Using Ampere's circuital law, derive an expression for the magnetic field along the axis of a toroidal solenoid.

Explanation

Ampere.s Circuital Law : This law states that line integral of magnetic field over a closed circuit is equal to Ho (absolute permeability of free space) times the current threading the circuit.
`oint vecB. vec(dl) = mu_0 I`
Magnetic field due to current carrying toroid : Consider a toroid which is an anchor ring of radius r with centre O. Large number of turns of metallic wire are wound over this anchor shaped ring.
Let I = Current flowing through this toroidal solenoid.
According to Ampere.s circuital law,
`oint vecB. vec(dl) = mu_0 I`
`oint vecB. vec(dl) = mu_0 xx` net current enclosed of the circle of radius r.
`oint vecB. vec(dl) = mu_0(n xx 2pi r) I" " ....(I)`
where `" "` n= Number of turns per unit length of the toroid
`n xx 2 pir` = Total number of turns in the toroid
Also at any point on the toroid, magnetic field is along the tangent to the ring.
It means angle between `vecB` and `vec(dl) = theta = 0^@`
From eqn. (1),
`oint B dl cos theta = mu_0 (2pi rn) I`
`implies B oint dl = mu_0(2 pi n)I`
`implies B (2pi r) = mu_0 (2pi n) I`
`B = mu_0 n I`
which is the expression for magnetic field due to a current carrying toroid.