance r from the center of a long straight wire of radius R carrying a uniform current I.
Case 1: Inside the wire ($r \le R$)
The enclosed current is $I_{enc} = I \left(\frac{\pi r^2}{\pi R^2}\right) = I \frac{r^2}{R^2}$.
Ampere's Law: $\oint \vec{B} \cdot d\vec{l} = B(2\pi r) = \mu_0 I_{enc} = \mu_0 I \frac{r^2}{R^2}$.
$B_{in} = \frac{\mu_0 I r}{2\pi R^2}$. The field increases linearly with r, so $B \propto r$.
Case 2: Outside the wire ($r>R$)
The enclosed current is the total current, $I_{enc} = I$.
Ampere's Law: $B(2\pi r) = \mu_0 I$.
$B_{out} = \frac{\mu_0 I}{2\pi r}$. The field decreases hyperbolically with r, so $B \propto \frac{1}{r}$.
The magnetic field is maximum at the surface of the wire ($r=R$), where $B_{max} = \frac{\mu_0 I}{2\pi R}$.
Now, let's compare wire A (radius a) and wire B (radius b), with $a<b$.
Both carry the same current I.
Maximum field for wire A: $B_{max, A} = \frac{\mu_0 I}{2\pi a}$ (at $r=a$).
Maximum field for wire B: $B_{max, B} = \frac{\mu_0 I}{2\pi b}$ (at $r=b$).
Since $a<b$, it follows that $\frac{1}{a}>\frac{1}{b}$.
Therefore, $B_{max, A}>B_{max, B}$. The peak magnetic field for the thinner wire (A) is greater than that for the thicker wire (B).
Analyzing the graphs:
- All graphs correctly show the linear increase inside and hyperbolic decrease outside.
- We need the graph where the peak for 'a' is higher than the peak for 'b', and the peaks are located at $r=a$ and $r=b$ respectively, with $a<b$.
- Option (A) is the only graph that satisfies both conditions: the peak for curve 'a' is higher than for curve 'b', and the peak position 'a' is to the left of 'b'.