A circular coil of radius 2.0 cm carries a current of 1.0 A. If the coil has 50 turns, the magnetic field at the centre of the coil will be

$1.57 × 10^{-3} Т$

The correct answer is Option (4) → $1.57 × 10^{-3} Т$

Radius of coil, R = 2.0 cm = 0.02 m

Current, I = 1.0 A

Number of turns, N = 50

Magnetic field at the center of a circular coil is given by:

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

Substitute values ($\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$):

$B = \frac{4\pi \times 10^{-7} \times 50 \times 1}{2 \times 0.02}$

$B = \frac{200 \pi \times 10^{-7}}{0.04}$

$B = 5000 \pi \times 10^{-7} = 5 \pi \times 10^{-4}\ \text{T} = 1.57 × 10^{-3} Т$

∴ Magnetic field at the center = $1.57 × 10^{-3} Т$