Let's break down the solution step by step: ### **Step 1: Understanding the Problem** We have an infinitely long wire carrying current \(I\) as shown in the figure. The wire has three segments: - A straight segment along \(x_1\) (left of the origin). - A quarter-circle segment of radius \(a\), from \((a,0)\) to \((0,a)\). - A straight segment along \(x_2\) (right of the origin). We are asked to **find the magnetic flux density** \(\vec{B}\) **at the origin** \(O\). --- ### **Step 2: Use Biot-Savart Law** The magnetic field at a point due to a current element is given by the **Biot-Savart Law**: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \] where: - \(d\vec{l}\) = current element vector - \(\vec{r}\) = vector from element to observation point (here, the origin) - \(r = |\vec{r}|\) - \(\mu_0\) = permeability of free space We will find the contribution to \(\vec{B}\) at the origin from each segment and **sum them up**. --- ### **Step 3: Contribution from Each Segment** #### **A) Infinite Straight Wires (\(x_1\) and \(x_2\) segments)** ##### **Left Segment (\(x_1\)):** - This is a semi-infinite wire, extending from \(x_1 = -\infty\) to \(x_1 = -a\). - The magnetic field at the origin due to a semi-infinite wire at perpendicular distance \(a\) is: \[ B = \frac{\mu_0 I}{4\pi a} \] The direction is **out of the page** (\(+\hat{x}_3\)) by the right-hand rule. ##### **Right Segment (\(x_2\)):** - This is a semi-infinite wire, extending from \(x_2 = a\) to \(x_2 = \infty\). - The perpendicular distance is \(a\) (since at \((a,0)\)). - The magnetic field is: \[ B = \frac{\mu_0 I}{4\pi a} \] The direction is **into the page** (\(-\hat{x}_3\)). #### **B) Quarter-Circle Arc (\(x_3\) direction)** - For a wire segment forming a circular arc of radius \(a\), subtending angle \(\theta\) at the center, the magnetic field at the center is: \[ B = \frac{\mu_0 I \theta}{4\pi a} \] Here, \(\theta = \frac{\pi}{2}\) (quarter circle). So, \[ B = \frac{\mu_0 I}{4a} \] The direction by the right-hand rule is **into the page** (\(-\hat{x}_3\)). --- ### **Step 4: Add up the Contributions (Vectorially)** Let’s assign directions: - Out of page: \(+\hat{x}_3\) - Into page: \(-\hat{x}_3\) So, - Left straight wire: \(+\frac{\mu_0 I}{4\pi a}\) (\(+\hat{x}_3\)) - Right straight wire: \(-\frac{\mu_0 I}{4\pi a}\) (\(-\hat{x}_3\)) - Arc: \(-\frac{\mu_0 I}{4a}\) (\(-\hat{x}_3\)) Add up: \[ B_{\text{total}} = \left[\frac{\mu_0 I}{4\pi a} - \frac{\mu_0 I}{4\pi a} - \frac{\mu_0 I}{4a}\right] \hat{x}_3 \] \[ B_{\text{total}} = -\frac{\mu_0 I}{4a} \hat{x}_3 \] --- ### **Step 5: Final Answer** \[ \boxed{ \vec{B}(O) = -\frac{\mu_0 I}{4a} \hat{x}_3 } \] - **Direction:** Into the page at the origin (\(-\hat{x}_3\)). - **Magnitude:** \(\frac{\mu_0 I}{4a}\). --- #### **Summary Table** | Segment | Magnitude | Direction | |------------------|----------------------------------|-------------------| | Left straight | \(+\frac{\mu_0 I}{4\pi a}\) | Out of page | | Right straight | \(-\frac{\mu_0 I}{4\pi a}\) | Into the page | | Arc | \(-\frac{\mu_0 I}{4a}\) | Into the page | **Total:** \(\boxed{-\frac{\mu_0 I}{4a} \hat{x}_3}\)