State Ampere’s circuital law and prove it for the magnetic field produced by a straight curre
Answers
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→=Bdlcos0o=Bdl
Hence the line integral of the magnetic field along the circular path is
∲B→.dl→=∲Bdl=B∲dl=μ