NAME:VIPUL KUMAR MISHRA (014)PARAS BRAHMBHATT (001)

ENROLMENT NUMBER:130950111014140953111001

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 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

Ampere’s Law of Force (Cont’d)

▪ Experimental facts:

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

–Two parallel wires carrying current in the opposite directions repel.

I1 I2

F12F21

I1 I2

F12F21

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

facts:

–A short current-carrying wire oriented perpendicular to a long current-carrying wire experiences no force.

I1

F12 = 0

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 currents carried by the two wires.

Ampere’s Law of Force (Cont’d)▪ The direction of the force established by the experimental facts can be mathematically represented by.

1212ˆˆˆˆ 12 RF aaaa

unit vector in direction of force on

I2 due to I1

unit vector in direction of I2 from I1

unit vector in direction of current I1

unit vector in direction of current I2

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

212

1122012

12ˆ

4 R

aldIldIF R

Permeability of free spacem0 = 4p 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

2 1

12

212

1221012

ˆ

4 C C

R

R

aldldIIF

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.

1221 FF

Ampere’s Law Applied to a Long Wire

Choosing path a, and integrating H around the circle of radius gives the enclosed current, I:

So that: as before.

Symmetry suggests that H will be circular, constant-valuedat constant radius, and centered on the current (z) axis.

In the coax line, we have two concentric solid conductors that carry equal and oppositecurrents, I.

The line is assumed to be infinitely long, and thecircular 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

Coaxial Transmission Line

Field Between ConductorsThe inner conductor can be thought of as made up of a

bundle of filament currents, each of which produces thefield 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 sameas that of filament conductor on the z axis that carries current,I. Specifically:

a < < b

Field Within the Inner ConductorWith 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 wehave circular symmetry, the field would have to be constant over the circular integration path, and so itmust be true that:

Field Inside the Outer Conductor

Inside the outer conductor, the enclosed current consistsof 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:

THANK YOU