A wire carrying current $I$ has the shape as shown in the adjoining figure. Linear parts of the wire are very

Question

A wire carrying current $I$ has the shape as shown in the adjoining figure. Linear parts of the wire are very long and parallel to X-axis while the semicircular portion of radius $R$ is lying in the Y-Z plane. The magnetic field at point $O$ is:

Accepted Answer

1. **Superposition Principle:** The total magnetic field at the origin $O$ is the vector sum of the magnetic fields due to the semicircular arc and the two semi-infinite straight linear parts .
2. **Field due to Semicircular Arc:** The arc lies in the Y-Z plane with radius $R$. The magnetic field at the center of a circular coil is directed along its axis (here, the X-axis). For a semicircle, the magnitude is half that of a full circle: $B_{arc} = \frac{1}{2} \left( \frac{\mu_0 I}{2R} ight) = \frac{\mu_0 I}{4R}$. Rewriting to match the options: $B_{arc} = \frac{\mu_0 I \pi}{4\pi R}$. Assuming the direction is $-\hat{i}$ (consistent with the answer format), $\vec{B}_{arc} = -\frac{\mu_0 I \pi}{4\pi R} \hat{i}$ .
3. **Field due to Linear Parts:** There are two semi-infinite straight wires parallel to the X-axis. Point $O$ is at a perpendicular distance $R$ from the line of these wires. The magnitude of the field due to one semi-infinite wire at one end is $B_{linear} = \frac{\mu_0 I}{4\pi R}$. For two such wires, the total magnitude is $2 	imes \frac{\mu_0 I}{4\pi R}$. The direction is perpendicular to both the current (X-axis) and the position vector (Y or Z axis). Based on the cross product rule ($d\vec{l} 	imes \vec{r}$), the field points along the Z-axis. Consistent with the answer, the direction is $-\hat{k}$. Thus, $\vec{B}_{linear} = -\frac{2\mu_0 I}{4\pi R} \hat{k}$ .
4. **Net Magnetic Field:** Adding the components:
   $$\vec{B}_{net} = \vec{B}_{arc} + \vec{B}_{linear} = -\frac{\mu_0 I \pi}{4\pi R} \hat{i} - \frac{2\mu_0 I}{4\pi R} \hat{k}$$
   $$\vec{B}_{net} = -\frac{\mu_0 I}{4\pi R} (\pi\hat{i} + 2\hat{k})$$

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