A long, straight wire carrying a current of 1.0 A is positioned...

The magnetic field produced by a long straight current-carrying wire at a distance $d$ is given by:

$B_{\text{wire}} = \dfrac{\mu_0 i}{2\pi d}$

Given $i = 1.0 \text{ A}$ and $d = 2.0 \text{ cm} = 0.02 \text{ m}$, we get:

$B_{\text{wire}} = \dfrac{4\pi \times 10^{-7} \times 1.0}{2\pi \times 0.02} = 10^{-5} \text{ T} = 10 \mu\text{T}$

According to the right-hand thumb rule, the direction of the magnetic field due to the wire at point $P$ (above the wire) is outwards ($\odot$), and at point $Q$ (below the wire) is inwards ($\otimes$).

The external uniform magnetic field is $B = 1.0 \times 10^{-5} \text{ T} = 10 \mu\text{T}$, directed outwards ($\odot$).

At point $P$, both magnetic fields are in the same direction. The resultant magnetic field is:

$B_P = B_{\text{wire}} + B = 10 \mu\text{T} + 10 \mu\text{T} = 20 \mu\text{T}$

At point $Q$, the magnetic fields are in opposite directions. The resultant magnetic field is:

$B_Q = B_{\text{wire}} - B = 10 \mu\text{T} - 10 \mu\text{T} = 0$

Therefore, the resultant magnetic fields at $P$ and $Q$ are $20 \mu\text{T}$ and zero, respectively.