6. THE STATIC MAGNETIC FIELD 6-1 LORENTZ FORCE EQUATION

Chapter 6 6-1 Proprietary of Prof. Lee, Yeon Ho

6. THE STATIC MAGNETIC FIELD 6-1 LORENTZ FORCE EQUATION

Lorentz force equation ( )e m q= + = + ×F F F E v B

Example 6-1 An electron has an initial velocity o o yv=v a in a uniform magnetic flux density o zB=B a . (a) Show that the electron moves in a circular path.

(b) Find the radius or of the circular path in terms of electron charge e , electron mass m and the initial speed ov .

Solution (a) The magnetic force on the electron is m e= ×F v B . Because mF is perpendicular to v , it

only changes the direction of v , and corresponds to the centripetal force on the electron. (b) The magnetic force on the electron m o oev B ρ=F a ( )0e <

The force on the electron, mass× acceleration d d dm mdt d dt

φ=

φv v

where tφ = ω and ω is angular velocity. Rewrite it using ov φ=v a and /d dφ ρφ = −a a

( ) ( ) ( )oo

d v d tm mv

d dtφ

ρ

ω= − ω

φ

aa


From two equations

oeBm

−ω =

Using o ov r= ω

oo

o

mvre B

= −

where 0e < . 6-2 BIOT-SAVART LAW

Biot-Savart law

24

Idd

π′×

=l a

H R

R : ′= −r rR

For a steady current flowing in a wire

24C

Id′

′ ×= ∫

l aH R

Rπ (6-7)


Id ′l in terms of J and sJ sId dv ds′ ′ ′= =l J J

For a current density J

24V

dvπ′

′×= ∫

J aH R

R (6-9)

For a surface charge density sJ

24s

S

dsπ′

′×= ∫

J aH R

R (6-10)

All the unit vectors in the unprimed coordinates The integration in the primed coordinates.

Example 6-2 H of a steady current in a straight wire

Determine H at a point ( ); , ,P x y z due to an infinitely long straight conducting wire of negligible thickness carrying a steady current I and lying along z -axis.

Solution The line current has cylindrical symmetry, and Biot-Savart law only gives dH in φa direction. So we expect ( )H φ φ= ρH a The translation vector, : ' zzρρ ′= − = −r r a aR The vector differential length, : zd dz′ ′=l a


From the definition

( )3 3/22 2

( )44

z zz z

z z

Idz z dzI

z

′ ′=+∞ =+∞ρ φ

′ ′=−∞ =−∞

′ ′ ′× ρ − ρ= =

ππ ′ρ +∫ ∫

a a a aH

R

(6-11)

or

2I

φ=πρ

H a

(6-12)

H also has cylindrical symmetry, being independent of φ and z .

• For a finite line current

( )

1

1

2

2

3/2 2 22 24 4

z zz

zz z

dzI I zzz

′ ′=′ φ

φ′′ ′=

′ ′ρ= =

π πρ ′ρ +′ρ +∫

aH a

Referring to Fig. 6-4

[ ]1 2cos cos4I

φ= θ + θπρ

H a (6-13)

This formula can be applied even for 1θ

or 2θ larger than 90o .


Example 6-3 Find H , at a point ;( , ,0)p a b in the first quadrant, due to an infinite right-angled wire carrying a steady current I .

Solution

For 0y < : 1θ = α , 2 0θ = and aρ =

[ ] ( )1 cos 1

4 zIa

= α + −π

H a

For 0x < : 1 0θ = , 2θ = β and bρ =

[ ] ( )2 1 cos

4 zIb

= + β −π

H a

In Fig. 6-5

( )2 2

cos sin 90o ba b

α = − α − = −+

( )2 2

cos sin 90o aa b

β = − β − = −+

.

The magnetic field intensity at p is

( ) ( )

( )

1 2 2 2 2 2

2 2

1 14 4

4

z z

z

I b I aa ba b a bI a b a bab

⎡ ⎤ ⎡ ⎤= + = − + − + − −⎢ ⎥ ⎢ ⎥π π+ +⎣ ⎦ ⎣ ⎦

⎡ ⎤= + − +⎣ ⎦π

H H H a a

a


Example 6-4 Determine H at a point ( ); 0,0,p b due to a circular wire of radius a centered at the origin in xy -plane, and carrying a steady current I.

Solution

In cylindrical coordinates : zb a ρ= −a aR

d ad φ′ ′= φl a . From Eq. (6-7)

2

2 30

22 2

3 30 0

( )4 4

4 4

z

C

z

Ia d b aI d

Iab Iad d

′φ = π φ ρ

′ ′φ =

′ ′φ = π φ = πρ

′ ′φ = φ =

′′ φ × −×= =

π π

′ ′= φ + φπ π

∫ ∫

∫ ∫

a a al aH

a a

R

R R

R R (6-14)

The first integral cancels to zero due to ( ) ( )ρ ρ′ ′φ = φ = − φ = φ + πa a .

2

3/22 22z

Ia

a b=

⎡ ⎤+⎣ ⎦H a

(6-15)


6-3 AMPERE’S CIRCUITAL LAW

The closed line integral of H is equal to the total current enclosed by the path of integration.

Cd I=∫ H li : Right-hand rule for C and I

• The total enclosed current I

The continuity equation is

0S

d =∫ J si (6-17)

The total current enclosed by C is equal to the total current across the surface bounded by C .

• Ampere’s circuital law

C Sd d=∫ ∫H l J si i

: Right-hand rule for C and S

Applying Stokes’s theorem

S Sd d∇× =∫ ∫H s J si i

(6-19)

The point form of Ampere’s circuital law ∇× =H J (6-20) To determine H by Ampere’s circuital law, the closed path C should be chosen such that the magnetic field intensity H is at least constant on the path C .


Example 6-5 Find H , by Ampere’s circuital law, due to an infinitely long straight wire lying along z -axis and carrying a steady current I.

Solution

The cylindrical symmetry → H independent of φ and z Biot-Savart law. → dH due to Id ′l only in φa direction We expect ( )Hφ φ= ρH a : H φ depends only on ρ On C of a radius ρ

( )2 2

0 0Cd H d H d I

π π

φ φ φ φφ= φ== ρ φ = ρ φ =∫ ∫ ∫H l a ai i

2IHφ φ φ= =πρ

H a a

The same result with that by Biot-Savart law


Example 6-6 An infinitely long coaxial cable carries a uniform steady current I . The radii of the cylindrical surfaces are , a b and c , respectively. Find H in the region 0ρ > by Ampere’s circuital law.

Solution The uniform current → Cylindrical symmetry Current elements at 1φ and 1−φ → Only H φ -component We can expect ( )H φ φ= ρH a : H φ is constant on the circle 1C In the region 0 a< ρ ≤ , ( )

1

2

02

Cd H d H

π

φ φ φ φφ== ρ φ = π ρ∫ ∫H l a ai i

The current enclosed by 1C , 2

2 Iaπρπ

The result

22IHaφ φ φ

ρ= =

πH a a

: ( 0 a< ρ ≤ )

The enclosed current I : (a b≤ ρ ≤ )

2 2

2 2

cIc b

− ρ−

: (b c≤ ρ ≤ )

0 : ( cρ ≥ ) The result

2I

φ=πρ

H a

(a b≤ ρ ≤ ) (6-23b)


2 2

2 22I cc b φ

− ρ=

πρ −H a

(b c≤ ρ ≤ ) (6-23c)

0=H ( cρ ≥ ) (6-23d) Example 6-7

Find H of a toroidal core having N turns of a wire carrying a steady current I . The toroid has the inner radius of a and the outer radius of b .

Solution

Rotation symmetry about z -axis → Constant H on 1C I across 1φ = φ plane → Hρ and zH , samll Large enclosed current NI → H φ , large An ideal toroidal coil → H φ φ=H a inside 0=H outside. In the region a b< ρ <

1

2

02

C Cd H d H d H

π

φ φ φ φ φ= ρ φ = ρ φ = π ρ∫ ∫ ∫H l a ai i

The total enclosed current isNI

2NIH φ = πρ

2NIHφ φ φ= =πρ

H a a : (a b< ρ < )

0=H (No current enclosed) : aρ < and bρ >


6-4 MAGNETIC FLUX DENSITY

In free space o= μB H : 74 10o

−μ = π × , permeability of free space The total magnetic flux

SdΦ = ∫ B si .

An infinite line current I → H , concentric circles An arbitrary distribution of I → H , always closed. ↑ Linear property Gauss’s law for the magnetic field

0S

d =∫ B si Point form of Gauss’s law 0∇ =Bi

• Maxwell’s equations for the static electric and magnetic fields

00

v∇ = ρ

∇ × =∇ =∇ × =

DEBH J

i

i

In integral forms

0

0

vS V

C

S

C S

d dv

d

d

d d

= ρ

=

=

=

∫ ∫∫∫∫ ∫

D s

E l

B s

H l J s

i

i

i

i i

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6. THE STATIC MAGNETIC FIELD 6-1 LORENTZ FORCE EQUATION