A circular coil of radius 0.08m consisting of 100 turns is carrying a current of 0.4A. Calculate the magnitude of the magnetic field i) at the center of the coil and ii) at a point 0.2m from the center of the coil on its axis.

Explanation

Given : r=0.08m, n = 100 turns, I = 0.4A
i) Magnetic field at the centre of the coil
`B = (mu_0 nI)/(2r)`
`B = (4pi xx 10^(-7) xx 100 xx 0.4)/(2 xx 0.08)`
`B = (502.4 xx 10^(-7))/(0.16)`
`B = 3.140 xx 10^(-7)`
`B = 3.14 xx 10^(-4) T`
ii) At a point 0.2m from the centre of the coil on its axis
`B = (mu_0)/(4pi) (2pi n I r^2)/((x^2 + r^2)^(3//2))`
`B = 10^(-7) xx (2 xx 3.14 xx 100 xx 0.4 xx (0.08)^2)/([(0.2)^2 + (0.08)^2]^(3//2))`
`B = (1.607 xx 10^(-7))/((0.04 + 0.0064)^(3//2))`
`B = (1.607 xx 10^(-7))/(0.3593)`
`B = 4.47 xx 10^(-7) T`