A circular coil of 120 turns has a radius of 18 cm and carries a current of 3 A. What is the magnitude of the magnetic field at a point on the axis of the coil at a distance from the centre equal to the radius of the circular coil .

Explanation

Given :
Number of turns, `N = 120 `
Current , `I = 3A`
Radius of coil , ` r = 18` cm
` = 0.18` m
distance from the centre to a point on axis,
` a = r = 0.18` m
`B = (mu_(0)NIR^(2))/(2[R^(2) + Z^(2)]^(3/2))`
` = (4 pi xx 10^(-7) xx 120 xx 3 xx (0.18)^(2))/(2[(0.18)^(2) + (0.18)^(2)]^(3/2))`
` B = 4pi xx 10^(-4) T`.