Question

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?

Answer 1

Given:

The number of turns on the coil (n) is 100

The radius of each turn (r) is  8 cm or 0.08 m

The magnitude of the current flowing in the coil (I) is 0.4 A

The magnitude of the magnetic field at the centre of the coil can be obtained by the following relation:

\( |\bar B| = \frac{\mu_{0}\; 2\pi nI}{4\pi r} \)

where \( \mu_{0} \) is  the permeability of free space =\( 4\pi \times 10^{-7}\; T\;m\;A^{-1} \)

hence,

\( |\bar B| = \frac{4\pi \times 10^{-7}}{4\pi }\times \frac{2\pi \times 100\times 0.4}{0.08} = 3.14 \times 10^{-4}\; T \)

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

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