Obtain an expression for the magnetic induction at a point due to an infinitely long straight conductor carrying current.

Magnetic induction due to infinitely long straight conductor carrying current: XY, is an infinitely long straight conductor carrying a current I(Figure). P is a point at a distance a from the conductor. AB is a small element of length dl. θ is the angle between the current element Idl And the line joining the element dl and the point P. According to Biot-Savart law, the magnetic induction at the point P due to the current element Idl Is. dB = μ o 4 π Idl r 2 ⋅ sinθ . . . . . . ( 1 ) AC is drawn perpendicular to BP from A. ∠ OPA = ϕ , ∠ APB = dϕ In ΔABC , sinθ = ABAC = dlAC ∴ AC = dlsinθ . . . . . . . . . . ( 2 ) From ΔAPC , AC = rdϕ . . . . . . . . . ( 3 ) From equations (2) and (3), rdϕ = dlsinθ . . . . . . . . . . . . ( 4 ) Substituting equation (4) in equation (1) dB = μ o 4 π Irdϕ r 2 = μ o 4 π Idϕ r . . . . . . . . . . . . . ( 5 ) In ΔOPA , cosϕ = a r ∴ r = a cosϕ . . . . . . . . . . ( 6 ) Substituting equation (6) in equation (5) dB = μ ο 4 π 1 a cosϕdϕ The total magnetic induction at P due to the conductor XY is B = ∫ − ϕ 1 ϕ 2 dB = ∫ − ϕ 1 ϕ 2 μ o 4 πa I cosϕdϕ B = μ o 4 πa I [ sinϕ 1 + sinϕ 2 ] For infinitely long conductor, ϕ 1 = ϕ 2 = 90 o ∴ B = μ o I 2 πa If the conductor is places in a medium of permeability μ 0 , B = μ o I 2 πa