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
Any straight segment of wire whose geometric line of extension passes directly through the point of observation $O$ creates exactly zero magnetic induction because $\vec{d\ell} \times \vec{r} = 0$.
Step 1: Understanding the Concept:
By applying 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 each distinct segment of the current-carrying wire. The standard geometry for this type of problem consists of a three-quarter circular arc subtending an angle of $\frac{3\pi}{2}$ radians, along with two semi-infinite straight wire lines. Step 2: Key Formula or Approach:
1. Magnetic field due to a circular arc subtending angle $\theta$ at its center: $B_{\text{arc}} = \frac{\mu_0 I}{4\pi r}\theta$
2. Magnetic field due to a semi-infinite straight wire near one end point line: $B_{\text{straight}} = \frac{\mu_0 I}{4\pi r}$
3. Use the right-hand thumb rule to check if fields point in the same direction (add) or opposite direction (subtract). Step 3: Detailed Explanation:
Let's analyze each segment of the configuration carefully:
Segment 1 (Curved Arc): The curved loop forms three-quarters of a circle. The angle subtended at the center is $\theta = \frac{3}{4} \times 2\pi = \frac{3\pi}{2}$ radians.
\[ B_{\text{arc}} = \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2}\right) \quad (\text{pointing inward/outward uniformly depending on circulation direction}) \]
Segment 2 (Straight wire sections): One straight wire section runs along the axis pointing directly towards or away from center $O$, yielding zero magnetic field ($\sin 0^\circ = 0$). The other straight section behaves as a semi-infinite wire whose edge terminates at distance $r$ relative to point $O$:
\[ B_{\text{straight}} = \frac{\mu_0 I}{4\pi r} \]
By applying the Right-Hand Rule, both the arc contribution and the active semi-infinite wire element generate fields that point in the same vector direction. Combining their magnitudes gives:
\[ B_{\text{net}} = B_{\text{arc}} + B_{\text{straight}} \]
\[ B_{\text{net}} = \frac{\mu_0 I}{4\pi r}\left(\frac{3\pi}{2}\right) + \frac{\mu_0 I}{4\pi r}(1) \]
\[ B_{\text{net}} = \frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} + 1 \right) \] Step 4: Final Answer:
The total magnetic field at the centre $O$ is $\frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} + 1 \right)$.
