Question

Two identical long current carrying wires are bent into the shapes shown. If the magnitude of magnetic fields at the centres \(P\) and \(Q\) of a semicircular arc are \(B_1\) and \(B_2\) respectively, then the ratio \( \dfrac{B_1}{B_2} \) is:

• A. \( \dfrac{2+\pi}{1+\pi} \)
• B. \( \dfrac{1+\pi}{1-\pi} \)
• C. \( \dfrac{2+\pi}{1-\pi} \)
• D. \( \dfrac{1+\pi}{2-\pi} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Magnetic field at the center of a semicircular loop is \(B_{semi} = \frac{\mu_0 I}{4R}\).
Magnetic field due to a semi-infinite wire at a point perpendicular to its end is \(B_{wire} = \frac{\mu_0 I}{4\pi R}\).
Step 3: Detailed Explanation:
In figure (I), for point P:
The field is due to two semi-infinite wires and one semicircular arc. All three fields are in the same direction (into the page).
\[ B_1 = \frac{\mu_0 I}{4R} + 2 \times \frac{\mu_0 I}{4\pi R} = \frac{\mu_0 I}{4\pi R} (\pi + 2) \].
In figure (II), for point Q:
The field is due to one semi-infinite wire and one semicircular arc. The lower straight wire points towards Q, so its magnetic field contribution is zero.
\[ B_2 = \frac{\mu_0 I}{4R} + \frac{\mu_0 I}{4\pi R} = \frac{\mu_0 I}{4\pi R} (\pi + 1) \].
The ratio \(\frac{B_1}{B_2}\) is:
\[ \frac{B_1}{B_2} = \frac{\pi + 2}{\pi + 1} \].
Step 4: Final Answer:
The ratio \(\frac{B_1}{B_2}\) is \(\frac{2+\pi}{1+\pi}\).