As illustrated in the figure, two parallel wires separated by a...

The magnetic field due to a long straight wire carrying current $I$ at a distance $r$ is given by $B = \dfrac{\mu_0 I}{2\pi r}$.

Given $I = 10 \text{ A}$, the constant $\dfrac{\mu_0 I}{2\pi} = \dfrac{4\pi \times 10^{-7} \times 10}{2\pi} = 2 \times 10^{-6} \text{ T}\cdot\text{m}$.

Let the left wire be $W_1$ (current into the page) and the right wire be $W_2$ (current out of the page). The distance between them is $4 \text{ cm}$.

For point $A_1$:
Distance from $W_1$ is $r_1 = 2 \text{ cm} = 0.02 \text{ m}$.
Distance from $W_2$ is $r_2 = 2 + 4 = 6 \text{ cm} = 0.06 \text{ m}$.
Since $A_1$ is outside the wires and the currents are in opposite directions, the magnetic fields oppose each other.
$B_{A_1} = B_1 - B_2 = \dfrac{2 \times 10^{-6}}{0.02} - \dfrac{2 \times 10^{-6}}{0.06} = 10^{-4} - \dfrac{1}{3} \times 10^{-4} = \dfrac{2}{3} \times 10^{-4} \approx 0.67 \times 10^{-4} \text{ T}$.

For point $A_2$:
Distance from $W_1$ is $r_1 = 1 \text{ cm} = 0.01 \text{ m}$.
Distance from $W_2$ is $r_2 = 4 - 1 = 3 \text{ cm} = 0.03 \text{ m}$.
Since $A_2$ is between the wires, the magnetic fields are in the same direction (downwards).
$B_{A_2} = B_1 + B_2 = \dfrac{2 \times 10^{-6}}{0.01} + \dfrac{2 \times 10^{-6}}{0.03} = 2 \times 10^{-4} + \dfrac{2}{3} \times 10^{-4} = \dfrac{8}{3} \times 10^{-4} \approx 2.67 \times 10^{-4} \text{ T}$.

For point $A_3$:
Distance from $W_1$ is $r_1 = 2 \text{ cm} = 0.02 \text{ m}$.
Distance from $W_2$ is $r_2 = 4 - 2 = 2 \text{ cm} = 0.02 \text{ m}$.
The magnetic fields are in the same direction.
$B_{A_3} = B_1 + B_2 = \dfrac{2 \times 10^{-6}}{0.02} + \dfrac{2 \times 10^{-6}}{0.02} = 10^{-4} + 10^{-4} = 2.0 \times 10^{-4} \text{ T}$.

For point $A_4$:
Distance from $W_1$ is $r_1 = \sqrt{2^2 + 2^2} = 2\sqrt{2} \text{ cm} = 0.02\sqrt{2} \text{ m}$.
Distance from $W_2$ is $r_2 = \sqrt{2^2 + 2^2} = 2\sqrt{2} \text{ cm} = 0.02\sqrt{2} \text{ m}$.
The magnitudes of the fields are $B_1 = B_2 = \dfrac{2 \times 10^{-6}}{0.02\sqrt{2}} = \dfrac{10^{-4}}{\sqrt{2}} \text{ T}$.
The magnetic field vectors $\vec{B}_1$ and $\vec{B}_2$ are perpendicular to their respective position vectors. Since the position vectors are at $90^{{}^{{}^{\circ}}}$ to each other, $\vec{B}_1$ and $\vec{B}_2$ are also at $90^{{}^{{}^{\circ}}}$ to each other.
$B_{A_4} = \sqrt{B_1^2 + B_2^2} = \sqrt{\left(\dfrac{10^{-4}}{\sqrt{2}}\right)^2 + \left(\dfrac{10^{-4}}{\sqrt{2}}\right)^2} = \sqrt{\dfrac{10^{-8}}{2} + \dfrac{10^{-8}}{2}} = 1.0 \times 10^{-4} \text{ T}$.

Answer: $0.67 \times 10^{-4} \text{ T}$, $2.7 \times 10^{-4} \text{ T}$, $2.0 \times 10^{-4} \text{ T}$, $1.0 \times 10^{-4} \text{ T}$