Lecture Note #5A Chapter 5. Magnetostatics & & & & v Belearning.kocw.net/KOCW/document/2015/inha/kimkyunghon/... · 2016. 9. 9. · Inha University 1 Chapter 5. Magnetostatics Lecture

Inha University 1

Chapter 5. MagnetostaticsLecture Note #5A

5.1 The Lorentz Force Law

5.2 The Biot-Savart Law

5.3 The Divergence and Curl of B

5.4 Magnetic Vector Potential

• 움직이는 전하 (= 전류) → 주위에 자기장 형성

- 전기장과 자기장 속에서 있는 전하에 작용하는 힘

- Cyclotron motion

- 자기장 속에서 움직이는 전하에 작용하는 Lorentz force와 전하가 움직이는방향은 수직이므로 자기장이 전하에 한 일은 영.

• 전하가 자기장 속에서 이동할 때 작용하는 힘 = Lorentz Force(즉, 자기장 속에서 전류에 작용하는 힘)

• 전류의 흐름: 연속방정식 (Continuity equation)

dt

dqI

BvQFmag

BvEQF

BldIFmag

dt

dJ

0 dtvBvQldFdW magmag

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from Eqs. (4.30), (4.32), (4.34), (4.22)Electrostatics (정전기학)

+

- 전기장의 공간적인 분포- 전하 간에 작용하는 힘- 전위차

- 전자의 이동 → 전류 → 유도 자기장

-

Electrodynamics (전기동력학)

Magnetostatics (정자기학)

ro

E

0

E

B

0v q

0v q

(일정한 전류가 흐르는 경우)

(자기장이 일정한 경우)

Inha University 3


Q

0v iq

vv iq

iq

Q

For stationary charges For moving charges

“Test” charge

“Test” charge

In this case, what kinds of forces are acting

on the test charge?

When there are moving charges along the wire,

what kinds of forces are acting on the test charge?

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5.1.1 Magnetic Fields 5.1.2 Magnetic Forces

Right-hand rule:

• Thumb (엄지) Direction of Current

(전류 방향)

• The other four fingers (나머지 4손가락)

Direction of Magnetic Field

(자기장 방향)

Q

B v

BvQFmag

When there is a moving charge q with velocity v

inside the magnetic field B, the force acting on the

charge is

F

“Lorentz Force Law”

When a moving charge q is placed in the

presence of both electric and magnetic fields, the

force acting on the charge is

Q

B

v

BvEQF

Fmag

E

(5.1)

(5.2)

Felect

Ftotal

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5.1.2 Magnetic Forces

I

BveFmag

ve

- continued

I

For two wires with the same current

directions

F

I

BveFmag

ve

I

For two wires with the opposite current

directions

F

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: the cyclotron formula

[Example 5.1]

where m = the particle’s mass, and p = mv.

(5.3)

Cyclotron motionThe archetypical motion of a charged particle in a magnetic field is circular, with the

magnetic force providing the centripetal acceleration.

B

Q

v

F

R

pv

R

vmQvBBvQFmag

2

RBQp

This is used to determine the momentum of the charged

particle by measuring the radius of the trajectory.

1) When the charged particle moves in a plan

perpendicular to B.2) When the charged particle moves in a plan

perpendicular to B but with some additional

speed v|| parallel to B

v

v cos

v||=v sin

BParticle path

The charged particle follows

a helical trajectory.Momentum of

the charged particle =

(자기장 속에서 움직이는 전하에 작용하는 힘에 의한 원운동)

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[Example 5.2]

The position of the particle at any time t = (0, y(t), z(t)).

Cycloid MotionWhen a uniform electric field is applied to a positively charged particle in addition

to a magnetic field at right angles, what path will it follow?

zzyymamzyByzBzEQBvEQF ˆ ˆ ˆ ˆ ˆ

Then, we apply the Newton’s 2nd law:

For the particle at rest initially, the electric field accelerates the charge in the z-direction.

E

B

Q

Q

v

Fmag

Felec

The velocity of the particle at any time t = (0, ý(t), ź(t)). where

,dt

tdyty

dt

tdztz

Thus,

zyByzB

B

zy

zyx

Bv ˆ ˆ

00

0

ˆˆˆ

,2

2

dt

tydty

2

2

dt

tzdtz

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zzyymzyBzEyzBQ ˆ ˆ ˆ ˆ ˆ

The solution of this equation becomes

For convenience, let

321 sincos CtBEtCtCty

[Example 5.2] - Continued (1)

(5.4)

, ymzBQ zmyBQEQ

m

BQ

: “cyclotron frequency”

, zy Thus, we get

y

B

Ez

(5.5a)

y

B

Ez

dt

yd

2

Differentiating Eq. (5.5a) & using Eq. (5.5b), we get

y

B

Ey 2

(5.5b)

(5.6a)

Eq. (5.6a) Eq. (5.5b) : 412 sincos CtCtCtz (5.6b)

, cossin 21 BEtCtCty

BEtCtC

B

Ey

B

Ez cossin 21

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We let

Since the particle started from rest,

321 sincos CtBEtCtCty


: The formula for a circle of radius R centered at (0, Rt, R)

at the origin

(5.7a)Thus, finally we get

B

ER

(5.7b)

(5.8)

Eqs. (5.6a) &

(5.6b)

412 sincos CtCtCtz

(5.9)

000 zy 000 zy

BEtCtCty cossin 21

tCtCtz cossin 12

00 31 CCy

00 42 CCz

00 2 BECy

00 1 Cz 01 CB

EC

2

13 CC

24 CC

ttB

Ety

sin

tB

Etz

cos1

Since , 1cossin 22 tt 222RRztRy

The y-coordinate of the center keeps moving with time t with a

constant speed u :

B

ERu

The moving curve is “cycloid”.

(5.10)

Q

v

FmagFelec

0222 R

dt

dRztRy

dt

d

,Rdt

dy 0

dt

dz

022

Rz

dt

dztRyR

dt

dy

2R

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because

If the charge Q is moved to dl = v dt in a magnetic field, the work done is

0 dtvBvQldFdW magmag

Further discussion will be covered in Chapter 8.

“Magnetic forces do no work on the charge ??”

(5.11)

5.1.2 Magnetic Forces - continued

Work done by Lorentz force :

vBv

i

q

B

1F

v

l

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i

q

B

1F

v

l


5.1.3 Currents

Current in a wire : The charge per unit time passing a given point

• The electric current in wire is based on electron

flow. (In some semiconductor materials, both

electrons and holes moves in the opposite directions.)

• The direction of the electron flow is opposite to the

direction of current flow.

dt

dqI

s

CA 1 1

dlBIdlBvdqBvFmag

(5.12)

(5.13)

Unit : Amperes = Coulmb/sec

vdt

dlI

Free electron

AV BV

BA VV

EeFe

Ei

: line chargev : charge flow speed

Magnetic force acting on a segment of current-carrying wire is

vI

( We consider that =q/L is positive charge.)

BldIFmag

Since the current is constant along the wire, BldIFmag

(5.14)

(5.15)

(5.16)

(5.17)

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[Example 5.3]

If I > Ib, the loop rises.

aBIF up

mag

ahBIhFW up

magmag

The magnetic force acting on the wire is

(Solution)

IThis magnetic force should be balanced with the weight :

mgaBIb Ba

mgIb

Then, the work done by the magnetic force over the distance of rise h is

i

q

B

1F

w

l

Let the charge flow speed be w for the current I.

wI

(5.18)

(5.19)

aBIBaBqF up

mag ww Thus,

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[Example 5.3]

During a time interval dt, the charges moves a horizontal distance dl

auBquBF horiz

mag

ahBIdtuaBdtuaBdlFW horiz

maghoriz ww

The speed of the upward rise due to the magnetic

force acting on the charge q is u

(Solution)

The magnetic force acting on the upward motion of the charge

becomes

dtdl w

Then, the work done by the horizontal magnetic force on the

charge over the distance dl is

i

q

B

1F

w

l

- continued (1)

: the same as the work done to flow the charge

along the wire in the magnetic field.

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I

I

I


5.1.3 Currents

Surface Current Density

The current per unit width ( = current ribbon density) :

vvdl

d

dl

dK

I

daBKdaBdqBFmag v v

(5.22)

where : surface charge density, v : speed of the charge flow

- Magnetic force acting on the current per unit width is

vvda

d

da

dJ

I

v dBvFmag

(5.24)

(5.25)

(5.27)

- continued

vK

(5.23)

The volume current density (= current per unit area) :

where : volume charge density, v : speed of the charge flow

vJ

(5.26)

- Magnetic force acting on the current per unit area is

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2a

I

A

IJ

,ksJ

(5.24)The volume current density (= current per unit area) :

Since the total current in the wire is

[Example 5.4]

(Solution)

(a)

(b)

3

22

3

0

2

0

2

0

kadsskddssksJdaI

aa

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5.1.3 Currents

Current Density vs. Current

(5.28)

From the divergence theorem,

According to charge conservation law

SS

adJdaJI

(5.29)

- continued

: the continuity equation (연속 방정식)

(The minus sign reflects the fact that an outward flow decreases the charge left of the volume v.)

v v

dJadJS

v v v v vvvv

ddt

dd

dt

ddvdJ

vJ

from Eq. (5.26)

dt

dJ

Thus,

dt

dqI

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Chapter 5. Magnetostatics





• 일정한 전류가 흐르는 도선 주위의 자기장을 구하는 방법 : Biot-Savart Law

• 일정한 전류가 흐르는 두 도선 사이에 작용하는 힘

P

2

0

2

0ˆ'

4'

Î

4 r

rldIdl

r

rrB

d

II

L

Ff

mag 210

2

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Electrostatics (정전기학)

- Stationary charges

Magnetostatics (정자기학)

ro

E

0

t

5.2.1 Steady Currents

+ - Constant electric fields

- Steady currents Constant magnetic fields

0

t

J

: charge density

J : current density (=I/A)

constant

constant J

I

0

t

Q

0 J

: the continuity equation (5.33)

(5.32a)

(5.32b)

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The Biot-Savart Law:

- The magnetic field of a steady line current is given as

2

0

2

0ˆ'

4'

Î

4 r

rldIdl

r

rrB

5.2.2 The Magnetic Field of a Steady Current

Unit of the magnetic field (B) :

0 = 4 x 10-7 N/A2 : the permeability of free space

(tesla)mA

N 1T 1

(5.34)

(5.35)

(5.36)

I

r

P

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From the Biot-Savart law,

[Example 5.5]

(5.37)

tan' sl

r

At the point P, the direction of (dl’ x r ) points out of the page.

(Solution)

cos'sin'ˆ' dldlrld

2

0

2

0ˆ'

4'

Î

4 r

rldIdl

r

rrB

ds

dl2cos

' dxx

dxxxd2

2

cos

1 sectan

cosrs 2

2

2

cos1

cos1

srsr

1200

22

2

0 sinsin 4

cos 4

coscos

scos

4

2

1

2

1

s

Id

s

Id

sIB

xyyxyx cossincossinsin

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In the case of an infinite wire,

(5.38)

r

Eq. (5.37)

,2

1

s

IB

2

0


22

: B is inversely proportional to

the distance s from the wire.

ˆ 2

0

s

IB

(5.39)

: The direction of B follows the right-hand rule.

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The magnetic field at the wire (2) due to the wire (1) is

(5.40)

Eq. (5.17)

Calculation of the electric force acting on two long and parallel wires

the force is repulsive.

The force per unit length becomes

The Lorentz force law acting on the wire (2) due to the wire (1) is

ydld

II

d

IldI

BldIFmag

ˆ 2

ˆ2

210

210

22

122

x

y

z

ˆ 2

10

d

IB

d

II

L

Ff

mag 210

2

The magnitude of the force :2

102

102

2

2L

d

IIdl

d

IIF

The direction of the force : toward the wire (1) “attractive force”

i) When the directions of the two currents are parallel,

ii) When the directions of the two currents are antiparallel,

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[Example 5.6]

• The magnetic field dB attributed to the segment dl’ points

as shown in the figure.

Since

P

z

I

R• As we integrate dl’ around the loop, dB sweeps out a cone.

Bd

horizontaldB

verticaldB

The horizontal components cancel, and the vertical components combine,

to give the following relation:

Rr

Idl

r

I

r

dlIzB

2

cos

4'

cos

4cos

'

4 2

0

2

0

2

0

r

Eq. (5.34)

2

0ˆ'

4 r

rldIrB

,and cos 22 zRrr

R

2322

2

0

2 zR

RIzB

(5.41)

Find the magnetic field a distance z above the center of a circular loop of radius R, which

carries a steady current I.

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For surface and volume current densities, the Biot-Savart law becomes

(5.34)

- continued

All are related to the “charge flow”

(A point charge cannot provide a steady current.)

I5.2.2 The Magnetic Field of a Steady Current

2

0

2

0ˆ'

4'

Î

4 r

rldIdl

r

rrB

'ˆ'K

4 2

0 dar

rrrB

For a current flowing in a wire, the Biot-Savart law is

The current per unit width (= current ribbon density) :

vvdl

d

dl

dK

I

(5.22)

'ˆ'J

4 2

0 dvr

rrrB

vvda

d

da

dJ

I

(5.25)

The volume current density (= current per unit area):

(5.13)vdt

ldI

The line current :

vI

vK

vJ

When there are a collection of many source currents,

the total magnetic field can be obtained by superposition principle.

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The magnetic force acting on a current-flowing wire due to the magnetic field B is

(Hint)

aSS

aI

aS

aI

sS

aIFFFFFFFtotal

222

22

0

2

0

2

0314321

BldIFmag

Eq. (5.17)

(1)

(2)

(3)

(4)

ˆ 2

0

s

IB

For side (1), the magnetic field is constant:

Eq. (5.39)

(1)

I

I

I

I(2)

(3)

(4)

S

aS

Idl

S

IIF

2 2

2

001

,

2

01

S

IB

(upward)

For side (2),

S

aSIds

s

IIdl

s

IIdlBIF

aS

S

aS

S

aS

Sln

2 2 2

2000

22

(toward right-hand side)

For side (3), the magnetic field is constant : aS

IB

2

03

a

aS

Idl

aS

IIF

2 2

2

003

(downward)

For side (4), 2

2000

44 ln 2 2 2

FS

aSIds

s

IIdl

s

IIdlBIF

S

aS

S

aS

S

aS

(toward left-hand side)

(upward)

Problem 5.10

Find the force on a square loop placed as shown in Figure near an infinite straight

wire. Both the loop and the wire carry a steady current I.

(opposite-direction to F(2))

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Next Class

Chapter 5. Magnetostatics





Documents

Lecture Note #5A Chapter 5. Magnetostatics & & & & v Belearning.kocw.net/KOCW/document/2015/inha/kimkyunghon/... · 2016. 9. 9. · Inha University 1 Chapter 5. Magnetostatics Lecture