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using ampere,s circuital law field at a point due to very long current carrying straight wire, when point is outside the wire.


Om kumar 21st Jan, 2020
Answer (1)
Inayath 23rd Feb, 2020
Hello Om,

Ampere's circuital law states that the integral lines of the magnetic field B around any closed circuit is equal to (permeability constant) times the total current 'I' passing through this closed circuit.

Mathematically;

B.dl= *I

Proof for a straight current carrying conductor:

Consider a long straight current carrying conductor 'I'. According to Biot-Savart law, the magnitude of the magnetic field B due to the current carrying conductor at any point at a distant 'r' from it is mathematically given by;

B= *I*2*r

The magnetic field B is directed along the circumference of the circle of radius 'r' with the wire as center. The magnitude of the field B is same all points on the circle. To evaluate the line integral of the magnetic field B along the circle, we consider a small current element dI along the circle. At every point on the circle, both B and dl are tangential to the circle so that the angle between them is zero.

B.dI= B*dl cos(0)= B*dl (1) = B*dl

Hence the line integral of the magnetic field along the circular path is

B.dl= B*dl= B (dl)= *I*2*r*I = *I*2r*2r

B.dl=*I

This proves Ampere's law. This law is valid for any assembly of current and for any arbitrary closed loop.

Hope it helped. Thank you.
Best of luck.

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