Obtain an expression for magnetic induction along the axis of toroid on the basis of Ampere's circuital law.

Explanation

(ii) Magnetic induction along the axis of toroid : Consider a toridal solenoid of radius r having centre O, carrying current I.
According to Apmere's circuital law,
`ointvec(B).vec(dl)=mu_(0)I`

Now, current I flows through the ring as many times as there are number of turns. If N is the total number of turns, then the tatal current flowing through toroid is NI.
`ointvec(B).vec(dl)=mu_(0)(NI)" "......(i)`
`vec(B)andvec(dl)` are in same direction, therefore `theta=0^(@)cos0^(@)=1`
`ointvec(B).vec(dl)=B.dl=B(2pir)" "......(ii)`
Comparing equations (i) and (ii) we get
`mu_(0)NI=B(2pir)`
`B=(mu_(0)NI)/(2pir)" "......(iii)`
Now, if n is number of turns per unit length of toroid, then `n=(N)/(2pir)`.
Putting the value of `n=(N)/(2pir)` in equation (iii), we get
`B=mu_(0)nI`.
which is the expression for magnetic induction along the axis of toroid.