Question

State ampere's circuital law. Derive magnetic field at a point due to current carrying straight conductor? plz immediately give answer.

Answer 1

Ampere's circuital law states that the line integral of the magnetic field B→B→around any closed circuit is equal to μ0μ0(permeability constant) times the total current II threading or passing through this closed circuit. Mathematically,  

      ∲B→.dl→=μ0I∲B→.dl→=μ0I 

Proof for a straight current  carrying conductor:

Consider an infinitely long straight conductor carrying a current II.From Biot -Savart law,the magnitude of the magnetic field B→B→ due to the current carrying conductor at a  point,distant rrfrom it is given by

  
      B=μ0I2πrB=μ0I2πr  

As shown in fig. the field B→B→ is directed along the circumference of the circle of radius rr with the wire as centre. The magnitude of the field B→B→ is same all points on the circle. To evaluate the line integral of the magnetic field B→B→ along the circle, we consider a small current element dl→dl→ along the circle. At every point on the circle, both B→B→ and dl→dl→ are tangential to the circle so that the angle between them is zero.  

    B→.dl→=Bdlcos0o=BdlB→.dl→=Bdlcos⁡0o=Bdl 

Hence the line integral of the magnetic field along the circular path is  

    ∲B→.dl→=∲Bdl=B∲dl=μ0I2πr.I∲B→.dl→=∲Bdl=B∲dl=μ0I2πr.I 
    =μ0I2πr.2πr=μ0I2πr.2πr 
    ∴∲B→.dl→=μ0I∴∲B→.dl→=μ0I 

This proves Ampere's law. This law is valid for any assembly of current and for any arbitrary closed loop.