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
277
279
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
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
283
ienc=i1-i2
Note: the right-hand rule
Application:
Case 1 B outside a long straight line with current
285
Case 2 uniformly distributed current i
Magnitude: 2.0X10-5(T) Direction: counterclockwise
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Only (n: the number of turns per unit length)
289
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
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
293
We note that
The current: an induced current
The work per unit charge in producing that current: an induced emf The process: induction
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:
295
To oppose the magnetic field increase being caused by the approaching magnet.
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(a)
Direction: Clockwise(to oppose B) (b)
299
Direction: counterclockwise
The same Æ 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.
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)
∫
∫
∫
EK•dSK = E•dS =E dS = E(2πr)= dΦdtBm 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
: 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= 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
( ) NΦ=Li
dt L di = −
307
(Like a RC circuit)
SÆb
309
(a) i(0)=0
Æ Inductors: broken wire
(b)
Long after the switch has been closed ÆEquilibrium
ÆInductors: 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?
(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
313
A l n
L 2
μ0
= (31-33) Æ
( )
(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!)
(a)
317
R1>>R2
ÆWe 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
Chapter 32
319
Surface 1: N pole
Surface 2: No magnetic dipole
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.
321
Two important issues:
Protons and neutrons ( 1/1000 !)
323
Classical analysis
Net component: upward
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
If T>TC, ferromagnetic materialsÆparamagnetic
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
effects are concerned.
=?
329
Recall ampere’s law
(a)
ienc=0
(For r≦R) r=0 Æ B=0 (b)
(c)
331
(a)
(b)
333
Exercise :17,33,39,51
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
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
337
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
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
=
L
e Rt
C Q2 / 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.
(v and i in phase)
343
(v and i 900 out of phase )
345
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
(b)
rad R
X XL C
513 . 0 ) 564 . 0 ( tan
564 . 0 tan
1 − =−
=
−
− =
= φ −
φ
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(d) With that change in capacitance, what would Pav be ?
(a)
(b)
(c)
Æ
(d)
I
rmsξ Z
rms=
P I cos(00) 72.0(W)rms rms
av = ξ =
I2R: Ohmic losses
351
Exercises:27,53,75,87
