Chapter 6. Static Magnetic Fields

Magnetism

Magnetism & EM force• Magnetism

– Discovered when pieces of magnetic loadestone were found to exhibit a mysterious attractive force.

– Found near the ancient Greek city called Magnesia

• A magnetic field– Caused by a permanent magnet, by moving charges (current)

• Electromagnetic force : electric force + magnetic force– Electric force : Fe = qE (N)– Magnetic force : Fm = q u×B (N), B : magnetic flux density [T] – Total electromagnetic force on a charge q :

F = q (E + u×B) Lorentz’s force Equation

Note! Magnetic forces do no work! ( ) 0!mag magdW F dl q u B udt

Magnetostatics

Stationary charges constant electric fields Electrostatics

Steady current constant magnetic fields Magnetostatics

0 0 t

JMagnetostatics Steady current

• Biot-Savart Law Coulomb’s law

'

0 R2 T=N/(A m)

4 C

I dlR

aB

7 20 permeability of free space 4 10 N/A

( , , )I x y z R

( , , )B x y z

Divergence of B:

0 R2 T=N/(A m)

4 C

I dlR

aB

Idl JSdl Jd ( , , )x y z J

R

S

( , , )B x y z

0 R2 '

4J d

R

aB

0 R2 '

4J d

R

aB

R R R2 2 2( )J J J

R R R

a a a

Divergence and curl are to be taken with respect to (x,y,z) coordinate.J is a function of (x’,y’,z’)

R2 0

R

a

0J

0 B

0 B

2R RR2 2 0 sind ds ds R d d r

R R

a a a

Curl of B:

0 R2 '

4J d

R

aB

0 R2 '

4J d

R

aB

R R R2 2 2( )J J J

R R R

a a a

0J B

300( ')4 ( ') ' ( )

4J R R R d J R

B

3R2 4 ( )R

R

a R2( ) ' 0J d

R

a

0J B Called Ampere’s law

0da dl J da B B0 encdl I B Integral version of Ampere’s law

R R2 2

3 3

3

3 3

R2

( ) ( ')

' '' ( ' )

''

' '' ' '

0 on the boundary (all current is safely inside)

( ) 0

J JR R

R R R RJ JR R

R R JR

R R R RJ d JdaR R

J

J dR

a a

a

Two postulates of magnetostatics in free space

• The divergence and the curl of B in nonmagnetic media

–

– No magnetic flow sources (no isolated magnetic charges)

– Divergence of the curl of any vector field = 0

0

0. (in nonmagnetic media)

BB J

70 permeability of free space 4 10 H/m

0 0V S

d d B B s

0 0 for steady currents B J

Comparison of Magnetostatics and Electrostatics

0

0

E

E

0

0

BB J

: Gauss’s law

: Ampere’s law

Coulomb’s law

Biot-Savart law

( )F q E B Lorenz force law

Ampere’s circuital law• Example 6-1.

– An infinite long, straight, solid, nonmagnetic conductor with a circular cross section of radius b carries a steady current I. Determine the magnetic flux density both inside and outside the conductor.

1

1

1 1

2

1 10

122

01 12 2

) Inside the conductor,,

2

The current through the area enclosed by C

, 2

C

aB

d rd

d B rd rB

Ir rI I I r r bb b b

B al a

B l

B a

Ampere’s circuital law• Example 5-1.

2

2

2

2

2 20

02 2

) Outside the conductor:

,,

2

1 , 2

C

b

Bd rd

d B rd rB

IB r br

2B al a

B l

B a a

Ampere’s circuital law• Example 6-2.

– Determine B inside a closely wound toroidal coil with an air core having N turns of coil and carrying a current I.

– The toroid has a mean radius b, and the radius of each turn is a.– Cylindrical symmetry B has only a φ component and is constant

along any circular path about the axis of the toroid.

0

0

A cicular contour with radius .

2 , -

, -2

0 for and .

C

C r

d r NI b a r b a

NIB b a r b ar

r b a r a b

B l B

B a a

B

Ampere’s circuital law• Example 6.3. Infinitely long solenoid with air core

– Direct application of Ampere’s law

– A special case of toroid (b)

0 0BL nLI B nI

00 02 2

NI NB I nIr b

6-3. Vector magnetic potential• B : divergence-free

–

– A : vector magnetic potential (Wb/m)– not sufficient definition. Its divergence is needed.

0 B

0 T B B A

02

0

2 2 2 2 2

20

Simplifying it, we choose =0 (Coulomb gauge). : vector Poisson's equation

o x x y y z zA A A

B J

A J A A A

A A J A a a a

AA J

0 (V: scalar potential)E E V

Ampere’s law

Vector potential:

0

20

Amp , , becomes vector Poisson's equation,

( Poisson's equations, one forthr each Cartesian component)

e

ee

re law

B J

A J

0 ? A

0 0If we add the gradient of any scalar , is also satisfied.B A A A A

0 0 0Suppose that satisfying is not divergenceless ( 0)B A A A

20The new divergence is A A

20If a function can be found that satisfies the Poisson's equation, ,

It is always possible to make divergenceless. A

A

0 A

Finding A from current density• Mathematical analogy with the scalar Poisson’s eqn.

2

0 0

, , 1, ,4 V

x y zV V x y z d

R

0 , ,, , Wb/m

4 V

x y zx y z d

R

JA

2 2 2( ) ( ) ( )R x x y y z z R

20Ampere's law in vector potential, , gives A J

If goes to zero at infinite,

by assuming goes to zero at infinite,JThe solution of the vector Poisson’s eqn.

For line current: 0 0, , 1, , ' '4 4

I x y z Ix y z dl dlR R

A

For surface current: 0 , ,, , '

4K x y z

x y z daR

A

For bulk current:

Magnetic flux• Magnetic flux Φ through a given area S

(Wb)s S

C

d d dl B s A s A

Biot-Savart Law• For thin wire with cross sectional area S, d’ equal to Sdl’

• Magnetic flux from the vector potential

( , , )x y z JR

RS

( , , )P x y z

0

0

1 (Wb/m)4 4V

C

Jd JSdl IdlI dld

R R

JA

0 0 4 4

The curl operation implies differentiations with respect to the space coordinates of the field point, and the integral operation is with res

C C

I Idl dlR R

unprimed

B A

pect to the .primed source coordinate

Biot-Savart Law• Magnetic flux from the vector potential (cont’d)

0

1/ 22 2 2

Using

1 1+ 4

1 ( ) ( ) ( )

1 1 1 1

( ) ( ) ( ) =

(

C

x y z

x y z

f f f

I dl dlR R

x x y y z zR

R x R y R z Rx x y y z z

G G G

B A

a a a

a a a

R3/ 2 3 22 2 2) ( ) ( ) R Rx x y y z z

aR

Biot-Savart Law• Biot-Savart Law

0 0 R2

0 0R2 3

1 (T)4 4

or 4 4

C C

C

I I dldlR R

I Idl dld d dR R

aB

a RB B B B

A current-carrying straight line

• Example 6-4. – Find magnetic flux density B from A

2 2

2 20 02 2

2 20

2 2

2 20 0

2 2 2 2

0

)

ln4 4

= ln4

1

ln4 2

When , (infin2

z

LL

z zL L

z

z zr

dl dz

a R z rI Idz z z r

z r

I L r LL r L

A Ar r

I ILL r Lr L r L r L r

Ir Lr

a

A a a

A a

B A a a

B a a

B a itely long wire carrying current )I

A current-carrying straight line

• Example 6-4.

0 0 0

3/23 2 22 2

)

4 4 2

z

r z

z r z

L

L

dl dz

b r z

dl z r z rdz

I I ILdl rdzdR r L rz r

a

R a a

R a a a a

RB B a a

A square loop (example 6-5) • Find B at the center of a planar square loop, with side w

carrying a direct current I.

• Exercise – 8(cm) × 6 (cm) rectangular conducting loop and 5 (A) direct

current– B at the center of the loop ?

0 0

The magnetic flux density at the center of the square loopis equal to four times that caused by a single side of length

/ 2

2 242z z

wL r w

I Iww

B a a

A circular loop (example 6.6)• Find B at a point on the axis

2 2

2

220 0

3/ 22 2 2 2

, ,

component is canceled by the contribution of the element locateddiametrically opposite to

4 2

z r

z r r z

r

z z

dl bd z b R z b

dl bd z b bzd b d

dlI Ibb d

z b z b

a R a a

R a a a a aa

B a a 2

3/ 20 T

A circular loop (example 6.7)• Uniform magnetic field

– Helmholtz coil– MRI Magnet

The magnetic dipole• Example 6.7

0

1

x

x y

2 /20 0x 0 /2

1 12 2 2

1

= 4

at is not the same as at the . In fact, at is -

sin cos '

sin sin= or 4 2

2 coscos sin cos 90

C

I dlR

dl P P

dl bd

I Ibb d dR R

R R b bRR R

A

a a a a

a a

A a a

2 2 2 2 21

1/22

21

sin sin

2 cos 2 sin sin

1 1 21 sin sin

R

R R b bR R b bR

b bR R R R

x

cos sin sinR R

The magnetic dipole• Example 6.7

2 2 2 2

1/ 2

1

1

/ 20/ 2

20

2

When , / can be neglected in comparison with 1

1 1 21 sin sin

1 1 1 sin sin 1 1 when 1.

= sin 1 sin sin2

sin, =4

A a

A a

B

p

R b b R

bR R R

b x px xR R R

Ib b dR R

IbRR

2

03 2cos sin

4 A aR

Ib aR

The magnetic dipole• Similarity between electric dipole and magnetic dipole at distant points.

• Simplest form of magnetic dipole

30

20

3

2cos sin ,4

2cos sin4

R

R

p qR

IbR

E a a p d

B a a

q

qd

20 2 2

2

02 2

0

sin A m4

Wb/m V 4 4

z z z

R R

I bI b IS m

R

VR R

A a m a a a

m a p aA

mI

Magnetization• All matters are composed of atoms, each with a positive charged nucleus and a

number of orbiting electrons.

• In addition, both electrons and the nucleus of an atom rotate (spin) on their own axes with certain magnetic dipole moments.

• In the absence of an external magnetic field, the magnetic dipoles of the atoms of most materials (excepts permanent magnets) have random orientations, resulting in no net magnetic moment. The application of an external magnetic fields cause both an alignment of the magnetic moments of spinning electrons and an induced magnetic moment due to a change in orbital motion of electrons

e-

m

nucleus because and nucleus electron nucleus electron electronm m M M

12qM

m L

Magnetization vector, M• To obtain a formula for determining the change in the magnetic flux density

caused by the presence of a magnetic material, we let mk be the magnetic dipole moment of an atom.

• The magnetic dipole moment dm of dv´ is dm =Mdv´will produce a vector potential.

• The total A is the volume integral of dA and the contribution of magnetization to B is A

0 1

1lim A/m , # of atoms within n

kk

n

M m

024

Rd dR

M aA

024

RV

dR

M aB A

Equivalent current densities• The total A

0 0 02 2

0 0

14 4 4

1 1Using the vector identity,

4 4

Using vector identity,

R RV V V

V V

VS

d d dR R R

R R R

d dR R

d d

M a M aA M

MM M

M MA

F F s

2

1 R

R R

a

0 0

4 4n

VS

d dsR R

M aMA

current

A/mn ms M a J 2 A/mm M J

Magnetization current density

• Magnetization surface current density

• Magnetization volume current density

Magnetization current density

A/mms n J M a

2 A/mm J M

P

P b P n

P

: out of paperM

msJ msJ

On the surface of the material, 0.If is uniform (space invariant) inside, the current of the neighboring atomic dipoles will cancelled averywhere.No net current in the interior ( 0),or,

ms n

ms

a

J MM

J0 if is space invariant.m M J M

Physical interpretation of Magnetic current

&

surface current: /ms

m at m IaI Mt

J I t M

M

For uniformly magnetized material,all the internal currents cancel.

0mJ M

However, at the edge there is a current.

J M ams n

Physical interpretation of Magnetic current

2 A/m J Mm

( ) ( )x z z

zx

zm x

I MtI M y dy M y dz

MI dydzy

MJy

For nonuniformly magnetized material,the internal currents no longer cancel.

yzm x

MMJy z

( ) ( )x y y

yx

ym x

I Mt

I M z dz M z dy

MI dzdy

zM

Jz

Magnetization current density• Example 6.8 – cylindrical magnet bar

– Axial magnetization : 0zMM a

magnet

b

L

0 0

0

20

3/22 2

Constant magnetization no magnetic volume current

To find at 0,0, , We consider a differential length dz with a current

and use Eq. (6-38) or to obain 2

ms z r

z

M M

P z M dz

Ib dz b

J a a a

B a

B a B

20 0

3/22 2

0 02 2 2 2

2

2

z

z

M b dz

z z b

M z z Ldz b z L b

a

B B a

0m M J

Magnetization current density• Example 6.8 – cylindrical magnet bar

0 0

2 2 2 22

B azM z z L

z b z L b

• Because the application of an external magnetic field causes both an alignment of the internal dipole moments and an induced magnetic moment in a magnetic material, the resultant magnetic flux density in the presence of a magnetic material will be different from its value in free space.

6-7. B&H and Ampere’s law

Bext BindBtotal = Bext + Bind

0

0

0

1m

B J J

B J M

B M J

2

Ampere's law

A/m

encdl I

H J

H

(magnetic field intensity)H

(free current)

0

A/m

BH M

B&H and Permeability• When the magnetic properties of the medium is linear and isotropic,

• B&H

• Permeability– The permeability of most materials is very close to that of free

space, 0– For ferromagnetic materials such as iron, nickel and cobalt,

, : magnetic susceptibilitym m M H

20 0 01 Wb/m

, 1m r

r m

B H M H H H

B H

Absolute permeability Relative permeability

1r

Analogous Relation• Electrostatics and Magnetostatics

Electrostatics MagnetostaticsE BD H 1/P - M JV A

6-9. Magnetic Materials• Magnetic materials

– Diamagnetic, if r 1 • m ~ -10-5

• the orbital motion of the electrons• Copper, germanium, silver, gold

– Paramagnetic, if r 1• m ~ 10-5

• Magnetic dipole moments of the spinning electrons• Aluminum, magnesium, titanium and tungsten

– Ferromagnetic, if r >>1• m >> 1 ( 100~ 100,000)• Magnetized domains (strong coupling forces

between the magnetic dipole moments of the atoms• Nickel, cobalt, iron (pure), mumetal

Magnetic Materials

Ferromagnetic material• Hysteresis loops

Residual flux density(permanent magnets)

Coercive field intensityH

VI

B

M

Magnetic recording

Ferromagnetic material• Hysteresis loss per unit volume per cycle

– The area of the hysteresis loop– The energy lost in the form of heat in overcoming the friction

encountered during the domain-wall motion and domain rotation– “Soft” material with tall and narrow hysteresis.

• Curie temperature– Demagnetization temperature (100 ~ 770 ◦C)

• Ferrites– Ceramic like compounds with very low conductivity– Low eddy current losses at HF– Extensive uses

• HF and microwave application such as FM antennas, HF transformers and phase shifters

• Computer magnetic core and disk memory devices and MRAM

6-10. Boundary conditions• The normal component of B

• The tangential component

1 2

1 2

1 1 2 2

TFor linear and isotropic media, and

=

n n

n n

B B

H H

1 1 2 2

BB H B H

1 2

1 2 0

2 1 2

w

lim

surface current density normal to the contour

A/m

C S

iabcda

t t snh

sn

n s

dl d

H dl J h

H H J h J

J abcda

H J H J s

H w H w

a H H J

21 2 (C/m )n n sD D D

1 20 t tE E E

6-11. Inductances and Inductors• Mutual flux:

2

12 1 2

1 1 12 1

WbS

d

I I

B s

B

2 2 12 12

12 2 12 12 12 1

In case has turns, the flux linkage due to is Wb

C NN L I

Mutual inductance between two circuits: 2

12 212 1 2

1 1

H; henryS

NL dI I

B s

Self-inductance of loop C1 : 1

11 111 1 1

1 1

HS

NL dI I

B s

Inductor: a conductor arranged in an appropriate shape (such as a conducting wire wound as a coil) to supply a certain amount of self-inductance

Procedure for determining self-inductance1. choose an appropriate coordinate system for the given geometry2. Assume a current I in the conducting wire3. Find B from I by Ampere’s circuital law if symmetry exists;

if not, Biot-Savart law must be used.4. Find the flux linkage with each turn, Φ from B by integration5. Find the flux linkage Λ by multiplying Φ by the number of turns6. Find L by taking the ratio L =Λ /I.

(Example 6-14) Find the self-inductance of a toroid.

2 00 00

0 0 0

Cylindrical coordinate system:For , and ;

By Ampere's law: 22

ln2 2 2

C S

b

S S a

a r b B dl rdNIdl d B rd rB NI Br

NI NIh NIhdr bd hdrr r a

B a a

B J s

B s a a

2

20

0

l2

ln

n

2N h bL

N IN

I a

h ba

Air Coaxial Cable

(Example 6-16) Determine the inductance per unit length of the line.Because of the cylindrical symmetry, B has only a φ-component and the current I is assumed to be uniformly distributed over the cross section of the inner conductor.

01 1 2

02 2

) Inside the inner conductor 0 : 2

) Between the inner and outer conductors : 2

Consider an annular ring in the inner conductor between and .The current

rIa r a B Ba

Ib a r b B Br

r r dr

a a

a a

2

2 20 0 0 01 2 2

2

in a unit length of this ring is linked by the flux

ln2 2 4 2

2 2The flux linkage for this annular ring is a fraction

a b a b

r a r a

I I I Idr bd B dr B dr rdr a ra r a a

rdr rdrd da

2

2 202 20 0 0

0 0

0 0

of the currents

1The total flux per unit length is ln2

ln8 2Inductance ln H/m

8 2

a a a

aI bd a r rdr rdr

a a aI I b

baLI I a

drr

b

a

Two wire transmission line• Example 6-17

– Calculate the internal and external inductances per unit length.

0 01 2

- 0 0 01 2

0

To get the mutual inductance per unit length,

, 2 2

The flux linkage per unit length 1 1 ln ln

2

ln H/m

Total in

y y

d a d a

y ya a

e

I IB Bx d x

I I Id a dB B dx dxx d x a a

dLI a

0

ductance per unit length of the two-wire line1 ln H/m4i e

dL L La

01 1 2

2 20 02 20 0

0 0

) Inside the wire conductor 0 : 2

1The total flux per unit length of the wire is 2 8

Self inductance of the two wire is 28 4

a a

i

rIa r a B BaI Id a r rdr

a a

L

a a

A solenoid with two windings• Example 6-18

2112 1

1

212 2 12 1 2 1

1

21212 1 2

1 1

N a Il

N N N a Il

L N N aI l

A conducting rectangular loop• Example 6-19

– -Determine the mutual inductance between a conducting rectangular loop and a very long straight wire

1

0 22

21 2 1 1

o

0 2 0 221

02121

2

2

where

tan 60 3

3 3 ln 12 2

The mutual inductance

3 ln 1 H2

B a

B s s aS

d b

d

Ir

d d zdr

z r d b r d b

r d b drI I bb d br d

bL b d bI d

I2

6-12. Magnetic Energy• Work needs to be expended in sending currents into conducting loops and it

will be stored as magnetic energy• A current generator is connected to the loop, which increases the current i1

from zero to I1 . • The work required to overcome emf is

• Consider two closed loops C1 and C2 carrying current i1 and i2 respectively. We keep i1 at I1 and increase i2 from zero to I2. Because of mutual coupling, some of the magnetic flux due to i2 will link with loop C1, giving rise to an induced emf that must be overcome by a voltage in order to keep i1constant at its value I1, the work involved is

• W22 must be done in loop C2 in order to counteract the induced emf as i2 is increased from 0 to I2.

• Total amount of work done

2

21 21 1 21 1 2 21 1 20

IW I dt L I di L I I

2 222 2 2 2 2 2 2 20

12

IW i dt L i di L I

2 21 1 21 1 2 2 2

1 12 2tW L I L I I L I

1 211 1 1 1 1 1 1 1 1 10

1 1: 2 2

Id dIemf L W i dt L i di L I Idt dt

Magnetic Energy in generalized form• Generalizing magnetic energy to a system of N loops carrying I1, I2, I3,,,.In,

we obtain

• For a current I flowing in a single inductor with inductance L, the stored magnetic energy is

1 1

1 1 , 2 2

N N

m jk j k k k k jk jj k k j

W L I I I L I

212mW LI

Magnetic energy & Fields• Magnetic energy in terms of field quantities

– Electrostatic Energy & Fields

• Magnetic energy

• Desirable to express the magnetic energy in terms of field quantities B and H instead of current density J and vector potential A.

212Eu E

12

is the total flux through from all sources including itself.

1 12 21 J2

j j

j

m j jj

j j

j n j jS C

N

m j j j j j j jC Vj

m V

W I

C

ds dl

W I dl d I dl J a dl

W d

B a A

A A J J

A J

Magnetic energy & Fields• Making use of the vector identity,

• Magnetic energy in B & H

• Self-inductance from stored magnetic energy

1 1 2 2m nV S

W d ds

A H H A A H A H H A A H

A J H B A H

H B A H a

0 2

2

1 1, ,A H S RR R

2

2

1 2

1 J or 2 2

m V

m m mV

W d

BW w d w H

H B

2

2 mWLI

1 J2m V

W d

A J

Magnetic energy & Fields• Example 6-20

– By using stored magnetic energy, determine the inductance per unit length of an air coaxial transmission line.

• Exercise 6.14 – A current I flows in the N turn toroidal coil in Fig. 6.23– Obtain an expression for the stored magnetic energy– Determine its self-inductance and check your result with Eq. 6-132.

1

2 22 30 0

1 40 00

The magnetic energy per unit length stored in the inner conductor1 2 J/m

2 4 16The magnetic energy per unit length stored in the region between the inner and outer

a a

mI IW B rdr r dra

2

2 22 0 0

20

0 01 22

conductor is1 12 ln J/m

2 4 4Therefore,

2 ln H/m8 2

b b

m a a

m m

I I bW B rdr drr a

bL W WI a

6-13. Hall Effect• In the steady state, the net force on the charge carrier is zero:

• This effect can be used for measuring the magnetic field and determining the sign of the predominant charge carriers ( n type or p type).

0 u B u Bq h hF q E E

0

0 0

0 00

For n-type semiconductors (electron carriers)

u aa

y

h xd

h h

uE u B

V E dx u B d

1 : Hall coefficient

x y z

x y z y z

x

y z

E u BNqE Nqu B J B

ENq J B

Magnetic forces

Nm CI dl F Bm q F u B

Magnetic forces • Magnetic forces

–

• Let us consider an element of conductor dl with a cross-sectional area S. If there are N charge carriers (electrons) per unit volume moving with a velocity u in the direction of dl,

• Magnetic force on a complete (closed) circuit of contour C.

• When we have two circuit carrying current I1 and I2, respectively, the force F12 on circuit C2.

m q F u B

Nmd NeS dl NeS dl Idl F u B u B B

I

Nm CI dl F B

12

1 2

10 112 2 2 12 12 2

12

, 4

R

C C

dlII dlR

aF B B

Magnetic Forces

• Ampere’s law of force between two current-carrying circuits.

12

2 1

2 10 1 212 2

12

N4

R

C C

dl dlI IR

aF

12 21 F F

(Example 6-21) Determine the force per unit length

12 2 12

12 1

0 1 0 21212

1

where B , the magnetic flux density at wire 2, set up by the current IUsing cylindrical sysmetry,

N/m2

The force between two wires carrying in the same di2y

z

x

I

I I Id d

F

F a

a B

B a

rection is one of attraction

Magnetic Forces • Rail gun

Magnetic Torques

• Small circular loop of radius b and carrying a current I in uniform B.

dl dl dl

B B B

F B B B B

,, no net force

z

r

dl Bdl

a B aF B a

1 1 1

1 2 1 2

2 2

2 2 2

0

sin sin sin

,

2 sin

The total torque acting on the loop

2 sin

x x

x

x x

d dF b Idl B b

dF d d dl dl dl bd

d Ib B d

d Ib B d I b B

T a a

F F

T a

T T a a

torque , d d d Idl T r F F B

F

F

2nI bm a 2

x I b B T a m BMagnetic moment:

Magnetic Torques

• Example 6-22– The force and torque on the loop

1 1 1 3

2 2 2 4

1 2

,

sum of (1)~(4) forces, all directed toward center is zero. no torgue is produced.

Net force on the loop,

z z

x x y y

x x x y y z y

y x x y y z x

BB B

dl

Ib B B Ib B

Ib B B Ib B

B aB a a

F B

F a a a a F

F a a a a F

F F F

3 4

213 1 3 1 1 2

24 2 4 1 2

13 24 1 2

0, however, they result in a net torque.

, 22

,

The total torque on the rectangular loop

= N m

x y x y

y x

x y y x

b Ib B Ib b B

Ib b B

Ib b B B

F

T F F a a

T F F a

T T T a a x x y yB B T m a a

Magnetic Torques

• Direct current motors

T m B

Electrostatics and magnetostatics