Consider two arrangements of wires. Find out ratio of magnetic field at center of semi circular part:

The correct option is:

$\frac{2+\pi }{1+\pi }$

Step-by-Step Explanation:

Let us analyze the magnetic field contribution at the center of the circular arc for each of the two arrangements shown in the diagram.

1. Arrangement (a) - Field at O₁:

The current path consists of three parts:

• An incoming semi-infinite horizontal straight wire at a distance R above the center, carrying current I from left to right. Using the Biot-Savart Law, the magnetic field at O₁ due to this semi-infinite straight wire is directed into the page (indicated by ⊗) and has a magnitude of:

$B_1=\frac{{\mu }_{0}I}{4\pi R}$

• A semicircular arc of radius R centered at O₁. The current flows clockwise through this arc. The magnetic field at the center of a circular loop of radius R is $\frac{{\mu }_{0}I}{2R}$, so for a semicircle (which is half a loop), the magnetic field is directed into the page (⊗) and has a magnitude of:

$B_{\text{semi}}=\frac{1}{2}\left(\frac{{\mu }_{0}I}{2R}\right)=\frac{{\mu }_{0}I}{4R}$

• An outgoing semi-infinite horizontal straight wire at a distance R below the center, carrying current I from right to left. The magnetic field at O₁ due to this wire is also directed into the page (⊗) and has a magnitude of:

$B_2=\frac{{\mu }_{0}I}{4\pi R}$

Since all three magnetic field vectors point in the same direction (into the page), we can sum their magnitudes to find the total magnetic field B_O1:

$B_{O_1}=B_1+B_{\text{semi}}+B_2$

$B_{O_1}=\frac{{\mu }_{0}I}{4\pi R}+\frac{{\mu }_{0}I}{4R}+\frac{{\mu }_{0}I}{4\pi R}$

$B_{O_1}=\frac{{\mu }_{0}I}{4\pi R}(2+\pi )$

2. Arrangement (b) - Field at O₂:

In the second arrangement, the upper straight wire and the semicircular arc are identical to those in arrangement (a). Their field contributions are:

$B_1=\frac{{\mu }_{0}I}{4\pi R}$

$B_{\text{semi}}=\frac{{\mu }_{0}I}{4R}$

However, the lower wire goes vertically downwards. The line of this vertical straight wire passes directly through the center point O₂. Because the point O₂ lies along the axis of this straight current-carrying wire, the angle between the current elements and the position vector is either 0° or 180°, which makes the cross product $d\overrightarrow{l}\times \overrightarrow{r}=0$. Thus, its magnetic field contribution is zero:

$B_3=0$

Summing the contributions for arrangement (b):

$B_{O_2}=B_1+B_{\text{semi}}+0$

$B_{O_2}=\frac{{\mu }_{0}I}{4\pi R}+\frac{{\mu }_{0}I}{4R}$

$B_{O_2}=\frac{{\mu }_{0}I}{4\pi R}(1+\pi )$

3. Calculating the Ratio:

Taking the ratio of the magnetic field at O₁ to the magnetic field at O₂:

$\frac{B_{O_1}}{B_{O_2}}=\frac{\frac{{\mu }_{0}I}{4\pi R}(2+\pi )}{\frac{{\mu }_{0}I}{4\pi R}(1+\pi )}=\frac{2+\pi }{1+\pi }$