Question

Two Concentric circular coils `X` and `Y` of radii `16 cm` and `10cm` lie in the same vertical plane containing `N-S` direction `X` has 20 turns and carries `16A .Y` has `25` turns & carries `18 A.X` has current in anticlockwise direction and `Y` has current in clockwise direction for the observer, looking at the coils facing the west. The magnitude of net magnetic field at their common center is?

Answer 1

Radius of coil `X, r_(1) = 16 cm = 0.16 m`
Radius of coil `Y, r_(2) = 10 cm = 0.1m`
Number of turns of on coil `X, n_(1) = 20`
Number of turns of on coil `Y, n_(2) = 25`
Current in coil `X, I_(1) = 16A`
Current in coil `Y, I_(2) = 18A`
Magnetic field due to coil X at their centre is given by the relation,
`B_(1) = (mu_(0)n_(1)I_(1))/(2r_(1))`
Where,
`mu_(0) =` Permeability of free space `= 4pi xx 10^(-7) TmA^(-1)`
`:. B_(1) = (4pi xx 10^(-7) xx 20 xx 16)/(2 xx 0.16)`
`= 4pi xx 10^(-4)T` (towards East)
Magnetic field due to coil Y at their centre is given by the relation,
`B_(2) = (mu_(0)n_(2)I_(2))/(2r_(2))`
`= (4pi xx 10^(-7) xx 25 xx 18)/(2 xx 0.10)`
`=9pi xx 10^(-4)T` (towards West)
Hence, net magnetic field can be obtained as:
`B = B_(2)-B_(1)`
`= 9pi xx 10^(-4) -4pi xx 10^(-4)`
`= 5pi xx 10^(-4)T`
`1.57 xx 10^(-3)T` (towards West)