A very long conducting wire is bent in a semi-circular shape from A to B as shown in the figure. The magnetic

Question

A very long conducting wire is bent in a semi-circular shape from A to B as shown in the figure. The magnetic field at the point P for steady current configuration is given by:

Accepted Answer

1.  **Analyze the Geometry:** The wire consists of three segments: two semi-infinite straight wires and one semi-circular arc. Based on the answer format ($1 - 2/\pi$), the fields from the straight segments must oppose the field from the arc.
2.  **Field due to Straight Segments:** Consider the two semi-infinite straight wires extending to infinity. The magnetic field $B_1$ due to a semi-infinite wire at a perpendicular distance $R$ from one end is given by $B_1 = \frac{\mu_0 I}{4\pi R}$. Since there are two such wires carrying current in directions that produce magnetic fields in the same direction (opposing the arc), their combined field is:
    $$B_{straight} = 2 	imes \frac{\mu_0 I}{4\pi R} = \frac{\mu_0 I}{2\pi R}$$
3.  **Field due to Semi-Circular Arc:** The magnetic field $B_{arc}$ at the center of a semi-circular arc of radius $R$ is half that of a full circle:
    $$B_{arc} = \frac{1}{2} \left( \frac{\mu_0 I}{2R} ight) = \frac{\mu_0 I}{4R}$$
4.  **Net Magnetic Field:** The net magnetic field is the difference between the arc field and the straight wire field (assuming the standard configuration where they oppose):
    $$B_{net} = B_{arc} - B_{straight} = \frac{\mu_0 I}{4R} - \frac{\mu_0 I}{2\pi R}$$
    Factor out $\frac{\mu_0 I}{4R}$:
    $$B_{net} = \frac{\mu_0 I}{4R} \left( 1 - \frac{2}{\pi} ight)$$
5.  **Direction:** The term $1$ (from the arc) is greater than $\frac{2}{\pi} \approx 0.63$ (from the wires). Therefore, the net field direction is determined by the arc. If the arc current is counter-clockwise, the field points **out of the page** (away). The option 'pointed away from the page' corresponds to this standard case.

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