Ampere's Circuital Law And Its Application 4p4q3i

NAME: VIPUL KUMAR MISHRA (014) PARAS BRAHMBHATT (001) ENROLMENT NUMBER: 130950111014 140953111001 DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING, ITM UNIVERSE, VADODARA. 5Th Semester. SUBJECT : ENGINEERING ELECTROMAGNETICS (2151002)

Ampere Circuit Law and Its Application

Ampère's Circuital Law ▪ Magneto statics is the branch of

electromagnetics dealing with the effects of electric charges in steady motion (i.e., steady current or DC).

▪ The fundamental law of magneto statics is Ampere’s law of force. ▪ Ampere’s law of force is analogous to Coulomb’s law in electrostatics.

Ampere’s Law of Force ▪ Ampere’s law of force is the “law of action” between current carrying circuits.

▪ Ampere’s law of force gives the magnetic force between two current carrying circuits in an otherwise empty universe. ▪ Ampere’s law of force involves complete circuits since current must flow in closed loops.

Ampere’s Law of Force (Cont’d)

▪ Experimental facts:

F21 F12 

– Two parallel wires carrying current in the same direction attract.

I1

F21

I2

F12  I1

– Two parallel wires carrying current in



 I2

Ampere’s Law of Force (Cont’d) ▪ Experimental facts:

F12 = 0 

– A short currentcarrying wire oriented perpendicular to a long currentcarrying wire

I1

I2

Ampere’s Law of Force (Cont’d)

▪ Experimental facts:

–The magnitude of the force is inversely proportional to the distance squared. –The magnitude of the force is proportional to the product of the

Ampere’s Law of (Cont’d) ▪Force The direction of the force

established by the experimental facts can be mathematically represented by. unit vector in direction unit vector in direction of current I2



of current I1

aˆ F12  aˆ 2  aˆ1  aˆ R12

unit vector in direction of force on I2 due to I1



unit vector in direction of I2 from I1

Ampere’s Law of Force (Cont’d) ▪ The force acting on a current element I2 dl2 by a current element I1 dl1 is given by

 0 I 2 d l 2   I1d l 1  aˆ R1 2  F 12  2 4 R12

Permeability of free space 0 = 4  10-7 F/m

Ampere’s Law of Force (Cont’d) ▪ The total force acting on a circuit C2 having a current I2 by a circuit C1 having current I1 is given by

d l 2   d l 1  aˆ R1 2   0 I1 I 2 F 12  2   4 C2 C1 R12

Ampere’s Law of Force (Cont’d) ▪ The force on C1 due to C2 is equal in magnitude but opposite in direction to the force on C2 due to C1.

F 21   F 12

Ampere’s Law Applied to a Long Wire

Symmetry suggests that H will be circular, constant-value at constant radius, and centered on the current (z) axis. Choosing path a, and integrating H around the circle of radius  gives the enclosed current, I:



So that:

as before.

Coaxial Transmission Line In the coax line, we have two concentric solid conductors that carry equal and opposite currents, I.

The line is assumed to be infinitely long, and the circular symmetry suggests that H will be entirely  - directed, and will vary only with radius . Our objective is to find the magnetic field for all values of 

Field Between The inner conductor can be thought of as made up of a Conductors bundle of filament currents, each of which produces the field of a long wire. Consider two such filaments, located at the same radius from the z axis, , but which lie at symmetric  coordinates, and -Their field contributions superpose to give a net H component as shown. The same happens for every pair of symmetrically-located filaments, which taken as a whole, make up the entire center conductor. The field between conductors is thus found to be the same as that of filament conductor on the z axis that carries current, I. Specifically: a <  < b

Field Within the Inner Conductor

With current uniformly distributed inside the conductors, the H can be assumed circular everywhere.

Inside the inner conductor, and at radius we again have:

But now, the current enclosed is

so that

or finally:

Field Outside Both Conductors Outside the transmission line, where > c, no current is enclosed by the integration path, and so

0

As the current is uniformly distributed, and since we have circular symmetry, the field would have to be constant over the circular integration path, and so it must be true that:

Field Inside the Outer Conductor

Inside the outer conductor, the enclosed current consists of that within the inner conductor plus that portion of the outer conductor current existing at radii less than  Ampere’s Circuital Law becomes

..and so finally:

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