276

Chapter 30

Calculating the magnetic field due to a current

In vector form

rrdqEd 3

041πε

=

Element ds

In vector form

277

278

279

280

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

282

Sol:

B(x)=Ba(x)+Bb(x)=)()(2)(2 22

000

xdid

xdi

xdi

−=

−+

+ πμ

πμ

πμ

283

ienc=i1-i2 Note: the right-hand rule

284

Application:

Case 1 B outside a long straight line with current

285

Case 2 uniformly distributed current i

286

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

287

288

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

289

290

291

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

20

2 ZiRμ

B(Z)= 30

2 ZNiA

πμ

(N turns A: the area of the loop)

In vector form 30

2)(

ZZB μ

πμ

=

Exercises:37,39,45,67

292

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 loop current + magnetic field torque

With i=0, torque + magnetic field current?

Let us consider two experiments.

First experiment

293

We note that

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

294

We note that

Switch on a current

Switch off a current

Only when there is a change in the current

Faraday’s law of induction:

295

296

To oppose the magnetic field increase being caused by the approaching

magnet.

297

(a)

298

Direction: Clockwise(to oppose B)

(b)

299

Direction: counterclockwise

300

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

Not a single loop!

301

New form of faraday’s law

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.

302

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

303

(a)

∫∫∫Φ

===•=•dt

drEdSEdSESdE B)2( π

mmVEcmrdtdBrE

dtdBrrE

rBBAB

/4.32.52

)2(

)(

2

2

=⇒=

=

=

==Φ

ππ

π

(b)

)!0(/8.35.12

2

)(2

2

rtransformeEmmVEcmr

dtdB

rRE

RBBAB

⇒≠=⇒=

=

==Φ π

304

: can be used to produce a desired magnetic field.

Def inductance

SI unit:

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

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= riNπ

μ2

0

hdrr

iNBhdrAdBb

a

b

a πμ2

0∫∫∫ ==•=Φ

)/ln(2

12

00 abiNh

hdrr

iNh b

a πμ

πμ

==Φ ∫

)/ln(2

/2

0 abhN

iNπ

μ=Φ L=

(b)

=2.45X10-3H~2.5mH L

306

( ) NΦ=Li

dtdiL− =

307

(Like a RC circuit)

308

S b

309

(a)

i(0)=0

Inductors: broken wire

(b)

Long after the switch has been closed

Equilibrium

310

Inductors: connecting wire

3R in parallel

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?

312

(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

LRt

eR

eLR

Rτξξ −−

=)(

PB=0.1907W~191mW

P= PB+ PR (energy conservation)

A solenoid: length l and cross-sectional area A

313

AnlL 2

0μ= (31-33)

( )

314

(a)

(b)

315

Mutual induction: two coils

Self induction: one coil

Def M21 of coil 2 with respect to coil 1

L (self inductance)

(Not simple!)

316

(a)

317

R1>>R2

We may take B1 to be the magnetic field at all points within the boundary of

the smaller coil

(b)

What about ?

Exercises: 5,63,91,98

318

Chapter 32

319

Surface 1: N pole

Surface 2: No magnetic dipole

320

M 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.

321

Two important issues:

Protons and neutrons ( 1/1000 !)

322

323

324

Classical analysis

Net component: upward

325

Electron atom material 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

326

If T>TC, ferromagnetic materials paramagnetic

327

Rowland ring: thin toroidal core of circular cross section.

(Iron core)

T<TC, strong 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

328

effects are concerned.

=?

329

Recall ampere’s law

(a)

ienc=0

330

(For r≦R) r=0 B=0

(b)

(c)

331

332

(a)

(b)

333

Exercise :17,33,39,51

334

Chapter 33

New physics – old mathematics

LC oscillations

See fig.33-1. Electric energy magnetic energy

335

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

conserved.

Q=CV

336

q x 1/c k i v L m

mk

=ω (Block-spring system) LC1

=ω (LC circuit)

(U is a constant)

(U is a constant)

337

338

UE+UB=C

Q2

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

339

UE =C

q2

2

and UE,max= CQ2

2

Cq2

2

=C

Q22

1 2

QQq 707.0~2

=⇒

(b)

0.707Q=Q tωcos

ωt=45。=4π

t=ωπ4

= ssXLC μπ 110~1012.14

4−=

340

LRte

CQ /

2

2−U= (decay)

(a)

(b)

341

An external emf device supplies enough energy to make up for the energy

dissipated as thermal energy in the resistance R.

342

(v and i in phase)

343

(v and i 900 out of phase )

344

345

346

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=−+= 184)( 22CL XXRZ

I= AZm 196.0=

ξ

348

(b)

radR

XX CL

513.0)564.0(tan

564.0tan

1 −=−=

−=−

=

−φ

φ

349

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

(a)

350

(b)

(c)

(d) ZI rms

rmsξ

= )(0.72)0cos( 0 WIP rmsrmsav == ξ

I2R: Ohmic losses

351

352

Exercises:27,53,75,87