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Step-by-Step Solution
Step 1: Understand the Geometry and the Point of Interest
• The wire consists of three parts:
1. A very long linear segment along the positive x-axis.
2. A semicircular arc of radius $R$ lying in the y-z plane.
3. Another very long linear segment along the negative x-axis.
• We wish to find the resultant magnetic field at the center O of the semicircle (which is also where the linear segments appear to meet).
• The current $I$ flows through all the segments of the wire as shown.
Step 2: Magnetic Field Contribution from Each Linear Segment
Each linear segment is infinitely long and parallel to the x-axis. We apply the right-hand rule and the known result for the magnetic field due to a long straight current-carrying wire at a point in the plane perpendicular to the wire.
Let’s label the two linear segments as Segment 1 and Segment 3.
2.1. Using the Biot–Savart Law for a Straight Conductor
For a straight wire and a point O at perpendicular distance $R$ from it, the magnetic field magnitude is given by:
$ B_{\text{straight}} \;=\; \frac{\mu_0\,I}{4\pi R}\,\bigl[\sin\alpha + \sin\beta\bigr],$
where $\alpha$ and $\beta$ are the angles subtended at point O by the two ends of the wire with respect to the perpendicular from O to the wire. For an infinitely long wire going in one direction, $\sin\alpha = 1$ and $\sin\beta = 0$ or vice versa, depending on the orientation.
2.2. Contribution from Each Linear Segment
• In this problem, each long linear segment effectively subtends angles of $90^\circ$ and $0^\circ$ at O. Hence for each segment,
$B_{\text{segment}} = \frac{\mu_0 I}{4\pi R}\bigl[1 + 0\bigr] = \frac{\mu_0 I}{4\pi R}.$
• The direction (given by the right-hand rule) for these segments is along $-\hat{k}$ for Segment 1 and also along $-\hat{k}$ for Segment 3 (because both are parallel to the x-axis and the center O lies in the y-z plane).
Therefore, each linear segment contributes
$\overrightarrow{B_1} = \overrightarrow{B_3} = -\,\frac{\mu_0 I}{4\pi R}\,\hat{k}.$
Step 3: Magnetic Field Contribution from the Semicircular Arc
Now consider the semicircular portion of radius $R$ lying in the y-z plane, centered at O, carrying the same current $I$.
By the known formula for the magnetic field at the center of a circular arc of radius $R$ carrying current $I$, the magnitude is
$B_{\text{arc}} \;=\; \frac{\mu_0 I}{4R}\,\theta,$
where $\theta$ is the angle subtended by the arc in radians. For a semicircle, $\theta = \pi$. Hence,
$B_{\text{arc}} = \frac{\mu_0 I}{4R}\,\pi = \frac{\pi \mu_0 I}{4R}.$
Next, we determine the direction of this field using the right-hand rule (thumb along the direction of current). For a semicircle in the y-z plane with current traveling in such a way that the wire extends along the x-axis at the ends, the direction of $B_{\text{arc}}$ comes out to be along $-\,\hat{i}$ (negative x-direction).
Step 4: Combine All Contributions
1. From Segment 1: $\overrightarrow{B_1} = -\,\frac{\mu_0 I}{4\pi R}\,\hat{k}.$
2. From Segment 3: $\overrightarrow{B_3} = -\,\frac{\mu_0 I}{4\pi R}\,\hat{k}.$
3. From the Semicircular Arc (Segment 2): $\overrightarrow{B_2} = -\,\frac{\pi\mu_0 I}{4\pi R}\,\hat{i} = -\,\frac{\mu_0 I\,\pi}{4\pi R}\,\hat{i}.$
Adding these vectors:
$\overrightarrow{B}
= \overrightarrow{B_1} + \overrightarrow{B_2} + \overrightarrow{B_3}
= -\,\frac{\mu_0 I}{4\pi R}\,\hat{k} \;+\; \bigl[-\,\frac{\mu_0 I\,\pi}{4\pi R}\,\hat{i}\bigr] \;+\; \Bigl(-\,\frac{\mu_0 I}{4\pi R}\,\hat{k}\Bigr).$
Combine like terms:
$\overrightarrow{B}
= -\,\frac{\mu_0 I\,\pi}{4\pi R}\,\hat{i} \;+\; \Bigl[-\,\frac{\mu_0 I}{4\pi R}\hat{k} -\,\frac{\mu_0 I}{4\pi R}\hat{k}\Bigr]
= -\,\frac{\mu_0 I}{4\pi R} \Bigl(\pi\,\hat{i} + 2\,\hat{k}\Bigr).
Step 5: State the Final Result
Thus, the magnetic field at point O is
$\displaystyle \overrightarrow{B}
= -\,\frac{\mu_0 I}{4\pi R}\,\bigl(\pi\,\hat{i} + 2\,\hat{k}\bigr).
Answer
$\overrightarrow{B} = -\,\frac{\mu_0 I}{4\pi R}\,\bigl(\pi\,\hat{i} + 2\,\hat{k}\bigr).$
