Question

A STRAIGHT WIRE CARRYING CURRENT OF 12A IS BENT IN TO A SEMICIRCULAR ARC OF RADIUS 2 CM WHAT IS THE MAGNETIC FIELD DUE TO STRAIGHT SEGMENTS

Answer 1

The magnitude of magnetic field at the centre of the current carrying coil is given by
$B=\frac{{\mu }_0\times 2\pi I}{4\pi r}$
Since the wire is bent into semicircular arc of radius 2 cm, the equation become

$B=\frac{1}{2}\times \frac{{\mu }_0}{4\pi }\times \frac{2\pi I}{r}=\frac{{\mu }_0}{4\pi }\times \frac{\pi I}{r}={10}^{-7}\times \frac{\pi \times 12}{2\times {10}^{-2}}=1.885\times {10}^{-4}T$