# Ampere’s Circuital Law

## What is Ampere’s Circuital Law?

Ampere’s Circuital Law is an approach to find the intensity of magnetic field at a point produced by a wire carrying a current. It states that

The line integral of magnetic field around any closed path in free space is equal to absolute magnetic permeability \left ( \mu_0 \right) multiplied by the net current passing through any surface enclosed by that closed path.

• Ampere’s circuital law is not an universal law.
• It directly deals with only the steady currents.
• This law holds good for a closed path of any size and shape around a current carrying conductor.
• It is independent of the distance between conductor and closed path around the conductor.

According to Ampere’s Circuital Law, \quad \oint {\vec {B}  \vec {dl}} = \mu_0 I

Where, ( \oint {\vec {B} \vec {dl}} ) represents the line integral of magnetic field around the closed path, ( \mu_0 ) is the permeability of free space and ( I ) is the net current enclosed by the closed path.

In general form if ( \theta ) is the angle between the current element vector ( \vec {dl} ) and magnetic field ( \vec {B} ) , then –

\oint B {dl} \cos \theta = \mu_0 I

• Importance of Ampere’s circuital law is similar to the Biot-savart law.
• This law helps to find the magnitude of magnetic field produced by a current carrying element.

### Expression for Ampere’s circuital law

Consider about an infinite long straight conductor placed vertical as shown in figure. Let –

• The conductor is carrying a current ( I ) .
• Due to this current, the magnetic field lines are produced around the conductor in the form of concentric circles in horizontal plane.
• As per right hand thumb rule, the direction of magnetic field lines will be anti clockwise when vied from top of horizontal plane of figure.

From Biot-Savart law, magnetic field at a distance ( a ) from the conductor will be –

B = \left ( \frac {\mu_0}{4 \pi} \right ) \left ( \frac {2 I}{a} \right ) ………. (1)

Now, consider a imaginary circle of radius a around the wire. This circle is called an Amperian loop. Let –

• Length of a small element XY of the amperian loop is {dl} .
• So, in this case directions of ( \vec {dl} ) and ( \vec {B} ) are in same line, because direction of ( \vec {B} ) is along the tangent to the circular field lines around the conductor.

Thus, angle ( \theta ) between ( \vec {dl} ) and ( \vec {B} ) will be –

• – either ( 0 ) when current is flowing in upward direction in the conductor.
• – or ( \pi ) when current is flowing in upward direction in the conductor.

For the given figure, current is flowing in upward direction, so ( \theta = 0 ) \ \text {and} \ ( \cos \theta = 1 )

Therefore, product \quad \vec {B}  \vec {dl} = B {dl} \cos \theta = B {dl}

Taking line integral for the entire closed path of Amperian loop, we get –

\oint {\vec {B}  \vec {dl}} = \oint B {dl} ………. (2)

Substituting the value of ( B ) from equation (1), we get –

\oint B dl = \oint \left ( \frac {\mu_0}{4 \pi} \right ) \left ( \frac {2I}{a} \right ) {dl}

= \left ( \frac {\mu_0 2 I}{4 \pi a} \right ) \oint {dl}

But, \quad \oint {dl} = ( 2 \pi a ) ( Circumference of the Amperian loop )

Therefore, \quad \oint B dl = \left ( \frac {\mu_0 2 I}{4 \pi a} \right ) \left ( 2 \pi a \right )

= \mu_0 I

Thus, Ampere’s Circuital law is expressed as –

\oint {\vec {B}  \vec {dl}} = \mu_0 I

## Magnetic Field of infinite long wire

Consider about a conductor of infinite length and radius ( a ) carrying a steady current ( I ) . The current in the wire gives rise to a magnetic field around it. The magnetic field lines are represented by concentric circles with their centres on the axis of the wire.

The intensity of the magnetic field at a point will be different depending upon the distance of that point from the axis of wire. Let, find the magnetic field at following locations –

### 1.Field at a point outside the wire

Consider about a small length of the current carrying wire as shown in figure. Let, P_1 is a point outside the wire at a distance ( r ) from the axis of the wire as shown in figure. Here ( r > a ) .

Assume an Amperian loop passing through point P_1 . Now consider a small element of length ( dl ) of loop.

According to Ampere’s circuital law –

\oint {\vec {B}} d \vec {l} = \mu_0 I

Or, \quad \oint B {dl} \cos \theta = \mu_0 I

In this case the directions of ( \vec {dl} ) and ( \vec {B} ) are same.

Therefore, \quad \oint B {dl} \cos 0 \degree = \mu_0 I

Or, \quad B \oint {\vec {dl}} = \mu_0 I

So, \quad B \times 2 \pi r = \mu_0 I

Or, \quad B = \left ( \frac {\mu_0 I}{2 \pi r} \right ) ………… (1)

Therefore, magnetic field intensity at a point outside the wire varies inversely to the distance of the point from the axis of wire.

B \propto \left ( \frac {1}{r} \right )

### 2.Field on surface of wire

Magnetic field intensity on the surface of wire can be obtained by putting ( r = a ) in above equation. Therefore, intensity of magnetic field on the surface of wire will be –

B = \left ( \frac {\mu_0 I}{2 \pi a} \right ) ………… (2)

This is the maximum possible value of magnetic field intensity.

B_{Max} = \left ( \frac {\mu_0 I}{2 \pi a} \right )

### 3.Field at a point inside the wire

Consider about a point P_2 inside the wire at a distance ( r ) from the axis of the wire. In this case ( r < a ) . Assume that the current is uniformly distributed throughout the cross-section of the wire.

Then current through the circular area of wire of radius ( r ) will be –

I' = I \times \left ( \frac {\text {Cross sectional area of Amperian circle}}{\text {Cross sectional area of wire}} \right )

= I \times \left ( \frac {\pi r^2}{\pi a^2} \right ) = \left ( \frac {I r^2}{a^2} \right )

According to Ampere’s circuital law –

\oint {\vec {B}} \vec {dl} = \mu_0 I'

Or, \quad \oint B {dl} \cos 0 \degree = \mu_0 I'

So, \quad B \oint {\vec {dl}} = \mu_0 \left ( \frac {I r^2}{a^2} \right )

= \left ( \mu_0 \frac {I r^2}{a^2} \right )

Hence, \quad B \times 2 \pi r = \left ( \frac {\mu_0 I r^2}{a^2} \right )

Or, \quad B = \left ( \frac {\mu_0}{2 \pi} \right ) \left ( \frac {Ir}{a^2} \right )