A long, straight wire of radius $$a$$ carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $$\frac{a}{3}$$ and $$2a$$, respectively from the axis of the wire is:

We consider a very long straight wire of radius $$a$$ that carries a steady current $$I$$. The current is said to be uniformly distributed over the circular cross-section, which means the current density $$J$$ (current per unit area) is the same at every point inside the wire.

Because the situation is cylindrically symmetric, we use Ampère’s circuital law in its integral form, which states

$$\oint \vec B \cdot d\vec l \;=\;\mu_0 I_{\text{encl}}$$

where $$I_{\text{encl}}$$ is the current enclosed by the chosen circular Amperian path and $$\mu_0$$ is the permeability of free space. For a circle of radius $$r$$ centred on the axis of the wire, $$\vec B$$ is tangential and has the same magnitude everywhere on the circle, so the left side becomes $$B\,(2\pi r)$$. Thus

$$B\,(2\pi r)=\mu_0 I_{\text{encl}} \quad\Longrightarrow\quad B=\frac{\mu_0 I_{\text{encl}}}{2\pi r}$$

Now we need the magnetic field at two different distances from the axis: (i) at $$r_1=\dfrac{a}{3}$$ (inside the wire) and (ii) at $$r_2=2a$$ (outside the wire).

1. Field at $$r_1=\dfrac{a}{3}$$ (inside the wire)

Because the point lies inside the conductor, only the current contained in the smaller circle of radius $$r_1$$ contributes. The current density is

$$J=\frac{I}{\pi a^{2}}\;,$$

so the enclosed current is

$$I_{\text{encl}}=J\;(\text{area within }r_1)=\frac{I}{\pi a^{2}}\;(\pi r_1^{2})=\frac{I r_1^{2}}{a^{2}}.$$

Substituting $$r_1=\dfrac{a}{3}$$ gives

$$I_{\text{encl}}=\frac{I}{a^{2}}\left(\frac{a}{3}\right)^{2}=\frac{I a^{2}}{9a^{2}}=\frac{I}{9}.$$

Putting this value into the expression for $$B$$, we get the magnetic field at $$r_1$$:

$$B_1=\frac{\mu_0 I_{\text{encl}}}{2\pi r_1} =\frac{\mu_0 \left(\dfrac{I}{9}\right)}{2\pi \left(\dfrac{a}{3}\right)} =\frac{\mu_0 I}{9}\;\frac{3}{2\pi a} =\frac{\mu_0 I}{6\pi a}.$$

2. Field at $$r_2=2a$$ (outside the wire)

For a point outside, the entire current $$I$$ is enclosed, i.e., $$I_{\text{encl}}=I$$. Using the same formula, we have

$$B_2=\frac{\mu_0 I}{2\pi r_2} =\frac{\mu_0 I}{2\pi (2a)} =\frac{\mu_0 I}{4\pi a}.$$

3. Required ratio $$\dfrac{B_1}{B_2}$$

We now divide $$B_1$$ by $$B_2$$:

$$\frac{B_1}{B_2} =\frac{\dfrac{\mu_0 I}{6\pi a}}{\dfrac{\mu_0 I}{4\pi a}} =\left(\frac{\mu_0 I}{6\pi a}\right)\left(\frac{4\pi a}{\mu_0 I}\right) =\frac{4}{6} =\frac{2}{3}.$$

Hence, the ratio of the magnetic fields at the two specified distances is $$\dfrac{2}{3}$$.

Hence, the correct answer is Option A.