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Question

(a) Using Biot-Savart's law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop
(b) What does a toroid consist of? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero.

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Solution

I current
R Radii
X Axis
x Distance of OP
dl Conducting element of the loop
According to Biot-Savart's law, the magnetic field at P is
db=μ0I|dl×r|4πr3
r2=x2+R2
|dl×r|=rdl(they are perpendicular)
dB=μ04πIdl(x2+R2)
dB has two components dBx and dB.dB is cancelled out and only the x-component remains dBx=dBcosθ cosθ=R(x2+R2)1/2
dBx=μ0Idl4πfrR(x2+R2)1/2
Summation of dl over the loop is given by 2πR
B=Bxi=μ0IR22(x2+R2)3/2i
(b) Toriod is a hollow circular ring on which a large number of turns of wire are closely wound.Three circular Amperian loops 1,2 and 3 are shown by dashed lines.
By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop.
Let the magnetic field inside the toroid be B. We shall now consider the magnetic field at S.By Ampere's Law,
¯B.¯dI=μ0IBL=μ0NI
Where L is the length of the loop for which B is tangential I be the current enclosed by the loop and N be the number of turns.
We find, L=2πr
The current enclosed I is NL
B(2πr)=μ0NI,therefore,B=μ0NI2πr
For a loop inside the toroid, no current exists thus, I=0 Hences, B=0
Exterior to the toroid :
Each turn of current coming out of the plane of the paper is cancelled exactly by the current going into it. Thus I=0,and,B=0

561065_502667_ans_5c6478eac35b49a6a05d408ff6507ec6.png

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