Post on 20-Mar-2018
Lecture 2: Magnetostatic Fields
Instructor:
Dr. Vahid Nayyeri
Contact:
nayyeri@iust.ac.ir
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)
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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)
2. Magnetostatic Fields
(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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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. Magnetostatic Fields
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)