Question

find out the expression for the magnetic field at a point on the axis of a toroid of N turns having average radius r and carrying a current I, show that the magnetic field in the open space inside and outside the toroid is zero.​

Answer 1

Answer:

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=  4πr3 μ0 I∣dl×r∣ r2=x2 +R2

 ∣dl×r∣=rdl(they are perpendicular)

∴dB=  4π μ0 (x2 +R2  )Idl

dB has two components −dBx  and dB⊥

dB⊥  is cancelled out and only the x-component remains ∴dBx  =dBcosθ cosθ=  (x2 +R2  ) 1/2 R

​dBx  =  4π μ0  Idl

​fr(x2 +R2  )  1/2 R

Summation of dl over the loop is given by 2πR

∴B=Bxi=  2(x2 +R2  )  3/2 μ0 IR2  i

(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,

∫Bˉ.dIˉ =μ0

IBL=μ0 NI

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)=μ0

NI,therefore,B= 2πr μ0  NL

​  

 

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

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