Four long parallel wires pass through the corners of a square with side $0.10 \mathrm{m}\(. All four wires carry the same magnitude of current \)I=10.0 \mathrm{A}\( in the directions indicated. Find the magnetic field at point \)R,$ the midpoint of the left side of the square.

In this problem, we will find the contribution of each wire at point R using the Biot-Savart law. The Biot-Savart law for a straight wire is given by: $$ d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{L}\times \vec{r}}{r^3} $$ where \(\mu_0\) is the permeability of free space, \(I\) is the current, \(d\vec{L}\) is a small segment of the wire, \(\vec{r}\) is the position vector from the segment to the point where we want to find the magnetic field, and \(r\) is the magnitude of the position vector. Since all four wires carry the same current and are symmetrically placed, the total magnetic field at point R will be the sum of the magnetic fields produced by each wire.

We will now calculate the magnetic field produced by the wires at point R. We shall use the following notation: \(B_1\) for the magnetic field produced by the top left corner wire \(B_2\) for the magnetic field produced by the top right corner wire \(B_3\) for the magnetic field produced by the bottom right corner wire \(B_4\) for the magnetic field produced by the bottom left corner wire Our goal is to find \(B_R = B_1 + B_2 + B_3 + B_4\). Due to symmetry, we can see that \(B_1\) and \(B_2\) will have horizontal components that cancel out with each other. Hence, we only need to find the vertical components of \(B_1\) and \(B_2\). Similarly, we can see that \(B_3\) and \(B_4\) will have vertical components that cancel each other out. Hence, we only need to find the horizontal components of \(B_3\) and \(B_4\).

At point R, both \(B_1\) and \(B_2\) are at a distance of \(0.1 \mathrm{m}\) and have an angle of \(45^{\circ}\) with respect to the horizontal axis. Using the Biot-Savart law, the magnetic field components can be calculated as follows: $$ B_{1_{ver}} = B_1\sin 45^{\circ} = \frac{\mu_0 I}{4 \pi (0.1\mathrm{m})} \cdot \sin 45^{\circ} $$ As \(B_2\) is a mirror image of \(B_1\), we have: $$ B_{2_{ver}} = B_{1_{ver}} $$

At point R, both \(B_3\) and \(B_4\) are at a distance of \(0.1 \mathrm{m}\) and have an angle of \(45^{\circ}\) with respect to the horizontal axis as well. However, for \(B_3\) and \(B_4\) we need to find the horizontal components, and so we have: $$ B_{3_{hor}} = B_3\cos 45^{\circ} = \frac{\mu_0 I}{4 \pi (0.1\mathrm{m})} \cdot \cos 45^{\circ} $$ And since \(B_4\) is a mirror image of \(B_3\), we have: $$ B_{4_{hor}} = B_{3_{hor}} $$

Now that we have all the components, we can find the total magnetic field at point R: $$ B_R = \sqrt{(B_{1_{ver}} + B_{2_{ver}})^2 + (B_{3_{hor}} + B_{4_{hor}})^2} $$ Plug in the values for the components and the permeability of free space (\(\mu_0 = 4\pi \times 10^{-7} \mathrm{T m/A}\)), we get: $$ B_R = \sqrt{2(\frac{4\pi \times 10^{-7} \mathrm{Tm/A} \cdot 10\mathrm{A}}{4 \pi (0.1\mathrm{m})})^2} $$ Solve for \(B_R\): $$ B_R = 2 \times 10^{-5} \mathrm{T} $$ So, the magnetic field at point R is \(2 \times 10^{-5} \mathrm{T}\).