A circular coil of wire consisting of 100 turns, each of radius `8*0cm` carries a current of `0*40A`.
What is the magnetude of the magnetic field `vecB` at the centre of the coil?

Explanation

Number of turns on the circular coil, n = 100 , Radius of each turn, r = 8.0 cm = 0.08 cm
Current flowing in the coil, `I=0.4A`
Magnitude of the magnetic field at the centre of the coil is given by the relation,
`|B|=(mu_(0))/(4pi)(2pinI)/r` , `mu_(0)=` Permeability of free space , `=4pixx10^(-7)TmA^(-1)`
`|B|=(4pixx10^(-7))/(4pi)xx(2pixx100xx0.4)/0.08=3.14xx10^(-4)T`
Hence, the magnitude of the magnetic field is `3.14xx10^(-4)T`