A rectangular loop of wire in free space joins point \(A(1,0,1)\) to point \(B(3,0,1)\) to point \(C(3,0,4)\) to point \(D(1,0,4)\) to point \(A\). The wire carries a current of \(6 \mathrm{~mA}\), flowing in the \(\mathbf{a}_{z}\) direction from \(B\) to \(C\). A filamentary current of 15 A flows along the entire \(z\) axis in the \(\mathbf{a}_{z}\) direction. \((a)\) Find \(\mathbf{F}\) on side \(B C\). \((b)\) Find \(\mathbf{F}\) on side \(A B\). \((c)\) Find \(\mathbf{F}_{\text {total }}\) on the loop.

First, we need to determine the magnetic field produced by the filamentary current along the z-axis at any point in space. To do this, we will apply the Biot-Savart Law formula, which relates the magnetic field \(\mathbf{B}\) produced by a current \(I\) to the infinitesimal length vector \(\mathrm{d}\mathbf{l}\) and the vector \(\mathbf{r}\) pointing from the infinitesimal length to the field point: \(\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \frac{I\, \mathrm{d}\mathbf{l} \times \mathbf{r}}{r^3}\), where \(\mu_0\) is the permeability of free space.

Since the filamentary current runs along the z-axis, we have \(\mathrm{d}\mathbf{l} = \mathrm{d}z\, \mathbf{a}_{z}\). Therefore, the magnetic field contribution at an arbitrary point \(\mathbf{r} = x \mathbf{a}_{x} + y \mathbf{a}_{y} + z \mathbf{a}_{z}\) is given by: \(\mathrm{d}\mathbf{B}(\mathbf{r}) = \frac{\mu_0 I_{\text {filament}}}{4\pi}\frac{\mathrm{d}z\, \mathbf{a}_{z} \times x \mathbf{a}_{x} + y \mathbf{a}_{y} + z \mathbf{a}_{z}}{(x^2 + y^2 + z^2)^{3/2}}\). To find the total magnetic field at \(\mathbf{r}\), we need to integrate this expression over the entire z-axis: \(\mathbf{B}(\mathbf{r}) = \int_{-\infty}^{+\infty} \mathrm{d}\mathbf{B}(\mathbf{r})\). We find that the magnetic field at \(\mathbf{r}\) is given by: \(\mathbf{B}(x, y, z) = -\frac{\mu_0 I_{\text {filament}}}{2\pi}\frac{y}{x^2 + y^2}\mathbf{a}_{x} + \frac{\mu_0 I_{\text {filament}}}{2\pi}\frac{x}{x^2 + y^2}\mathbf{a}_{y}\).

Now we will calculate the force on segment BC. The current flows in the az direction, and the magnetic field due to the filamentary current is given by the expression above. To find the force on BC, we use Ampere's law: \(\mathbf{F}_{BC} = I_{BC}\, \mathbf{l}_{BC} \times \mathbf{B}_{BC}\), where \(I_{BC} = 6\, \mathrm{mA}\) is the current in segment BC, \(\mathbf{l}_{BC} = (0, 0, 3-1)\, \mathbf{a}_{z}\) is the length vector of BC, and \(\mathbf{B}_{BC}\) is the magnetic field at the position of BC. Plugging in the given coordinates and calculating the force, we get: \(\mathbf{F}_{BC} = 0.006\, (0, 0, 2)\, \times \left(-\frac{\mu_0 \times 15}{2\pi}\frac{1}{10}, \frac{\mu_0 \times 15}{2\pi}\frac{3}{10}, 0\right) = -0.012\frac{\mu_0 \times 15}{2\pi}\frac{1}{10}\, \mathbf{a}_{y}\).

Similarly, for segment AB, we have: \(\mathbf{F}_{AB} = I_{AB}\, \mathbf{l}_{AB} \times \mathbf{B}_{AB}\), where \(I_{AB} = 6\, \mathrm{mA}\), \(\mathbf{l}_{AB} = (3-1, 0, 0)\, \mathbf{a}_{x}\), and \(\mathbf{B}_{AB}\) is the magnetic field at the position of AB. Plugging in the given coordinates and calculating the force, we get: \(\mathbf{F}_{AB} = 0.006\, (2, 0, 0)\, \times \left(-\frac{\mu_0 \times 15}{2\pi}\frac{1}{10}, 0, 0\right) = 0\). The force on AB is zero because the magnetic field is in the same plane as AB, so the cross product is zero.

Now we can find the total force on the loop by summing the forces on the individual segments: \(\mathbf{F}_{\text {total}} = \mathbf{F}_{AB} + \mathbf{F}_{BC} = 0 -0.012\frac{\mu_0 \times 15}{2\pi}\frac{1}{10}\, \mathbf{a}_{y}\). The total force on the loop is given by: \(\mathbf{F}_{\text {total}} = -0.012\frac{\mu_0 \times 15}{2\pi}\frac{1}{10}\, \mathbf{a}_{y}\).