Two parallel wires are separated by a distance of 2.0 cm and carry...

Let the two wires be at points A and B, and the observation point be P.

Since the distance between the wires is $2.0 \text{ cm}$ and point P is $2.0 \text{ cm}$ away from both wires, triangle PAB is an equilateral triangle with side length $a = 2.0 \text{ cm} = 0.02 \text{ m}$.

The magnitude of the magnetic field at P due to each wire is:

$B_1 = B_2 = \dfrac{\mu_0 I}{2 \pi a}$

Substituting the given values:

$B_1 = \dfrac{4 \pi \times 10^{-7} \times 10}{2 \pi \times 0.02} = 10^{-4} \text{ T}$

The magnetic field vectors $\vec{B}_1$ and $\vec{B}_2$ are perpendicular to the position vectors connecting the wires to point P. Since the angle between these position vectors is 60^{}^{{}^{\circ}} and the currents are in the same direction, the angle between $\vec{B}_1$ and $\vec{B}_2$ is also 60^{}^{{}^{\circ}}.

The resultant magnetic field $B_{net}$ is given by vector addition:

B_{net} = \sqrt{B_1^2 + B_2^2 + 2 B_1 B_2 \cos 60^{}^{{}^{\circ}}}

$B_{net} = \sqrt{B_1^2 + B_1^2 + 2 B_1^2 \left(\dfrac{1}{2}\right)} = \sqrt{3} B_1$

$B_{net} = \sqrt{3} \times 10^{-4} \text{ T} \approx 1.7 \times 10^{-4} \text{ T}$