The radius of the arc BD representing a wire is 60 cm. The magnitude of the magnetic field produced at centre C, if a current of 3.0 A flows in the wire, will be

$\frac{5\pi}{6} × 10^{-7}T$

The correct answer is Option (1) → $\frac{5\pi}{6} × 10^{-7}T$

Given: $R=60\ \text{cm}=0.6\ \text{m}$, $I=3.0\ \text{A}$, $\theta=30^\circ=\frac{\pi}{6}\ \text{rad}$

$B=\frac{\mu_0 I \theta}{4\pi R}=\frac{4\pi\times10^{-7}\cdot 3\cdot (\pi/6)}{4\pi\cdot 0.6}=\frac{\pi}{1.2}\times10^{-7}=\frac{5\pi}{6}\times10^{-7}\ \text{T}$

∴ Magnetic field at C = $\frac{5\pi}{6}\times10^{-7}\ \text{T}$