Ampere Circuital Law
Ampere Circuital Law: In classical electromagnetism, this concept of law gives a relation between the magnetic field and the electric current passing through the closed loop. A French scientist André-Marie Ampere proposed the discovery of Ampere’s Circuital Law in the year 1826. The expression which shows the relation between the magnetic field and current is called Ampere Circuital Law.
Statement of Ampere Circuital law
Ampere Circuital law says that the line integral of the magnetic field (B) around any closed loop is equal to μ0 times the total current (I) passing through the area enclosed by the curve. Mathematically it can be written as-
∮B. dl = μ0. I
Proof of Ampere Circuital Law
Take a long infinite wire through which I current is passing. Consider a randomly shaped closed loop around the infinite long wire. Consider an elementary portion dl1 and the distance from the centre of the loop to the elementary portion is r1 and an angle is formed, that is dθ1 and through this, a magnetic field is generated B1. Again let us consider a distance r2 from the centre to the elementary portion and an angle is formed, that dθ2 which generates magnetic field B2. Same as this elementary portion there are other no. of elementary portions around the loop.
So as we know according to ampere circuital law -
∮B. dl = μ0. I
For the elementary portion around the loop be- ∮B1. dl1 + ∮B2. dl2 + ∮B3. dl3+........(1)
As we know, the magnetic field on any point is-
B= μ0.2I/4π. r……(2)
Substituting the value of (2) in (1)
μ0.2I/4π. r1 + μ0.2I/4π. r2 + μ0.2I/4π. r3……..(3)
Substituting (3) in (1) we get,
∮B. dl = ∮B1. dl1 + ∮B2. dl2 + ∮B3. dl3+........
∮B. dl = μ0.2I/4π. r1 + μ0.2I/4π. r2 + μ0.2I/4π. r3+.......
∮B. Dl = μ0.2I/4π[1. dl1/r1 + 1. dl2/r2+ 1. dl3/r3+........] ……..(4)
As we know, θ = arc/ radius
So, dθ1 = dl1/r1
dl1= dθ1. r1………(5)
Subsitutuing the value of (5) in equation (4), we get
∮B. Dl = μ0.2I/4π[1. dl1/r1 + 1. dl2/r2+ 1. dl3/r3+........]
= μ0.2I/4π[ 1. dθ1. r1/r1 + 1. dθ2. r2/r2 + 1. dθ2. r2/r2+......]
∮B. Dl = μ0.2I/4π[ dθ1 + dθ2 + dθ3 +..……] ……(6)
From above we can say, dθ is a small angel covering an elementary angle
Therefore, we can write it as,
dθ = 2π …….(7)
Substitute the value of (7) in (6) we get,
∮B. Dl = μ0.2I. 2π/4π
∮B. Dl = μ0.I
Application of Ampere Circuital Law
The application of Ampere circuital law are-
- It helps to calculate the magnetic field along the current-carrying wire.
- It helps to calculate the magnetic field inside a toroid.
- It helps to calculate the magnetic field created by a long current-carrying conducting cylinder.
- It helps to calculate the magnetic field inside the conductor.
- It helps to find forces between currents.