1. The problem is to calculate the magnetic field $B(x)$ at a distance $x = -0.12$ m using the Biot-Savart law formula: $$B(x) = \frac{\mu_0 \cdot i \cdot r^2}{2 \left( r^2 + x^2 \right)^{3/2}}$$ where: - $\mu_0 = 4\pi \times 10^{-7}$ T·m/A (the permeability of free space) - $i = 1$ A (current) - $r = 0.066$ m (radius) - $x = -0.12$ m (distance) 2. Substitute the values into the formula: $$B(x) = \frac{4\pi \times 10^{-7} \times 1 \times (0.066)^2}{2 \left( (0.066)^2 + (-0.12)^2 \right)^{3/2}}$$ 3. Calculate $r^2$ and $x^2$: $$r^2 = (0.066)^2 = 0.004356$$ $$x^2 = (-0.12)^2 = 0.0144$$ 4. Sum inside the root: $$r^2 + x^2 = 0.004356 + 0.0144 = 0.018756$$ 5. Calculate the denominator: $$2 \times \left(0.018756\right)^{3/2} = 2 \times \left(0.018756\right)^{1.5}$$ Calculate $0.018756^{1.5}$: $$0.018756^{1.5} = 0.018756 \times \sqrt{0.018756} \approx 0.018756 \times 0.1369 = 0.002567$$ So denominator: $$2 \times 0.002567 = 0.005134$$ 6. Calculate numerator: $$4\pi \times 10^{-7} \times 0.004356 \approx 4 \times 3.1416 \times 10^{-7} \times 0.004356 = 12.5664 \times 10^{-7} \times 0.004356 = 5.47 \times 10^{-9}$$ 7. Finally, calculate $B(x)$: $$B(x) = \frac{5.47 \times 10^{-9}}{0.005134} \approx 1.065 \times 10^{-6} \text{ Tesla}$$ **Answer:** $$B(x) \approx 1.07 \times 10^{-6} \text{ T}$$ This means the magnetic field at $-12$ cm from the wire loop is approximately $1.07$ microtesla.