#### Question

State Ampere’s circuital law.

#### Solution 1

**Statement**: The line integral of the magnetic field `(vecB)` around any closed path is equal to μ_{0 }times

the total current (I) passing through that closed path.

`:.ointvecB.vec(dl)=mu_0I`

#### Solution 2

**Statement**: The line integral of the magnetic field `(vecB)` around any closed path is equal to μ_{0 }times

the total current (I) passing through that closed path.

`:.ointvecB.vec(dl)=mu_0I`

#### Solution 3

Ampere’s law states that the path integral or line integral over a closed loop of the magnetic field produced by a current distribution is given by `oint vec("B") . vec("dl") = µ_0"l"`

where I refers to the current enclosed by the loop.

Ampere’s law is a useful relation that is analogous to Gauss’s law of electrostatics. It is a relation between the tangential component of magnetic field at points on a closed curve and the net current through the area bounded by the curve.

To evaluate the expression for `oint vec("B") . vec("dl")` let us consider a long, straight conductor carrying a current I, passing through the centre of a circle of radius r in a plane perpendicular to the conductor.

According to Biot-Savart law of magnetism, the field has a magnitude `(µ_0"l")/(2pir)` at every point on the circle, and it is tangent to the circle at each point.

The line integral of `vec("B")` around the circle is

`oint vec("B") . vec("dl") = oint(µ_0"I")/(2pir)"dl" = (µ_0"l")/(2pir) oint"dl"`

Since `ointvec("dI") = 2pir` is the circumference of the circle,

Therefore , `oint vec("B") . vec("dl") = µ_0"l"`