A wire $A$, bent in the shape of an arc of a circle, carrying a current of $2\ \text{A}$ and having radius $2\ \text{cm}$ and another wire $B$, also bent in the shape of an arc of a circle, carrying a current of $3\ \text{A}$ and having radius of $4\ \text{cm}$, are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and $B$ at the common centre $O$ is:
Answer: (D) $6 : 5$
Field at the centre of an arc subtending angle $\theta$: $B = \dfrac{\mu_0I\theta}{4\pi r}$ (the straight radial leads give no field at $O$).
Arc $A$ subtends $360^\circ - 90^\circ = 270^\circ$; arc $B$ subtends $360^\circ - 60^\circ = 300^\circ$.
$$\frac{B_A}{B_B} = \frac{I_A\theta_A/r_A}{I_B\theta_B/r_B} = \frac{2\times270/2}{3\times300/4} = \frac{270}{225} = \frac65$$
Solution by Sreeraj P, M.Sc Physics