Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius '*r*', having '*n*' turns per unit length and carrying a steady current *I*.

#### Solution

A toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. Consider an air-cored toroid (as shown above) with centre O.

Given:*r =* Average radius of the toroid*I = *Current through the solenoid*n* = Number of turns per unit length

To determine the magnetic field inside the toroid, we consider three amperian loops (loop 1, loop 2 and loop 3) as show in the figure below.

For loop 1:

According to Ampere's circuital law, we have

`ointvecB.vec(dl)=mu_0(`

Total current for loop 1 is zero because no current is passing through this loop.

So, for loop 1

`oint.vecB.vec(dl)=0`

For loop 3:

According to Ampere's circuital law, we have

`ointvecB.vec(dl)=mu_0(`

Total current for loop 3 is zero because net current coming out of this loop is equal to the net current going inside the loop.

For loop 2:

The total current flowing through the toroid is *NI,* where *N* is the total number of turns.

`ointvecB.vec(dl)=mu_0(NI)`

Now, `vecB `

`ointvecB.vec(dl)=Bointdl`

`=>ointvecB.vec(dl)=B(2pir)`

Comparing (i) and (ii), we get

B(2πr)=μ_{0}NI

`=>B=(mu_0NI)/(2pir)`

Number of turns per unit length is given by

`n=N(2pir)`

** ∴B=μ _{0}nI**

This is the expression for magnetic field inside air-cored toroid.