1. **Problem statement:** A straight wire carrying current $I_1 = 5$ A is brought close to a circular wire of radius 10 cm carrying current $I_2 = 5$ A. Find the magnitude of the magnetic induction at the center of the circular wire. 2. **Magnetic field due to circular wire at its center:** The magnetic field by a circular loop carrying current $I$ at its center is given by $$B_{circle} = \frac{\mu_0 I_2}{2R}$$ where $\mu_0 = 4\pi \times 10^{-7}$ T\cdot m/A, $R = 10$ cm = 0.1 m. Substitute values: $$B_{circle} = \frac{4\pi \times 10^{-7} \times 5}{2 \times 0.1} = 3.14 \times 10^{-5} \text{ T}$$ 3. **Magnetic field due to straight wire at the center of circle:** The distance from the straight wire to the center of circle is 10 cm = 0.1 m. The magnetic field by a long straight conductor at distance $r$ is $$B_{wire} = \frac{\mu_0 I_1}{2\pi r}$$ Substitute values: $$B_{wire} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.1} = 1.0 \times 10^{-5} \text{ T}$$ 4. **Net magnetic field at the center:** Assuming the fields are perpendicular or opposing, the problem typically assumes vector addition. But no angle is given, we'll assume the fields add algebraically for magnitude estimation. Add the two magnetic inductions: $$B_{net} = B_{circle} + B_{wire} = 3.14 \times 10^{-5} + 1.0 \times 10^{-5} = 4.14 \times 10^{-5} \text{ T}$$ 5. **Final answer:** The magnitude of magnetic induction at the center of the circular wire is $$\boxed{4.14 \times 10^{-5} \text{ T}}$$ This corresponds to option A.