A coaxial cable consists of an inner wire of radius $$a$$ surrounded by an outer shell of inner and outer radii $$b$$ and $$c$$ respectively. The inner wire carries an electric current $$i_0$$ which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance $$x$$ from the axis when (i) $$x \lt a$$ and (ii) $$a \lt x \lt b$$?

We have a long straight coaxial system. Its inner solid conductor has radius $$a$$ and carries a total current $$I_0$$ uniformly over its cross-section. Its outer cylindrical shell occupies the space from $$r=b$$ to $$r=c$$ and carries an equal current $$I_0$$ but in the opposite direction, also distributed uniformly through its own metal. Because the currents are steady and the geometry is perfectly coaxial, we take every magnetic field line to be a circle centred on the common axis, so the magnitude of the field depends only on the radial distance $$r$$ from that axis.

For such circular symmetry we apply Ampère’s circuital law, first stating it explicitly:

$$\oint \vec B \cdot d\vec l = \mu_0 I_{\text{enclosed}}.$$

Choosing a circular Amperian path of radius $$r$$ (so $$d\vec l$$ is everywhere tangential and of magnitude $$r\,d\theta$$), the left side becomes $$B(2\pi r)$$. Hence

$$B(2\pi r)=\mu_0 I_{\text{enclosed}}\quad\Longrightarrow\quad B=\dfrac{\mu_0 I_{\text{enclosed}}}{2\pi r}.$$

We now find the enclosed current in the two separate radial zones asked for and then form their ratio.

Case (i)  $$r=x\lt a$$

The current density $$J$$ in the inner wire is uniform, so

$$J=\dfrac{I_0}{\text{cross-sectional area}}=\dfrac{I_0}{\pi a^2}.$$

The area enclosed by our path of radius $$x$$ is $$\pi x^2$$, hence the enclosed current is

$$I_{\text{enc}}^{(i)} = J(\pi x^2)=\dfrac{I_0}{\pi a^2}\,(\pi x^2)=I_0\,\dfrac{x^2}{a^2}.$$

Substituting this into Ampère’s result gives

$$B_{(i)}=\dfrac{\mu_0}{2\pi x}\,I_0\,\dfrac{x^2}{a^2}=\dfrac{\mu_0 I_0\,x}{2\pi a^2}.$$

Case (ii)  $$a\lt r=x\lt b$$

Now the entire inner conductor lies inside the Amperian loop, while none of the outer shell (which starts only at $$r=b$$) is enclosed. Therefore

$$I_{\text{enc}}^{(ii)}=I_0.$$

Inserting this into Ampère’s expression, we find

$$B_{(ii)}=\dfrac{\mu_0 I_0}{2\pi x}.$$

Forming the required ratio

$$\dfrac{B_{(i)}}{B_{(ii)}}=\dfrac{\dfrac{\mu_0 I_0\,x}{2\pi a^2}}{\dfrac{\mu_0 I_0}{2\pi x}} =\dfrac{x^2}{a^2}.$$

This matches Option D in the given list.

Hence, the correct answer is Option D.