Magnetic Field at Origin Due to Current Carrying Wire Segments in XY Plane

To find the net magnetic field $\vec{B}$ at the origin $O$ due to the current-carrying wire configuration in the $xy$-plane, we evaluate the contribution of each individual wire segment using the Biot-Savart law.

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1. Straight Wire Segments Passing Through or Directed Towards Origin $O$

For any straight wire segment lying along a line that passes directly through the origin $O$, the current element vector $\mathrm{d}\vec{l}$ is parallel or antiparallel to the position vector $\vec{r}$ relative to $O$. Therefore: $\mathrm{d}\vec{l} \times \vec{r} = 0 \implies \vec{B}_{\text{radial}} = 0$

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2. Upper Semi-Circular Arc

• Radius: $R_1 = \frac{L}{2}$
• Angle subtended at origin: $\theta_1 = \pi\text{ rad}$ (semi-circle)
• Direction of current: Clockwise in the upper half-plane ($y > 0$).

By the right-hand rule, a clockwise current produces a magnetic field directed into the page ($-\hat{k}$).

The magnitude of the magnetic field due to this semi-circular arc is: $B_{\text{arc1}} = \frac{\mu_0 I}{4\pi R_1} \theta_1 = \frac{\mu_0 I}{4\pi \left(\frac{L}{2}\right)} (\pi) = \frac{\mu_0 I}{2L}$

Thus, the magnetic field vector is: $\vec{B}_{\text{arc1}} = -\frac{\mu_0 I}{2L} \hat{k}$

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3. Lower Quarter-Circular Arc

• Radius: $R_2 = \frac{L}{4}$
• Angle subtended at origin: $\theta_2 = \frac{\pi}{2}\text{ rad}$ (quarter-circle)
• Direction of current: Clockwise in the fourth quadrant.

Using the right-hand rule, the magnetic field is also directed into the page ($-\hat{k}$).

The magnitude of the magnetic field due to this quarter-circular arc is: $B_{\text{arc2}} = \frac{\mu_0 I}{4\pi R_2} \theta_2 = \frac{\mu_0 I}{4\pi \left(\frac{L}{4}\right)} \left(\frac{\pi}{2}\right) = \frac{\mu_0 I}{2L}$

Thus, the magnetic field vector is: $\vec{B}_{\text{arc2}} = -\frac{\mu_0 I}{2L} \hat{k}$

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4. Vertical Straight Wire Segment

For the finite straight vertical wire segment, the magnetic field magnitude at a perpendicular distance $d$ subtending angles $\theta_1$ and $\theta_2$ at the ends is given by: $B_{\text{straight}} = \frac{\mu_0 I}{4\pi d} (\sin\theta_1 + \sin\theta_2)$

With perpendicular distance $d = L$, $\theta_1 = 0^\circ$, and $\theta_2 = 45^\circ$ ($\sin 45^\circ = \frac{1}{\sqrt{2}}$): $B_{\text{straight}} = \frac{\mu_0 I}{4\pi L} \left(0 + \frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{4\sqrt{2}\pi L}$

By the right-hand rule, the magnetic field due to this straight segment is directed along $-\hat{k}$: $\vec{B}_{\text{straight}} = -\frac{\mu_0 I}{4\sqrt{2}\pi L} \hat{k}$

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5. Net Magnetic Field at Origin $O$

Summing the individual contributions from all segments: $\vec{B}_{\text{net}} = \vec{B}_{\text{arc1}} + \vec{B}_{\text{arc2}} + \vec{B}_{\text{straight}}$

$\vec{B}_{\text{net}} = -\left( \frac{\mu_0 I}{2L} + \frac{\mu_0 I}{2L} + \frac{\mu_0 I}{4\sqrt{2}\pi L} \right) \hat{k}$

$\vec{B}_{\text{net}} = -\frac{\mu_0 I}{L} \left( 1 + \frac{1}{4\sqrt{2}\pi} \right) \hat{k}$

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Conclusion

The correct option is C.