Explain Ampere's Circuital Law with Proof and Limitations - Param Himalaya

Ampere's Circuital Law : 

Ampere Circuital Law states that the line integral of the magnetic field around any closed path in free space is equal to absolute permeability ($\mu_{0}$) times the net current passing through any surface enclosed by the closed path. 

Mathematically : 

$$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0}I$$

Where , $\overrightarrow{B}$ is the magnetic field , $\overrightarrow{dl}$ is the small element , $\mu_{0}$ is the absolute permeability of free space and I is the current enclosed by the closed path.

Proof : Consider an infinitely long straight conductor carrying current I. The magnetic field lines are produced around the conductor as concentric circles.

The magnetic field due to this current carrying infinite conductor at a distance a is given by

$$B = \frac{\mu_{0}}{4\pi }(\frac{2I}{a})$$                           

Consider a circle of radius a around the wire (called Amperian loop). Let XY be a small element of length dl. $\overrightarrow{dl}$  and  $\overrightarrow{B}$ are in the same direction because direction of  $\overrightarrow{B}$ is along the tangent to the circle.

$\therefore  \overrightarrow{B} .\overrightarrow{dl} = Bdlcos\theta  = Bdl cos0^{0} = Bdl$

Taking line interval over the closed path, we get

$\oint \overrightarrow{B} .\overrightarrow{dl} = \oint Bdl$    

$\oint \overrightarrow{B} .\overrightarrow{dl} = \oint \frac{\mu_{0}}{4\pi }\frac{2I}{a}dl = \frac{\mu_{0}}{4\pi }\frac{2I}{a}\oint dl$

But  $\oint dl$ = Circumference of the circle = $2\pi a$

$\therefore \oint \overrightarrow{B}.\overrightarrow{dl} = \frac{\mu_{0}I }{2\pi a} \times 2\pi a$

$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu _{0}I$

Which is Ampere's Circuital Law.

Sign assigned to current enclosed by the Amperian loop.

Consider an Amperian loop encloses current $I_{1}$ and $I_{1}$ 

According to Ampere's law : 

$$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0}I_{enc}$$

If the integration around the closed loop is anti clockwise , then right hand rule determines the signs for the currents enclosed by the loop. According to this rule , current $I_{1}$ is taken as positive and $I_{2}$ is taken negative. Thus $I_{enc} = I_{1}-I_{2}$

$$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0} I_{enc} = I_{1}-I_{2}$$

Limitations of Ampere's circuital law :

1. Ampere's law is not a universal law.

2. Ampere's law directly deals with the study current only.

Previous Post Next Post