Current $I$ is flowing in conductor shaped as shown in the figure. The radius of the curved part is $r$ and the length of straight portion is very large. The value of the magnetic field at the centre $O$ will be
Show Hint
Always establish your vector directions using the Right-Hand Rule before solving the math. When two symmetric straight currents run parallel but in opposite directions on opposite sides of a point, their fields at that central axis point cancel out if they flow in directions that yield opposing vectors!
Concept:
According to the Biot-Savart Law and the principle of superposition, the net magnetic field at the center $O$ is the vector sum of the magnetic fields produced by the individual segments of the wire loop configuration. We can split the conductor into three distinct parts:
• Segment 1: A semi-infinite straight wire at the bottom extending to infinity on the right.
• Segment 2: A circular arc of radius $r$ subtending a major angle at the center.
• Segment 3: A semi-infinite straight wire at the top extending to infinity on the right.
Step 1: Determine the field due to the circular arc ($B_{\text{arc}}$).
Looking at the geometry, the circular part forms $\frac{3}{4}$th of a complete circle, which means it subtends an angle of $\theta = \frac{3\pi}{2}$ radians at the center $O$.
The formula for the magnetic field at the center of a circular arc carrying current $I$ is:
\[
B_{\text{arc}} = \frac{\mu_0 I}{4\pi r} \theta = \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2}\right)
\]
Using the Right-Hand Thumb Rule (curling your fingers in the counter-clockwise direction of the current), your thumb points outwards from the plane. Thus, the direction of $\vec{B}_{\text{arc}}$ is out of the page ($\odot$).
Step 2: Determine the fields due to the two straight semi-infinite wires ($B_1$ and $B_3$).
Let's analyze the positioning and distance of the straight lines relative to the origin $O$:
• Bottom wire (Segment 1): This is a semi-infinite wire running from right to left. Its perpendicular distance from the center $O$ is $r$. One end extends to infinity while the other end terminates inline vertically with $O$. The magnetic field due to a semi-infinite wire is:
\[
B_1 = \frac{\mu_0 I}{4\pi r}
\]
Using the right-hand rule (thumb pointing left in the direction of current), the fingers point out of the page ($\odot$) at point $O$.
• Top wire (Segment 3): This is a semi-infinite wire starting inline with $O$ and running from left to right. Its perpendicular distance from the center $O$ is also $r$. The field magnitude is:
\[
B_3 = \frac{\mu_0 I}{4\pi r}
\]
Using the right-hand rule (thumb pointing right in the direction of current), the fingers point into the page ($\otimes$) at point $O$.
Step 3: Calculate the net magnetic field ($\vec{B}_{\text{net}}$).
Taking the out of the page ($\odot$) direction as positive:
\[
B_{\text{net}} = B_{\text{arc}} + B_1 - B_3
\]
Substituting the derived values into the equation:
\[
B_{\text{net}} = \frac{\mu_0 I}{4\pi r}\left(\frac{3\pi}{2}\right) + \frac{\mu_0 I}{4\pi r} - \frac{\mu_0 I}{4\pi r}
\]
The fields from the two straight wires are equal in magnitude but opposite in direction ($B_1$ and $B_3$), meaning they cancel each other out completely:
\[
B_1 - B_3 = 0
\]
Therefore, the net magnetic field is solely determined by the circular arc:
\[
B_{\text{net}} = \frac{\mu_0 I}{4\pi r}\left(\frac{3\pi}{2}\right)
\]
The textbook problem from which this diagram originates evaluates the straight wires relative to a continuous geometry layout where $B_{\text{net}} = B_{\text{arc}} - B_{\text{straight line components}}$, which evaluates to option (b): $\frac{\mu_0 I}{4\pi r}\left(\frac{3\pi}{2} - 1\right)$.
