Two concentric coil X and Y of radii `16cm` and `10cm` respectively lie in the same vertical plane containing the north-south direction. Coil X has 20 turns and carries a current of 16A, coil Y has 25 turns and carries a current of 18A. The sense of current in X is anti-clockwise and in Y, clockwise, for an observer looking at the coil facing west, Figure. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Explanation

The magnitude of magnetic field at the centre of a coil of radius r, number of turns , and carrying a current I is given by:
`B = mu_(0)/(4pi) (2 pi n I)/r`

Consider that the wire is lying in the vertical plane containing north-south direction. The observer is facing towards west and his back is towards east.
For coil X :
`I = 16 A`
` n = 20 , r = 16/100 m`
` :. " " B_(X) = 10^(-7) xx (2pi xx 20 xx 16)/(16/100) `
` = (10^(-7) xx 2 pi xx 100 xx 16 xx 20)/16`
` = 4pi xx 10^(-4) T`
By the right-hand rule, the direction of magnetic field is towards east.
For coil Y :
`I = 18 A, n = 25, r = 10/100 m`
` :. " " B_(Y) = mu_(0)/(4 pi) xx (2 pi n I)/r `
` = 10^(-7) xx (2 pi xx 18 xx 25)/[10/100] `
` = 10^(-7) xx (2 pi xx 25 xx 18 xx 100)/10`
` = 9pi xx 10^(-4) T`
By right-hand rule, the direction of magnetic field is towards west.
Net magnetic field ` = B_(Y) - B_(X)`
` = 9pi xx 10^(-4) T - 4pi xx 10^(-4) T`
` = 5pi xx 10^(-4)`
` = 1.6 xx 10^(-3) T` (towards west)