Question

) explain how biot-savart law enables one to express the ampere's circuital law in the integral form, viz., ∮ ????⃗⃗ . ???????? ⃗⃗⃗⃗ = ???????????? where i is the total current passing through the surface

Answer 1

Let any arbitrary closed path perpendicular to the plane of paper around a long straight wire XY , it is carrying current from X to Y lying in the plane of paper. 
   Let the closed path is formed by  large number of small elements 
e.g., AB = dL₁, BC = dL₂, CD = dL₃ .......
angles subtended by all elementary lengths are dθ₁ , dθ₂ , dθ₃.....
also you can see that,
       dθ₁ + dθ₂ + dθ₃ + ......... = 2π ---------------------------(1)
you know one important things, 
         θ = l/r , here θ is angle , r is the radius and l is the length 
⇒ dL₁/r₁ + dL₂/r₂ + dL₃/r₃ + ............ = 2π
 
now, B₁, B₂ , B₃.......  are magnetic field induced at a point along the elements dL₁ , dL₂ , dL₃ .........
  now, use Biot - savart law,
       we know, according to this law, magnetic field for infinite length conductor  is B = μ₀i/2πr, where r is the seperation between conductor and point of observation.
  now, here, B₁ = μ₀i/2πr₁ , B₂ = μ₀i/2πr₂ , B₃ = μ₀i/2πr₃ .........
  now, line integral of B around closed path is ,
                 ∮B.dL = B₁.dL₁ + B₂.dL₂ + B.dL₃ +..........
                            = μ₀i/2πr₁.dL₁ + μ₀i/2πr₂.dL₂ + μ₀i/2πr₃.dL₃ 
                            =  μ₀i/2π [dL₁/r₁ + dL₂/r₂ + dL₃/r₃ +.......] 
                            = μ₀i/2π  [ dθ₁ + dθ₂ + dθ₃ + .........]
                 now, from equation (1)
 ∮B.dL = μ₀i/2π × 2π = μ₀i
e.g.,   ∮B.dL = μ₀i , this is the expression of Ampere circuital law,
  hence, biot-savart law enables one to express the Ampere's circuital law in the integral form