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Step-by-Step Solution
Step 1: Identify the Current-Carrying Segments
The conducting wire is bent in a semi-circular shape from point A to point B, with a steady current $i$ flowing through it. The point P is at the center of the semi-circle. We need to calculate the net magnetic field at point P due to all segments of the wire.
Step 2: Magnetic Field Contribution from the Semi-Circular Arc
1. The radius of the semi-circular arc is $R$.
2. The magnetic field at the center of a full circular loop carrying current $i$ and having radius $R$ would be
$
B_{\text{full circle}} \;=\; \frac{\mu_0\;i}{2\,R}.
$
3. Because we only have a semi-circular arc, its contribution is half of that of a full circle:
$
B_{\text{semi-circle}} \;=\; \frac{\mu_0\;i}{4\,R}.
$
4. The direction of this magnetic field (using the right-hand rule) is into or out of the page, depending on the direction of current. We will combine directions at the end.
Step 3: Magnetic Field Contribution from the Remaining Straight Segments
1. The wire extends infinitely on both sides of the semi-circular portion, effectively giving two infinite straight wire segments (one from A outward and the other from B outward, each meeting at the center in the semi-circular portion).
2. Each infinitely long straight conductor contributes a magnetic field at point P. However, because point P is exactly at the center of the semicircle, the geometry for these infinite segments is such that each segment subtends some finite angle at P.
3. By standard results or by using the Biot–Savart law for an infinite straight wire, the magnetic field at a perpendicular distance $R$ from the wire is
$
B_{\text{infinite wire}} \;=\; \frac{\mu_0\,i}{2\,\pi\,R}.
$
But here, due to the geometry, these two wires do not act as complete infinite wires for the point P; rather, each subtends specific angles. A detailed calculation shows that the net field from these two straight segments together is
$
B_{\text{straight segments}} \;=\; \frac{\mu_0\,i}{4\,R}\left( \frac{2}{\pi} \right).
$
(The negative sign of this contribution will emerge when combining directions if it happens to oppose the arc's field.)
Step 4: Combine the Contributions
1. The field from the semi-circular arc is
$
\frac{\mu_0\,i}{4\,R}.
$
2. The field from the two straight segments combined is
$
\frac{\mu_0\,i}{4\,R}\left( \frac{2}{\pi} \right)
$
in the opposite direction to the arc’s contribution (based on the right-hand rule).
3. Hence, the net magnetic field at P becomes
$
B_{\text{net}}
\;=\; \frac{\mu_0\,i}{4\,R}
\;-\; \frac{\mu_0\,i}{4\,R}\left(\frac{2}{\pi}\right)
\;=\; \frac{\mu_0\,i}{4\,R}\left[\,1 \;-\;\frac{2}{\pi}\right].
$
Step 5: Determine the Direction
By the right-hand rule for the given current direction, this resultant field direction is away from the page (outward). Therefore, the final magnetic field at point P is
$
\displaystyle
\frac{\mu_0\,i}{4\,R}
\left[\,1 \;-\;\frac{2}{\pi}\right]
\text{ pointed away from the page.}
$
