Magnetic field due to currentMedium
Question
Two identical long current carrying wires are bent into the shapes shown in the following figures. 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 $\frac{B_{1}}{B_{2}}$ is $\_\_\_\_$ .
Options
A.$\frac{2 + \pi}{1 + \pi}$
B.$\frac{1 + \pi}{1 - \pi}$
C.$\frac{2 + \pi}{1 - \pi}$
D.$\frac{1 + \pi}{2 - \pi}$
Solution
Sol. [Key Concept: Add the straight-segment and semicircular-arc field contributions at each centre.]
$$\ B_{1} = \frac{\mu_{0}I}{4\pi r} \times 2 + \frac{\mu_{0}I}{4r} = \frac{\mu_{0}I}{4r}\left( \frac{2}{\pi} + 1 \right) $$
$$B_{2} = \frac{\mu_{0}I}{4\pi r} + \frac{\mu_{0}I}{4r} = \frac{\mu_{0}I}{4r}\left( \frac{1}{\pi} + 1 \right) $$
$$\frac{B_{1}}{{\text{ }B}_{2}} = \frac{2 + \pi}{1 + \pi} $$
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