A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

The number of coil turns (n) is 100.

Each turn’s radius (r) is 8 cm (0.08 m).

The magnitude of the current (I) passing through the coil is 0.4 A.

The magnitude of the magnetic field at the coil’s centre may be calculated using the following equation:

|\overline B | = \frac{{{\mu _0}2\pi nI}}{{4\pi r}}

Where ​{\mu _0} is the permeability of free space = 4\pi  \times {10^{ - 7}}Tm{A^{ - 1}}

hence,

|\overline B | = \frac{{4\pi  \times {{10}^{ - 7}}}}{{4\pi }} \times \frac{{2\pi  \times 100 \times 0.4}}{{0.08}}

= 3.14 \times {10^{ - 4}}T

The magnitude of the magnetic field is 3.14 \times {10^{ - 4}}T