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 $\dfrac{B_1}{B_2}$ is ______.
Answer: (A) $\dfrac{2+\pi}{1+\pi}$
Use: a semi-infinite straight wire ending at the foot of the perpendicular from the point gives $\dfrac{\mu_0I}{4\pi r}$; a semicircle gives $\dfrac{\mu_0I}{4r}$ at its centre; a straight wire whose line passes through the point gives zero.
**Figure (I):** two semi-infinite wires and the semicircle. All three fields at P point the same way (into or out of the page together):
$$B_1=2\cdot\frac{\mu_0I}{4\pi r}+\frac{\mu_0I}{4r}=\frac{\mu_0I}{4\pi r}(2+\pi)$$
**Figure (II):** the horizontal semi-infinite wire and the semicircle contribute; the vertical wire lies along a line through Q, so it contributes nothing:
$$B_2=\frac{\mu_0I}{4\pi r}+\frac{\mu_0I}{4r}=\frac{\mu_0I}{4\pi r}(1+\pi)$$
$$\frac{B_1}{B_2}=\frac{2+\pi}{1+\pi}$$
Solution by Sreeraj P, M.Sc Physics