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Ampere’s circuital Law in differential form with Applications


What is ampere’s law?

According to this law “The line integral of magnetic field B along a closed path due to current is equal to the product of the permeability of free space and the current enclosed by the closed path”.

Mathematically it is expressed as:



μ0=permeability of free space

i=current flowing through the conductor.

Proof:ampere's law

Consider a straight conductor in which current i is flowing.The current produces the magnetic field B around the conductor.The magnetic field lines are in the form of concentric circles.

Ampere showed that the flux density B at any point near the conductor is directly proportional to the current i and inversely proportional to the distance ‘r’ from the conductor ,so:der of amp lawamp

Where  is the length of the path called circumference of the circle.

Divide the circle representing the magnetic field line into large number of small elements each of length dl.The quantity B.dl is calculated for each element as:

B.dl=Bdlcos= Bdlcos0=Bdl

For complete circle:


Integral form of ampere’s law:

integral form of ampere law

Differential form of ampere law:

Since the integral form of ampere’s law is:

diff form of ampere law

The above relation is known as differential form of ampere’s circuital law.

Applications of ampere’s circuital law:

  • Field due to a solenoid:

Consider a solenoid having n turns per unit length.When current passes,through the solenoid,magnetic field is produced inside the solenoid which is directed along the axis of solenoid.The magnetic field in space outside is so weak that it is considered as zero.

In order to calculate the value of magnetic field B inside the solenoid using ampere’s law,we consider a closed path abcda in the form of rectangle.This closed path is known as Amperian path,as seen in the figure.


[latex]\\oint { B.dl={ \mu  }_{ 0 } } \times Current\quad enclosed\quad ————(1)\\ But\quad the\quad total\quad line\quad integral\quad of\quad B\quad along\quad the\quad closed\quad path\quad is\quad given\quad by:\\ \oint { B.dl= } \int _{ a }^{ b }{ B.dl } +\int _{ b }^{ c }{ B.dl } +\int _{ c }^{ d }{ B.dl } +\int _{ d }^{ a }{ B.dl } ——-(2)\quad \\ Now\quad inside\quad the\quad solenoid\quad ,B\quad is\quad parallel\quad to\quad path\quad \quad ab\quad \quad of\quad length\quad l,so\quad \theta ={ 0 }^{ o }\\ \int _{ a }^{ b }{ B.dl } =\int _{ a }^{ b }{ Bdl } cos\theta \\ \int _{ a }^{ b }{ B.dl } =\int _{ a }^{ b }{ Bdl } cos{ 0 }^{ o }=\int _{ a }^{ b }{ Bdl } \\ =B\int _{ a }^{ b }{ dl } \\ =Bl\\ For\quad path\quad bc\quad ,B\quad and\quad dl\quad are\quad perpendicular\quad to\quad each\quad other,so\quad \\ \int _{ b }^{ c }{ B.dl } =\int _{ b }^{ c }{ Bdl } cos\theta \\ =\int _{ b }^{ c }{ Bdl } cos90=\int _{ b }^{ c }{ Bdl } (0)\\ =0\\ As\quad the\quad field\quad inside\quad the\quad solenoid\quad is\quad zero,for\quad path\quad cd,B=0\\ Hence:\\ \int _{ c }^{ d }{ B.dl } =0\\ For\quad path\quad da,magnetic\quad field\quad B\quad and\quad dl\quad are\quad perpendicular\quad to\quad each\quad other,so\quad \theta =90\\ \int _{ d }^{ a }{ B.dl } =0\\ Putting\quad the\quad values\quad in\quad equation\quad 2\quad we\quad get:\\ \oint { B.dl= } Bl\quad +0+0+0\\ \oint { B.dl= } Bl\\ Now\quad equation\quad (1)\quad will\quad be\quad given\quad as:\\ Bl={ \mu  }_{ 0 }\times Current\quad enclosed\quad \quad ——(3)\\ If\quad n=number\quad of\quad turns\quad per\quad unit\quad length\\ Then\quad nl=number\quad of\quad turns\quad in\quad length\quad l\\ If\quad i\quad =current\quad flowing\quad through\quad each\quad turn\\ Then\quad nli=current\quad flowing\quad through\quad nl\quad turns.\\ Now\quad equation\quad (3)\quad becomes\quad as:\\ B={ \mu  }_{ 0 }i\quad \quad —-(4)\\ Which\quad is\quad the\quad relation\quad of\quad \quad magnetic\quad field\quad inside\quad the\quad solenoid.\\ \\ \\ \\ [/latex]

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  1. muy interesante y fascinante publicación de ciencia física.

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