# Ampere’s Circuital Law

Like Gauss’s law in electrostatics, this law provides us a simple method find magnetic fields in cases of symmetry.

According to this law, the line integral of a magnetic field over a closed path is equal to μtimes the net current linked by the area enclosed by that path.
Mathematically,

$\displaystyle \oint \vec{B}.\vec{dl} = \mu_0(I_{net})$

Positive direction of current and the direction of the line integral are given by the right hand thumb and curling fingers respectively.

##### Magnetic field due to an infinitely long straight conductor

Consider a long straight wire perpendicular to the plane of paper.  The direction of current is shown by the cross. There is a point P at a distance r from the wire where the magnetic field is to be calculated.

Draw a circle of radius r symmetrical around the wire. This circular loop is known as an Amperian loop.

This loop also represents magnetic lines of force due to the wire. The sense of lines of force can be given by right hand thumb rule.

At any point on this loop is directed along the tangent , $\vec{dl}$ is the elementary segment of this loop. This is also directed along the tangent.

Hence , $\vec{B}. \vec{dl} = B dl cos0^o$

$\displaystyle \oint \vec{B}.\vec{dl} =$ line integral of $\vec{B}. \vec{dl}$ over closed path.

I = Net current enclosed by the Amperian loop .

From symmetry magnitude of the magnetic field at any point on the Amperian loop is same, hence

$\displaystyle \oint \vec{B}.\vec{dl} = \mu_0(I )$

$\displaystyle B\oint dl = \mu_0(I )$

$\displaystyle B (2 \pi r) = \mu_0(I )$

$\displaystyle B = \frac{\mu_0}{2\pi}\frac{I}{r}$

Solved Example : Suppose that the current density in a wire of radius a varies with r according to J = Kr2 , where K is a constant and r is the distance from the axis of the wire. Find the magnetic field at a point distance r from the axis when

(i) r < a and (ii) r > a

Solution : Choose a circular path centred on the axis of the conductor and apply Ampere’s law.

(i) To find the current passing through the area enclosed by the path integrate

dI = JdA = (Kr2) (2πrdr)

$\displaystyle I = \int dI = K\int_{0}^{r}2\pi r^3 dr$

$\displaystyle I = \frac{K \pi r^4}{2}$

$\displaystyle B\oint dl = \mu_0(I )$

$\displaystyle B (2 \pi r) = \mu_0(\frac{K \pi r^4}{2})$

$\displaystyle B = \frac{\mu_0 K r^3}{4}$

(ii) If r > a

Then net current through the Amperian loop is

$\displaystyle I = K\int_{0}^{a} K r^2 2\pi r dr$

$\displaystyle I =\frac{\pi K a^4 }{2}$

$\displaystyle B =\frac{\mu_0 K a^4 }{4 r}$