### Related Formula The magnetic field contributions from unique structural line elements are given by: * **Semi-infinite straight wire segment** at a distance perpendicular to its \end tip: B_textstraight = fracmu_0 I4pi a$B_{\text{straight}} = \frac{\mu_0 I}{4\pi a}$ * **Circular arc path segment** subtending \angle theta$\theta$ at the center: B_textarc = fracmu_0 I4pi a theta$B_{\text{arc}} = \frac{\mu_0 I}{4\pi a} \theta$ ### Core Logic Let us decompose the structure into three functional parts as mapped out below[cite: 763, 764]:

1. **Segment 1 (Incoming semi-infinite line)**: The straight line extends to infinity, with its terminating tip at a perpendicular distance a$a$ from origin O$O$. Using the \right-hand grip rule, the direction points into the page: B_1 = fracmu_0 I4pi a quad (otimes)$B_1 = \frac{\mu_0 I}{4\pi a} \quad (\otimes)$ 2. **Segment 2 (Three-quarter circular loop)**: The loop forms an \angle of theta = frac3pi2$\theta = \frac{3\pi}{2}$ radians around O$O$. The field points into the page: B_2 = fracmu_0 I4pi a left(frac3pi2right) quad (otimes)$B_2 = \frac{\mu_0 I}{4\pi a} \left(\frac{3\pi}{2}\right) \quad (\otimes)$ 3. **Segment 3 (Outgoing semi-infinite line)**: This line aligns perfectly with the origin O$O$ along its vector axis, making sintheta = 0$\sin\theta = 0$ : B_3 = 0$B_3 = 0$ Summing the total fields via superposition: B = B_1 + B_2 + B_3 = fracmu_0 I4pi a + fracmu_0 I4pi aleft(frac3pi2 ight)$B = B_1 + B_2 + B_3 = \frac{\mu_0 I}{4\pi a} + \frac{\mu_0 I}{4\pi a}\left(\frac{3\pi}{2} ight)$ B = fracmu_0 I4pi a left[frac3pi2 + 1right]$B = \frac{\mu_0 I}{4\pi a} \left[\frac{3\pi}{2} + 1\right]$ ### Pattern Recognition Always check the axis alignment first. Any straight wire segment whose extended line passes directly through the field point contributes exactly zero to the total magnetic field value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetic Effects of Current and Magnetism