Figure A and B shown two long straight wires of circular cross-section ($$a$$ and $$b$$ with $$a < b$$), carrying current $$I$$ which is uniformly distributed across the cross-section. The magnitude of magnetic field $$B$$ varies with radius $$r$$ and can be represented as:

Upon reviewing the mathematical syntax in the previous derivation, a formatting error was found in the "Inside the Wire" bullet point where the enclosed current fraction failed to render properly. Here is the fully corrected, clean solution:

1. Derive the Magnetic Field Equations using Ampere's Law

For a long straight wire of radius $$R$$ carrying a uniformly distributed current $$I$$:

• Inside the Wire ($$r \le R$$):
  The current enclosed by an Amperian loop of radius $$r$$ is proportional to its cross-sectional area:

  $$I_{\text{enc}} = I \left( \frac{\pi r^2}{\pi R^2} \right) = I \frac{r^2}{R^2}$$

  Applying Ampere's Law ($$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$$):

  $$B_{\text{in}} \cdot 2\pi r = \mu_0 I \left(\frac{r^2}{R^2}\right) \implies B_{\text{in}} = \left(\frac{\mu_0 I}{2\pi R^2}\right) r$$

  This shows that inside the wire, the magnetic field increases linearly with radius ($$B \propto r$$).
• Outside the Wire ($$r > R$$):
  The entire current is enclosed ($I_{\text{enc}} = I$). Applying Ampere's Law:

  $$B_{\text{out}} \cdot 2\pi r = \mu_0 I \implies B_{\text{out}} = \frac{\mu_0 I}{2\pi r}$$

  This shows that outside the wire, the magnetic field decreases hyperbolically ($$B \propto \frac{1}{r}$$).
• At the Surface ($$r = R$$):
  The magnetic field reaches its maximum value:

  $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R}$$

  Since $$B_{\text{max}}$$ is inversely proportional to the wire's radius ($$B_{\text{max}} \propto \frac{1}{R}$$), the thinner wire ($$a$$) will reach a higher peak magnetic field than the thicker wire ($$b$$) because $$a < b$$.

2. Analyze the Graph Options

Let's compile the features our graph must show:

1. From $$r = 0$$ up to the respective surfaces, both lines must be straight lines starting from the origin ($$B = 0$$ at $$r = 0$$).
2. Since $$B_{\text{max}} \propto \frac{1}{R}$$, the linear peak for wire $$a$$ must be higher than the linear peak for wire $$b$$.
3. Beyond their respective surfaces ($$r > a$$ and $$r > b$$), both curves must decay non-linearly following a hyperbolic curve ($$\propto \frac{1}{r}$$). Outside both wires ($$r > b$$), the two curves merge because the total current $$I$$ is identical.

Reviewing the choices

• Option A: Correctly shows the peak for wire $$a$$ occurring earlier and higher than the peak for wire $$b$$, with both curves sloping linearly from the origin and merging asymptotically in the outer region.

Conclusion

The correct graphical representation of the magnetic field variation is given by Option A.