#### Question

Obtain an expression for magnetic induction along the axis of the toroid.

#### Solution

**Magnetic induction along the axis of toroid:**

The toroid is a solenoid bent into a shape of the hollow doughnut.

Consider a toroidal solenoid of average radius ‘r’ having center carrying the current I. In order to find magnetic field produced at the center along the axis of toroid due to the current flowing through the coil, imagine an Amperial loop of radius ‘r’ and traverse it in the clockwise direction.

According to Ampere’s circuital law,

`ointvecB.vec(dL)=mu_0I`

Here current I flow through the ring as many times time as there are the number of turns. Thus the total current flowing through toroid is N I , where N is the total number of turns.

`:.ointvecB.vec(dL)=mu_0NI" ---------(1)"`

Now, and are in same direction `:.ointvecB.vec(dL)=BointdL`

`:.ointvecB.vec(dL)=B(2pir)" ------(2)"`

Comparing equation (1) and equation (2 )

μ_{0}NI=B(2πr) `:.B=(mu_0NI)/(2pir)` .......(3)

If ‘n’ is the number of turns per unit length of toroid then `n=N/(2pir)`

Substituting this value in equation No (3) we get B = μ_{0} n I