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CHAPTER 3

ELECTROMAGNETIC FIELDS THEORY

- Magnetostatic Field Ampere Circuital Law

In this chapter you will learn

- Biot-Savats Law
- Ampere Circuital Law
- Magnetic flux density vector
- Magnetic potential vector and magnetic force
- Magnetic circuit
- Faradays Law
- Maxwells Equation

Analogy between Electric Magnetic Fields

Amperes Circuit Law

- Similar to Gausss law
- Amperes law states that the line integral of the

tangential components of H around a closed path

is the same as the net current Ienc enclosed by

the path - Eq 1 is the integral form of Amperes Circuit Law
- Amperes Circuit Law is used when we want to

determine H when the current distribution is

symmetrical

Amperes Circuit Law

- If we apply Stokes theorem to eq 1, we obtain
- But since
- We compare (2) and (3) to obtain
- Third maxwells Equation Amperes law in point

form

Amperes Circuit Law

- Magnetostatic field is not conservative. That is

Applications of Amperes law

- Some symmetrical current distributions
- Infinite line current
- Infinite Sheet of current
- Infinitely long coaxial transmission line

Infinite Line Current

- Consider an infinitely long filamentary current l

along the z-axis

To determine H at an observation point P, we

allow a closed path pass through P known as an

Amperian path (analogous to Gaussian surface)

Infinite Line Current

- Since the amperian path encloses the whole

current I, according to Amperes law - Thus

Infinite Sheet of current

- Consider an infinite current sheet in the z 0

plane with a uniform current density K Kyay A/m

Infinite Sheet of current

- Applying Amperes Law, we get
- Regard the infinite sheet as comprising of

filaments. The resultant dH has only an

x-component - Also, H on one side of the sheet is the negative

of that on the other side.

------------(1)

------------(2)

Infinite Sheet of current

- Evaluating the Amperes law along the closed path

we obtain - Compare eq (1) with eq (3), we get
- Subtitute H0 in eq (2)

------------(3)

Infinite Sheet of current

- Thus now we can say
- In general, for an infinite sheet of current

density K A/m,

Infinitely long coaxial transmission line

- Consider an infinitely long transmission line

consisting of two concentric cylinders having

their axes along the z-axis

Infinitely long coaxial transmission line

- The inner conductor has radius a and carries

current I while the outer conductor has inner

radius b and thickness t and carries return

current -I. - Since the current distribution is symmetrical, we

apply Amperes law for the Amperian path for each

of the four possible regions - 0 p a,
- a p b,
- b p b t,
- a n d p b t .

region 0 p a

- For region 0 p a, we apply Ampere,s law to

path L1 - Since the current is uniformly distributed over

the cross section,

region 0 p a

- Thus

or

Region a p b

- For region a p b, we use L2

or

Region b p b t

- For region b p b t, we use L3
- J in this case is the current density (current

per unit area) of the outer conductor and is

along az

Region b p b t

- Thus
- Then we can get H,

Region p b t

- For region p b t we use L4

Infinitely long coaxial transmission line

- Putting it all together,

Infinitely long coaxial transmission line

- Plot of HF against .

Example 2

- A toroid whose dimensions are shown below has N

turns and carries current I. determine H inside

and outside the toroid.

2a

0

0

Solution

- Apply Amperes circuit law to the Amperian path,

which is a circle of radius p. Since N wires cut

through this path each carrying current I, the

net current enclosed by the path is NI. Hence, - Where p0 is the mean radius of the toroid. Thus

the approximate value of H is

Solution

- Outside the toroid, the current enclosed by the

Amperean path is - NI - NI 0
- Hence H 0

Practice Exercise 7.6

- A toroid of circular cross section whose center

is at the origin and axis the same as the z-axis

has 1000 turns with p0 10 cm, a 1cm. if the

toroid carries a 100-mA current, find H at - (3cm, -4cm, 0)
- (6cm, 9cm, 0)

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