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

Concept: Magnetic field due to:

• Long straight wire: \[ B = \frac{\mu_0 I}{2\pi r} \]
• Circular arc (angle \(\theta\)): \[ B = \frac{\mu_0 I \theta}{4\pi r} \]

For a semicircle: \[ \theta=\pi \] Thus \[ B_{\text{arc}}=\frac{\mu_0 I}{4r} \]
Step 1:Field at point \(P\)} Contribution from two straight sections: \[ B_s=\frac{\mu_0 I}{2\pi r} \] From semicircle: \[ B_c=\frac{\mu_0 I}{4r} \] Total \[ B_1=\frac{\mu_0 I}{4r}+\frac{\mu_0 I}{\pi r} \]
Step 2:Field at point \(Q\)} Here only one straight section contributes with semicircle. \[ B_2=\frac{\mu_0 I}{4r}+\frac{\mu_0 I}{2\pi r} \]
Step 3:Take ratio} \[ \frac{B_1}{B_2} = \frac{\frac{1}{4}+\frac{1}{\pi}} {\frac{1}{4}+\frac{1}{2\pi}} \] Multiply numerator and denominator by \(4\pi\): \[ \frac{B_1}{B_2} = \frac{\pi+4}{\pi+2} \] \[ = \frac{2+\pi}{1+\pi} \] \[ \boxed{\frac{2+\pi}{1+\pi}} \]