Question

find magnetic field at a point inside a current carrying solenoid by using ampere's circuital law

Answer 1

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 PP in 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 B3. 
 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