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 = μ₀ie.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
Test
Yogita Ingle 5 years, 11 months ago
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 = μ₀ie.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
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