Chapter 30

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Chapter 30

Calculating the magnetic field due to a current

In vector form r r E dq

dK K

3

4 0

1

= πε

Element ds

In vector form

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277

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279

(5)
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281

Two long parallel wires a distance 2d apart carry equal currents i in opposite directions, as shown in Fig30-10a. Derive an expression for B(x), the magnitude of the resultant magnetic field for points at a distance x from the midpoint of a line joining the wires.

x

a p b d d

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B(x)=Ba(x)+Bb(x)=

) (

) ( 2 ) (

2 2 2

0 0

0

x d

id x

d i x

d i

= +

+ π

μ π

μ π

μ

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283

ienc=i1-i2

Note: the right-hand rule

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Application:

Case 1 B outside a long straight line with current

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285

Case 2 uniformly distributed current i

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Magnitude: 2.0X10-5(T) Direction: counterclockwise

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287

(13)

Only (n: the number of turns per unit length)

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289

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291

If Z>>R , B(Z)~ 3

2 0

2 Z μ iR

B(Z)= 3

0

2 Z

NiA π

μ (N turns A: the area of the loop)

In vector form 0 3 ) 2

(Z Z

B μ

π

μ K

K =

Exercises:37,39,45,67

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Chapter 31

In section29-8,we saw that if we put a closed conducting loop in a B and then send current through the loop, forces due to the magnetic field create a torque to turn the loopÆ current + magnetic field Ætorque

With i=0, torque + magnetic fieldÆcurrent?

Let us consider two experiments.

First experiment

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293

We note that

The current: an induced current

The work per unit charge in producing that current: an induced emf The process: induction

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We note that

Switch on Æ a current Switch off Æ a current

Only when there is a change in the current Faraday’s law of induction:

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295

(21)

To oppose the magnetic field increase being caused by the approaching magnet.

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297

(a)

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Direction: Clockwise(to oppose B) (b)

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299

Direction: counterclockwise

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The same Æ A small increase in the temperature of the loop.

Not a single loop!

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301

E: induced electric field

*Induced electric fields are produced not by static charges but by a changing magnetic flux. Although electric fields produced in either way exert forces on charged particles, there is an important difference between them.

Induced electric field: field lines form closed loops.

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0 or 5V?

In figure 31-14b, take R=8.5cm and dB/dt=0.13 T/S

(a) Find a expression for the magnitude E of the induced electric field at points within the magnetic field, at radius r from the center of the magnetic field. Evaluate the expression for r=5.2cm

(b) Find an expression for the magnitude E of the induced electric field at points that are outside the magnetic field. Evaluate the expression r=12.5cm

Copper ring removed

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303

(a)

∫

EKdSK = EdS =E dS = E(2πr)= dΦdtB

m mV E

cm r

dt dB E r

dt r dB r

E

r B

B BA

/ 4 . 3 2

. 5

2 ) 2 (

) (

2 2

=

=

=

=

=

= Φ

π π

π

(b)

)

! 0

(

/ 8 . 3 5

. 12

2

) (

2

2

r transforme E

m mV E

cm r

dt dB r E R

R B

B BA

=

=

=

=

=

Φ π

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: can be used to produce a desired magnetic field.

Def inductance

SI unit:

Inductance-like capacitance-depends only on the geometry of the device.

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305

Figure 31-16 shows a cross section, in the plane of the page, of a toroid of N turns like that in fig.30-21a but of rectangular cross section; its dimensions are as indicated.

(a) What is its inductance L?

(b) The toroid shown in fig.31-16 has N=1250turns, a =52mm, b=95mm,and h=13mm what is the inductance?

(a)

B= r

iN π μ

2

0

r hdr Bhdr iN

A d

B b

a b

a π

μ 2

0

= =

=

Φ K K

) / 2 ln(

1 2

0

0 iNh b a

rhdr iNh b

a π

μ π

μ =

=

Φ

∫

) / 2 ln(

/

2

0N h b a

i

N π

= μ

L= Φ

(b)

=2.45X10-3H~2.5mH L

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( ) NΦ=Li

dt L di =

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307

(Like a RC circuit)

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SÆb

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309

(a) i(0)=0

Æ Inductors: broken wire

(b)

Long after the switch has been closed ÆEquilibrium

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ÆInductors: connecting wire 3R in parallel

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311

and are both in units of power (work (energy)/time )

A 3.56H inductor is placed in series with a 12.8Ω resistor, and an emf of 3.24V is then suddenly applied across the RL combination.

(a)At 0.278s (which is one inductive time constant) after the emf is applied, what is the rat P at which energy is being delivered by the battery?

(b)At 0.278s, at what rate PR is energy appearing as thermal energy in the resistor?

(c)At 0.278s, at what rate PB is energy being stored in the magnetic field?

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(a)

i=32.4/12.8(1-e-1)=0.16A P=ξi=0.5184W~518mW (b)

PR=i2R=(0.16)2(12.8)=0.3277W~328mW (c)

PB=dUB/dt=Li(di/dt)

di/dt= L

t L

Rt

Re L e

R R

ξ τ

ξ =

) (

ÆPB=0.1907W~191mW

P= PB+ PR (energy conservation)

A solenoid: length l and cross-sectional area A

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313

A l n

L 2

μ0

= (31-33) Æ

( )

(39)

(a)

(b)

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315

Mutual induction: two coils Self induction: one coil

Def M21 of coil 2 with respect to coil 1

L (self inductance)

(Not simple!)

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(a)

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317

R1>>R2

ÆWe may take B1 to be the magnetic field at all points within the boundary of the smaller coil

(b)

Exercises: 5,63,91,98

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Chapter 32

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319

Surface 1: N pole

Surface 2: No magnetic dipole

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MK

lies along

Magnetic materials are magnetic because of the electrons within them.

We have already seen one way in which electrons can generate a magnetic field:

send them through a wire as an electric current, and their motion produces a magnetic field around the wire.

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321

Two important issues:

Protons and neutrons ( 1/1000 !)

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323

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Classical analysis

Net component: upward

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325

ElectronÆatomÆmaterial

If the combined magnetic dipole moments produce a magnetic field, then the material is magnetic.

*A diamagnetic material placed in an external magnetic field Bext develops a magnetic dipole moment directed opposite Bext . If the field is non-uniform, the diamagnetic material is repelled from a region of greater magnetic field toward a region of lesser field.

* A paramagnetic material placed in an external magnetic field Bext develops a magnetic dipole moment in the direction of Bext. If the field is non-uniform, the paramagnetic material is attracted toward a region of greater magnetic field from a region of lesser field.

Def. Magnetization M: the vector quantity, the magnetic dipole moment per unit volume

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If T>TC, ferromagnetic materialsÆparamagnetic

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327

Rowland ring: thin toroidal core of circular cross section.

(Iron core)

T<TC,Æstrong alignment of adjacent atomic dipoles in a ferromagnetic material.

Q: An iron nail, a naturally strong magnet?

A: No. There are magnetic domains.

Regions of the crystal throughout which the alignment of the atomic dipoles is essentially perfect. For the crystal as a whole, however, the domains are so oriented that they largely cancel each other as far as their external magnetic

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effects are concerned.

=?

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329

Recall ampere’s law

(a)

ienc=0

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(For r≦R) r=0 Æ B=0 (b)

(c)

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331

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(a)

(b)

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333

Exercise :17,33,39,51

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Chapter 33

New physics – old mathematics LC oscillations

See fig.33-1. Electric energy magnetic energy

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335

In contrast to RC and RL circuits, the total energy (UB+UE) in a LC circuit is conserved.

Q=CV

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qÆx 1/cÆk iÆv LÆm

m

= k

ω (Block-spring system) Æ

LC

= 1

ω (LC circuit)

(U is a constant)Æ

(U is a constant)Æ

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337

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UE+UB=

C Q 2

2

(a) In an oscillating LC circuit, what charge q, expressed in terms of the maximum charge Q, is present on the capacitor when the energy is shared equally between the electric and magnetic fields? Assume that L=12mH and C=1.7μF.

(b) When does this condition occur if the capacitor has its maximum charge at time t=0?

Sol:

(a) UE=1/2UE,max

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339

UE =

C q 2

2 and UE,max=

C Q 2

2

C q 2

2 =

C Q 2 2

1 2

Q Q

q ~0.707

= 2

(b)

0.707Q=Qcos

ω

t ωt=45=

4 π

t= πω

4 =π LC X s μs

110

~ 10 12 . 4 1

4

=

(65)

L

e Rt

C Q2 / 2

U= (decay)

(a)

(b)

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341

An external emf device supplies enough energy to make up for the energy dissipated as thermal energy in the resistance R.

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(v and i in phase)

(68)

343

(v and i 900 out of phase )

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(70)

345

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347

In an RLC circuit, Let R=160Ω, c=15.0μF,L=230mH, fd=60.0Hz and ξm=36.0V

(a) What is the current amplitude?

(b) What is the phase constant Φ?

Sol:

(a)

Ω

=

+

= R2 (XL XC)2 184 Z

I= A

Z

m =0.196 ξ

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(b)

X XL C

513 . 0 ) 564 . 0 ( tan

564 . 0 tan

1 =

=

=

= φ

φ

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349

(d) With that change in capacitance, what would Pav be ?

(a)

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(b)

(c)

Æ

(d)

rms

rms

=

P I cos(00) 72.0(W)

rms rms

av = ξ =

I2R: Ohmic losses

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351

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Exercises:27,53,75,87

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