lecture 4: magnetostatic fields -...

Lecture 2: Magnetostatic Fields

Instructor:

Dr. Vahid Nayyeri

Contact:

[email protected]

Class web site:

http://webpages.iust.ac.

ir/nayyeri/courses/BEE

Magnetic loops at the surface of the Sun, as seen with the TRACE

solar spacecraft. (©TRACE operation team, Lockheed Martin)

mailto:[email protected]

http://webpages.iust.ac.ir/nayyeri/courses/BEE

2. Magnetostatic Fields

2.1. Fundamentals of Magnetic Fields

Magnetic field lines are continuous, don’t originate nor terminate at a point.

There is no “magnetic monopole”…

0B ds

B is a magnetic flux density, [T] = [Wb/m2]

0B

By applying the divergence theorem to (2.1):v

B ds Bdv

(2.1)

(2.2)


2.2. Ampere’s Law

Magnetic field can be created by the electric current:

0 encB dl Ielectric current enclosed within a closed loop

77

0: 14 10 2.566 10 H mPermiability of free space

A cylindrical wire caring a current

creates a magnetic field.

“Right Hand Rule” (RHR).

Ampere’s Law: (3.1)


2.2. Ampere’s Law (cont)

By applying Stokes’s theorem to, we arrive to

0

s s

B dl B ds J ds

Therefore, the differential

form of Ampere’s Law is: 0B J

An axial view of the cortically-generated magnetic

field of a human listener, measured using whole-

head magnetoencephalography (MEG) – from the

journal “Cerebral Cortex”

It appears that even very small currents generate

magnetic fields…

(4.1)

(4.2)


2.2. Ampere’s Law (Example)

A symmetry in the system greatly simplifies

evaluation of integrals of Ampere's law

2

0

2B dl B u d u B

The right-hand side of Ampere’s law for the

radius greater than a is just 0I.

0 ,2

IB a

Assuming a uniformly distribution of current within the wire, the current density

inside the wire is

2 z

IJ u

a

(5.1)


2.2. Ampere’s Law (Example, cont)

22

2

0 0

enc

s r

II J ds r dr d I

a a

The total current enclosed within the circle of radius a

Therefore: 0

2,

2

IB

aa

We notice that at the edge of the

wire, two solutions are equal.

Can you further explain the

dependence of magnetic flux

on the radius?

(6.1)


2.3. Biot-Savart Law

If a current is passing through a wire

0

2

'( )

4RI u

B rR

dl(7.1)

For a surface current:

0

2( ) '

4Ru

B r dsR

sJ

(7.1)

& for a volumetric current:

0

2( ) '

4Ru

B r dvR

J(7.1)


2.3. Biot-Savart Law (Example)

Find the magnetic field on the axis perpendicular to

the loop of current. Use the Biot-Savart Law.

We identify the terms appearing in (7.1):

2 2' ' , ,R zdl a d u u a u z u R R a z

2

0 0

2 2 3 2 2 2 3 2

( ' ) ( ) ' '( )

4 ( ) 4 ( )

z za d u a u z u a d u a z d uI IB z

a z a z

Due to symmetry, the terms with the unit vector u are zero.

2

0 0

2 2 3 2 3( )

2 ( ) 2z

I a mB z u

a z R

2the magnetic dipole momentzm I a u

(8.1)

(8.2)


2.4. Classification of the methods

We have learned the following analytical methods to find the magnetic flux

density at a point in space from a current element:

1. Application of Ampere’s Law, which requires considerable

symmetry.

2. Application of the Biot-Savart law. No symmetry is required.


2.5. Magnetic forces

If a charged particle is moving with a constant velocity v in a region that ONLY

contains a magnetic field with the density B, the force that acts upon the particle is

mF q v B

Direction of the force

- RHR!

F+ stands for a

positively charged

particle;

F- represents a

negatively charged

one.

(10.1)


2.5. Magnetic forces (cont)

When a charged particle is going through an area with both: uniform electric field

and uniform magnetic field, the force exerted on it would be the Lorentz Force:

[ ]( ) NF q E v B

Recall that the work done by a charged particle moving in a field isb b

a a

W F dl q E dl

A differential charge dQ = vdv moving at a constant velocity creates a current. If

this current flows in a closed loop:

( ) ( )m vdF dQ v B v B dv J B ds dl Idl B

mF B I dl The total magnetic force:

(11.1)

Why?(11.2)

(11.3)


2.5. Magnetic forces (Example)

Evaluate the force existing

between two parallel wires caring

currents.

B1 will go up at the location of wire 2.

So force on the wire 2:

2 1 2y zF Bu I dl u to the left

2 1 2( )y zF Bu I dl u to the right

a)

b)


2.5. Magnetic forces (Example, cont)… ”Alternative approach”

Let’s re-state the force on wire 1 caused by the magnetic field generated by

the current in wire 2

1

12 1 12

L

F I B dl

From the Biot-Savart law:

21

2

20 212 2

214

R

L

u dlIB

R

21

1 2

2 10 1 212 2

214

R

L L

u dl dlI IF

R

Finally:

This is what’s called as Ampere’s force.

(13.1)


2.5. Magnetic forces (Example 2)

Consider a current-caring loop in a constant

magnetic field B = B0 uz.

We assume the separation between the In/Out

wires to be infinitely small.

Parallel wires carry the same current in the

opposite direction. Therefore, the net force will

be a vector sum of all forces, which is zero!

However, there will be a torque on the loop

that will make it to rotate (say, about x for

simplicity).


2.5. Magnetic forces (Example 2, cont)

The torque on the loop is given by

1 3sin sin2 2

y yT F F

1 0 3 0,where F IB x F IB x

0 sinT IB x y

Finally:

T m B

magnetic moment = IB0S

(15.1)


2.6. Magnetic materials

Two sources of magnetism inside an atom:

1) an electron rotating around a nucleus;

2) an electron spinning about its own axis.

Types of material:

1. Diamagnetic: 1) and 2) cancel each other almost completely, magnetic

susceptibility m -10-5.

2. Paramagnetic: 1) and 2) do not cancel each other completely, but

magnetic dipoles oriented randomly, magnetic susceptibility m 10-5.

3. Ferromagnetic: domain structure (magnetic dipoles in each domain are

oriented), very high m (hundreds and higher)

Paramagnetic Ferromagnetic


2.6. Magnetic materials (cont)

Total magnetization (magnetic dipole

moment per unit volume):

External magnetic field may change dipole orientation “permanently” - HDD.

0

1

1lim

N

jv

j

A mM mv

(17.1)

There is a current created inside domains (magnetization current):

m m

s s

I M dl J ds M ds

, mTherefore J M

We may modify the Ampere’s Law by adding the magnetization current:

0 0

1m

BB J J J M J M

(17.2)

(17.3)

(17.4)


(18.1)

2.6. Magnetic materials (cont2)

We introduce a new quantity, the Magnetic Field Intensity:

0

A mB

H M

Therefore, the Ampere’s circular law is

encH dl I (18.2)

: mMagnetizatio M Hn

Therefore: 0 0(1 )m rB H H H

(18.3)

(18.4)

where r is the relative permeability.


2.6. Magnetic materials: Ferromagnetic

Example: a magnetic flux density B = 0.05 T appears in a material with r = 50.

Find the magnetic susceptibility and the magnetic field intensity.

1 50 1 49m r 7

0

0.05796 [ ]

50 4 10r

BH A m

Hysteresis

Magnetic flux density B exhibits nonlinear

dependence on the magnetic field intensity H.

Because of hysteresis, magnetic materials

“remember” the magnitude and direction of

magnetic flux density. They can be used as

memory elements.

saturation

saturation


2.7. Inductance (an ability to create magnetic flux)

When j = k – self –inductance; otherwise – mutual inductance.

where is a magnetic flux linkage.

j

jk

k

H Wb ALI

Ex. 1: Let us consider a

solenoid of the length d,

cross-section area A,

and having N turns.

It may also have a core

made from a magnetic

material.

z is the solenoid’s axis.

(20.1)


2.7. Inductance (cont)

The magnetic flux density at the center of the solenoid is:

z

NIB

d

The total magnetic flux: m

NIA

d

The magnetic flux linkage:2

m

N IN A

d

Therefore, the self-inductance of a solenoid is

2NL A

d

(21.1)


2.7. Inductance (cont 2)

Example 2: a self-inductance of a coaxial cable

Here, the magnetic flux linkage equals to the

total magnetic flux.

What is the main difference as compared to a

solenoid?

2

IB

r

mfd:

0

ln2 2

z b

m

z r a

I I bdr dz z

r a

ln2

bL z

a

Therefore: (22.1)


2.7. Inductance (cont 3)

Example 3: a mutual inductance between two

circular solenoids, whose individual lengths are

d and areas S1 and S2, separated by x; x << dx

First coil:

0 1 11

N IB

d

,1 1 1m B S

Assuming that the magnetic flux has the

same value in the second solenoid:

2 2 1 1N B S

Therefore, the mutual inductance:

0 1 2 12

1

N N SM

I d

(23.1)

lecture 4: magnetostatic fields -...

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