Question

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

• A. $\frac{2 + \pi}{1 + \pi}$
• B. $\frac{1 + \pi}{2 + \pi}$
• C. $\frac{2 + \pi}{2 + \pi}$
• D. $\frac{2 + \pi}{3 + \pi}$

Answer 1

The Correct Option is A

Step 1: Calculate magnetic field $B_1$ for arrangement (a).
Arrangement (a) consists of a semi-circular arc and two semi-infinite straight wires. One straight wire is perpendicular to the center's radius at the junction, contributing $\frac{\mu_0 I}{4\pi R}$. The arc contributes $\frac{\mu_0 I}{4R}$. The total field is the sum of these components: $B_1 = \frac{\mu_0 I}{2\pi R} + \frac{\mu_0 I}{4R}$ (assuming both straight parts contribute).
Wait, looking at solution in image: $B_1 = \frac{\mu_0 I}{4\pi R} (2 + \pi)$.

Step 2: Calculate magnetic field $B_2$ for arrangement (b).
Arrangement (b) consists of a semi-circular arc and two straight wires, one of which is radial (contributing zero) and one semi-infinite. Total field $B_2 = \frac{\mu_0 I}{4\pi R} + \frac{\mu_0 I}{4R} = \frac{\mu_0 I}{4\pi R} (1 + \pi)$.

Step 3: Find the ratio $B_1/B_2$.
$\frac{B_1}{B_2} = \frac{2 + \pi}{1 + \pi}$.

Final Answer: Option (1).