The magnetic flux at two points on the axis of a circular coil at distances 5 cm and 20 cm from the centre are in the ratio 8:1. Find the diameter of the coil.

Explanation

Given `x_(1)=5 cm, x_(2)=20 cm`
`" "B_(1)/B_(2)=8/1`
But, `" "B=(mu_(0)/(4pi))((2pinir^(2))/(r^(2)+x^(2))^(3//2))`
Hence, `" "B_(1)/B_(2)=8/1=((r^(2)+x_(2)^(2))/(r^(2)+x_(1)^(2)))^(3//2)`
Taking cube root on both sides we write
`" "2=((r^(2)+x_(2)^(2))/(r^(2)+x_(1)^(2)))^(1//2)`
Squaring both sides we get,
`" "(r^(2)+x_(2)^(2))/(r^(2)+x_(1)^(2))=4`
`" "r^(2)+x_(2)^(2)=4r^(2)+4x_(1)^(2)`
`" "3r^(2)=(20)^(2)-4(5)^(2)`
`" "3r^(2)=300`
`therefore " "r=pm 10cm`
Diameter of the T.G. coil = 0.2 m.