Ampere's circuital law : In free space, the line integral of magnetic induction around a closed path in a magnetic field is equal to p0 times the net steady current enclosed by the path.
In mathematical form,
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ0 I …………… (1)
where \(\vec{B}\) is the magnetic induction at any point on the path in vacuum, \(\overrightarrow{d l}\) is the length element of the path, I is the net steady current enclosed and μ0 is the permeability of free space.
Explanation : Figure shows two wires carrying currents I1 and I2 in vacuum. The magnetic induction \(\vec{B}\) at any point is the net effect of these currents.
To find the magnitude B of the magnetic induction :
We construct an imaginary closed curve around the conductors, called an Amperian loop, and imagine it divided into small elements of length \(\overrightarrow{d l}\). The direction of dl is the direction along which the loop is traced.
(ii) We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.
For each length element of the Amperian loop, \(\vec{B} \cdot \overrightarrow{d l}\) gives the product of the length dl of the element and the component of \(\vec{B}\) parallel to \(\overrightarrow{d l}\) . If θ r is the angle between \(\overrightarrow{d l}\) and \(\vec{B}\) ,
\(\vec{B} \cdot \overrightarrow{d l}\) = (B cos θ) dl
Then, the line integral,
\(\oint \vec{B} \cdot \overrightarrow{d l}=\oint B \cos \theta d l\) …………. (2)
For the case shown in Fig., the net current I through the surface bounded by the loop is
I = I2 – I1
∴ \(\oint B \cos \theta d l\) = μ0 I
= μ0(I2 – I1) …………… (3)
Equation (3) can be solved only when B is uniform and hence can be taken out of the integral.
