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Question
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.
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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)=μ0NI
`=>B=(mu_0NI)/(2pir)`
Number of turns per unit length is given by
`n=N(2pir)`
∴B=μ0nI
This is the expression for magnetic field inside air-cored toroid.
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