The magnetic field at the centre of a closely wound circular coil of 100 turns, radius 22 cm, carrying a current of 3.5 A will be

$1 × 10^{-3} T$

The correct answer is Option (3) → $1 × 10^{-3} T$

The magnetic field at the centre of a circular coil is

$B = \frac{\mu_0 N I}{2R}$

Given:
$N = 100$, $I = 3.5\,A$, $R = 22\,cm = 0.22\,m$, $\mu_0 = 4\pi \times 10^{-7}\,T\,m/A$

$B = \frac{4\pi \times 10^{-7} \times 100 \times 3.5}{2 \times 0.22}$

$B = \frac{4\pi \times 10^{-7} \times 350}{0.44}$

$B = 4\pi \times 10^{-7} \times 795.45$

$B = 3.18 \times 10^{-4}\pi\,T$

$B = 1.0 \times 10^{-3}\,T$

∴ Magnetic field at the centre = $1.0 \times 10^{-3}\,T$