sources of magnetic field - university of hawaiiplam/ph272_summer/l8/28_lecture_lam.pdf · magnetic...

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures for

University Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 28

Sources of MagneticField


Topics for Chapter 28

• Magnetic field generated by a moving charge

• Magnetic field generated by a current-carrying

conductor, current loop and infinitely long,

straight conductor.

• Magnetic force between current-carrying

conductors

Intermission

• Ampere’s Law

• iClicker questions


Magnetic field generated by a moving charge

We know that two charges moving in the same direction attracts,

while two charges moving in opposite directions repel.

Let’s work backward to deduce the magnetic field generated by

q1.

+ v1q1

+ v2q2

F2

r F 2 = q2

r v 2

r B 1( )

r B 1 points out of the page.

+ v1q1

+v2q2

F2

r F 2 = q2

r v 2

r B 1( )

r B 1 points out of the page.

=>The magnetic field lines form circles around q1


The magnetic field of a moving charge

r B =

μo

4

q

r v ˆ r ( )r2 (for apoint charge moving at constant

r v )

μo = Universal constant = 4 x10 7 Tesla meter

Ampere

(associates with magnetic field)

o = 8.854x10 12 C2

N m2

(associates with electric field)

c1

oμo

= 3x108 m

s (= speed of light)


Example

Follow Example 28.1

Find the ratio of FE/FB

1

1

2

2


Magnetic field of a current element

qr v Id

r l (current segment)

dB =μo

4

I d

r l ˆ r ( )r2

The total magnetic generated by

the entire current loop is :

B =μo

4

I dr l ˆ r ( )r2

This is NOT an easy integral; we'll

consider special cases.


Magnetic field of a straight current-carrying conductor

dB =μo

4

I d

r l ˆ r ( )r2

dr l = dyˆ j

r r = xˆ i yˆ j ˆ r =

xˆ i yˆ j

x 2 + y 2

dr l ˆ r =

(dy)x

x 2 + y 2( ˆ j ˆ i ) =

(dy)x

x 2 + y 2

ˆ k ( )

B =μo

4

I dr l ˆ r ( )r2

=μoI

4

(dy)x

(x 2 + y 2)3 / 2ˆ k ( )

a

a

=μoI

4

2a

x x 2 + a2

B =μoI

2 x

for a

This is a simple case to integrate

because dB contributed from all the

segments point in the same direction.


Fields generated by multiple long wires


Magnetic field of a circular current loop

We will focus on a special case where point P is at the center of the circle.

Again, this is a simple case where dB from all the segments points in the samedirection (x -direction)

dB =μo

4

I d

r l ˆ r ( )r2

dr l ˆ r = dlˆ i

r2 = a2 (same for all segments)

B =μo

4

I dr l ˆ r ( )r2

=μoI

4 a2

( dl)ˆ i

=μoI

4 a2

(2 a)ˆ i

=μoI

2a

i


Forces and parallel conductors

• Two parallel currents in the same

direction attract.

• Assuming “infinitely long wires”,

we’ll calculate the force per unit

length.

Magnetic field generated by I1

at the location of I2 is given by :r B 1 =

μoI12 r

(direction indicated in figure)

Force per length on I2 is given by :r F /l =

r I 2

r B 1 =

μoI1I2

2 r (direction indicated in figure)


Intermission


Guass’s Law for magnetic field

Gaussian

surface

(r B • ˆ n )dA =μo enclosed "magnetic charge"( ) = 0

since "magnetic charges" always occurs in pair;

north and south poles.

See figure, B - field lines enter and exit the

Gaussian surface, consistent with (r B • ˆ n )dA =0.


Ampere’s Law

Amperian loop can

be any shape

r B • d

r l = positive

r B • d

r l = negative

No current enclosed.

r B • d

r l = μoIenclosed


Ampere’s Law - Examples

• Another right-hand-rule todetermine whether thecurrent is “positive” or“negative” in application ofAmpere’s Law


Use Ampere’s Law to find magnetic field generated by current with cylindricalsymmetry

• An infinitely long cylinder carrying a current I.


Use Ampere’s Law to find magnetic field of a(infinitely long) solenoid

• A helical winding of wire on a cylinder.

• Refer to Example 28.9 and Figures 28.22–28.24.


Use Ampere’ Law to find magnetic field of a toroidalsolenoid

• A doughnut-shaped solenoid.

• Refer to Example 28.10 and Figure 28.25.


Magnetic materials

• Consider Figure 28.27at right.

• Consider Figure 28.28below.


Magnetic materials II

• Consider Figure 28.29 below.

• Follow Example 28.12.

sources of magnetic field - university of hawaiiplam/ph272_summer/l8/28_lecture_lam.pdf · magnetic...

Documents