An infinite wire has a circular bend of radius $a$ and carries a current $I$ as shown in the figure. The magnitude of the magnetic field at the origin $O$ of the arc is given by
Answer: (D) $\dfrac{\mu_0}{4\pi}\dfrac{I}{a}\left[\dfrac{3\pi}{2} + 1\right]$
Three parts contribute:
- The lower straight wire comes in from infinity and ends at the point directly below $O$ (distance $a$). $O$ lies on the perpendicular through its end, so it gives half of an infinite wire: $\dfrac{\mu_0I}{4\pi a}$.
- The three-quarter circular arc of radius $a$: $\dfrac{3}{4}\cdot\dfrac{\mu_0I}{2a} = \dfrac{\mu_0}{4\pi}\dfrac{I}{a}\cdot\dfrac{3\pi}{2}$.
- The upper straight wire lies on a line through $O$, so it gives zero field at $O$.
The current goes round the arc clockwise and flows leftward along the lower wire below $O$; both give fields into the page, so they add:
$$B = \frac{\mu_0}{4\pi}\frac{I}{a}\left[\frac{3\pi}{2} + 1\right]$$
Solution by Sreeraj P, M.Sc Physics