Question

How to find magnetic field at axis using ampere's circuital law?

Answer 1

A solenoid bent into the form of a closed ring is called a toroidal solenoid. Alternatively, it is an anchor ring around which a large number of turns of a metallic wire are wound. We shall see that the magnetic field B→B→ has a constant magnitude everywhere inside the toroid while it is zero in the open space interior (point PP) and exterior (point QQ) to the toroid.  

The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Three circular Amperean loops as shown by dashed lines. By symmetry, the magnetic field should be tangential to them and constant in magnitude for each of the loops. 

1.For Points in the open space interior to the toroid. Let B1B1 be the magnitude of the magnetic field along the Amperean loop 1 of the radius r1r1 

   Length of the loop 1 L1=2πr1L1=2πr1 
  As the loop encloses no current, so I=0I=0 
  Applying Ampere's circuital law, 
                  B1L1=μ0IB1L1=μ0I 
  Or           B1×2πr1=μ0×0B1×2πr1=μ0×0 
  Or           B1=0B1=0 
  
Thus the magnetic field at any point PPin the open space interior to the toroid is zero. 

2.For points inside the toroid. Let BB be the magnitude of the magnetic field along the Amperean loop 22 of radius rr. 

   Length of loop 22, L2=L2= 2πr2πr 

If NN is the total number of turns in the toroid and II the current in the toroid,then total current enclosed by the loop 2=NI2=NI


Applying Ampere’s circuital law, 
    
    B×2πr=μ0×NIB×2πr=μ0×NI 
or        B=μ0NI2πrB=μ0NI2πr 

If rr be the average radius of the toroid and nn the number of turns per unit length, then 

           N=2πrnN=2πrn 
    ∴B=μ0nI∴B=μ0nI 

3.For points in the open space exterior to the toroid. 
Each turns of the toroid passes twice through the area enclosed by the Amperean loop 33. 
 But for each turns, the current coming out of the plane of paper is cancelled by the current going into the place of paper.  
Thus I=0I=0 and hence B3=0B3=0