In the figure shown, the magnetic field induction at the point $O$ will be
Show Hint
Check the vector directions before performing any algebraic addition. Since the top wire flows left, the arc goes counter-clockwise, and the bottom wire flows right, every single segment creates an additive field pointing out of the page at point $O$. Because the fields work together rather than canceling out, you know immediately that the terms must be linked by a plus sign, which narrows down your choices to options (b) and (c) instantly!
Concept:
Using the Biot-Savart Law and the principle of superposition, the total magnetic field induction ($\vec{B}_{\text{net}}$) at point $O$ equals the vector sum of the individual fields generated by the constituent segments of the conductor.
The configuration can be structurally divided into three geometric parts:
• Segment 1: A top semi-infinite straight wire carrying current $i$ pointing toward the left.
• Segment 2: A semicircular wire arc of radius $r$ carrying current $i$ running counter-clockwise.
• Segment 3: A bottom semi-infinite straight wire carrying current $i$ pointing toward the right.
Step 1: Determine the dimensions and radius of the configuration.
The distance between the two parallel straight wires is explicitly marked as $2r$. Because the curved section is a smooth semicircle connecting these two lines, the diameter of the semicircle is equal to this separation distance ($d = 2r$). Therefore, the radius of the semicircular arc is exactly $r$, and the center of this arc is point $O$. This means point $O$ lies at a perpendicular distance of $r$ from both the top and bottom straight wire segments.
Step 2: Calculate the magnetic field due to the semicircular arc ($B_{\text{arc}}$).
The arc forms a complete semicircle, meaning it subtends an angle of $\theta = \pi$ radians at its center $O$.
The magnetic field at the center of a circular arc is given by:
\[
B_{\text{arc}} = \frac{\mu_0 i}{4\pi r}\theta = \frac{\mu_0 i}{4\pi r}(\pi)
\]
Using the Right-Hand Thumb Rule (curling your right-hand fingers counter-clockwise along the path of the current arrow), your thumb points straight out of the page. Hence, the direction of $\vec{B}_{\text{arc}}$ is out of the page ($\odot$).
Step 3: Calculate the fields due to the two semi-infinite straight wires ($B_1$ and $B_3$).
Let us analyze the position of point $O$ relative to the two straight lines:
• Top wire (Segment 1): The current flows from right to left. The wire acts as a semi-infinite wire whose right end extends to infinity and whose left end terminates directly above point $O$. The field magnitude at a distance $r$ from its edge is:
\[
B_1 = \frac{\mu_0 i}{4\pi r}
\]
Using the Right-Hand Rule (pointing your right thumb to the left along the current arrow), your fingers curl and point out of the page ($\odot$) at point $O$.
• Bottom wire (Segment 3): The current flows from left to right. This is also a semi-infinite wire starting directly below point $O$ and extending to infinity on the right. Its field magnitude at a distance $r$ is:
\[
B_3 = \frac{\mu_0 i}{4\pi r}
\]
Using the Right-Hand Rule (pointing your right thumb to the right along the current arrow), your fingers curl and point out of the page ($\odot$) at point $O$.
Step 4: Sum the components to find the net magnetic field ($\vec{B}_{\text{net}}$).
Since all three individual magnetic field vectors point in the exact same direction (out of the page ($\odot$)), their magnitudes add together directly:
\[
B_{\text{net}} = B_{\text{arc}} + B_1 + B_3
\]
Substituting the expressions we calculated:
\[
B_{\text{net}} = \frac{\mu_0 i}{4\pi r}(\pi) + \frac{\mu_0 i}{4\pi r} + \frac{\mu_0 i}{4\pi r}
\]
Combining the identical straight-wire terms:
\[
B_{\text{net}} = \frac{\mu_0 i}{4\pi r}(\pi) + 2\left(\frac{\mu_0 i}{4\pi r}\right)
\]
Factoring out the common multiplier $\left(\frac{\mu_0}{4\pi}\right)\left(\frac{i}{r}\right)$ to align exactly with the format of the options:
\[
B_{\text{net}} = \left(\frac{\mu_0}{4\pi}\right)\left(\frac{i}{r}\right)(\pi + 2)
\]
