A long, straight wire of radius r carrying a uniformly distributed...

The external magnetic field is $B$ directed vertically upward.

The magnetic field due to the current-carrying wire forms concentric circles. In the horizontal plane passing through the axis of the wire, the magnetic field is vertical. On one side of the wire, it is directed upward, and on the other side, it is directed downward.

The resultant magnetic field will have a minimum magnitude on the side where the wire's magnetic field is directed downward, opposing the external field.

The magnitude of the magnetic field due to the wire at a distance $x$ from its axis is $B_{wire} = \dfrac{\mu_0 i x}{2\pi r^2}$ for $x \le r$ and $B_{wire} = \dfrac{\mu_0 i}{2\pi x}$ for $x \ge r$.

The maximum value of $B_{wire}$ occurs at the surface of the wire ($x = r$), which is B_\max = \dfrac{\mu_0 i}{2\pi r}.

If B_\max \ge B, there exists a point where $B_{wire} = B$. At this point, the resultant magnetic field is zero. This happens when $\dfrac{\mu_0 i}{2\pi r} \ge B \Rightarrow r \le \dfrac{\mu_0 i}{2\pi B}$.

If B_\max < B, the maximum downward field the wire can produce is B_\max. The minimum resultant magnetic field will be B - B_\max = B - \dfrac{\mu_0 i}{2\pi r}. This happens when $\dfrac{\mu_0 i}{2\pi r} < B \Rightarrow r > \dfrac{\mu_0 i}{2\pi B}$.

Therefore, the minimum magnitude is $0$ if $r \le \dfrac{\mu_0 i}{2\pi B}$, and $B - \dfrac{\mu_0 i}{2\pi r}$ if $r > \dfrac{\mu_0 i}{2\pi B}$.