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# shows0 volts for
each channel
screen menu,calcs & info
menu buttons
cursorcontrol
cursor
triggeringcontrol
usualchannelcontrols
measure
autoset
# shows0 volts for
each channel
screen menu,calcs & info
menu buttons
cursorcontrol
cursor
triggeringcontrol
usualchannelcontrols
measure
autoset
B
BB
B
Boutside=0 Boutside=0
uniformmagnetic
field inside
Ideal Solenoid
BIN
Use Ampere’s Lawto Find Magnetic Field
Am
peria
n Lo
op
L
B d
s
whole loop
oItotalenclosed
B d
s
insidesolenoid
oNI
Bindy0
L oNI
Bin dy0
L oNI
BinL oNI
Bin oNI
LBin onIN solenoid loops
enclosed, eachwith current I.
where n is “loop density” N/L of solenoid.
(Explain each step in your report.)
voltage
t
VR=RIR=RIL
Bampere
L
The current in the solenoid creates a magnetic field inside the solenoid due to Ampere’s Law.
The changing magnetic field inside the solenoid causes a back EMF (voltage) due to Faraday’s Law. Notice that dI/dt causes a phase shift.
Direction of currentinside the resistor?
NS
velocity
R
B
dB
dt 0Inside the solenoid:
NS
velocity
R
B Binduced
I
ReceiverTransmitter
Oscillating voltagereceived is measurable.
Oscillating transmittingmagnetic fields.
Oscillating transmitting voltage.
Transmitting magnetic fields reach inside coils.
voltage
The LRC Circuit - AC Driven
VLVC
VR
voltage
The LRC Circuit - AC Driven: Source from Addition
VLVC
VR
Vsource
A B C
0.1 F50 mH
100
Iamplitude
fdrive
fresonance
Iamplitude
fdrive
fresonance
Large R Small R
VR(t)
VS(t)
VR(t)
VS(t)
out of phase in phase
45o
(r) A
r
System: Charged hollow spherewith inner radius a and outer radius b.
Charges: nonuniform charge distribution in between (so not a conductor):
Problem solving strategy: 1) Draw non-physical Gaussian sphere at distance r where you want to find Er. 2) Use Gauss’s law to write equation for Er in terms of other parameters. 3) Solve for Er. In this case solve in 3 places, inside hollow region (rI), inside charged region (r2) and outside (r3).
rI
r2
r3
Problem: The electric field is a radial vectorfield due to the symmetry of thesystem. Findthe electricfield magnitudein the radial direction atevery distancefrom the origin.
dQcharge
dVvolume
d sind r2dr
0
0
2
4 A
r
r2dr
4 Ardr
Required vectorcalculus knowledge:
(r) Ar
System: Charged infinite cylinder with radius a.
Charges: Nonuniform charge distributioninside cylinder (so not a conductor):
rI
r2
Problem: The electric field is a radial vector field dueto the symmetry of thesystem. Find theelectric field magnitudein the radial directionat every distancefrom the origin.
dQcharge
dVvolume
d dz rdr
0
2
(hmm...)
Required vector calculus knowledge:
dQcharge
dVvolume
d dz rdr
0
zo
0
2
2zo Ar rdr
2zo Ar2dr
Try solving over a finite height zo:
Problem solving strategy: 1) Draw non-physical Gaussian cylinder at distance r where you want to find Er. 2) Use Gauss’s law to write equation for Er in terms of other parameters including an arbitrary height zo. 3) Solve for Er. In this case solve in 2 places, inside region (rI), and outside (r2). You will need to have the arbitrary height zo cancel in the end.
zo
Another view of drawing a Gaussian cylinder of radius r andfinite length zo around an infinite cylinder of charge (this one outside).
r
System: Charged infinite slab of width w in x-y direction.
Charges: Uniform slab of charge density :
Problem: The electric field is a vector fieldpointing perpendicular to the plane of the slabdue to the symmetry of the system. Find theelectric field magnitude in the perpendiculardirection at a given distance from the middleof the slab.
dQcharge
dVvolume
dx dy dz
(hmm...)
Required vector calculus knowledge:
dQcharge
dVvolume
dx dy dz0
z
0
yo
0
xo
xoyo dz0
z
Try solving over a finite box xo and yo:
Problem solving strategy: 1) Draw non-physical Gaussian rectangular prism from center of slab to height z where you want to find Ez. 2) Use Gauss’s law to write equation for Ez in terms of other parameters including arbitrary length and width xo and yo. 3) Solve for Er. In this case solve in 2 places, inside region (zI), and outside (z2). You will need to have the arbitrary xo and yo cancel in the end.
Set z=0 in middle of slab.
z2
xo
yo
z1
xo
yo
V V
Three representations of the same circuit:
BA
TT
ER
Y
+
BA
TT
ER
Y
+
BA
TT
ER
Y
+
BA
TT
ER
Y
+
V
3.0
Circuit Position0
(Note: bulb shape
distorted.)
3 V
b
c
d
3 V
a b
c
d
BA
TT
ER
Y
+
BA
TT
ER
Y
(Note: bulb shape
distorted.)
c d
BA
TT
ER
Y
+
BA
TT
ER
Y
+a b
c
d
1.5 V1.5 V
1.5 V
1.5 V
1.5 V
A. B. C.
1.5 V1.5 V
1.5 V
A. B. C.
1.5 V1.5 V
1.5 V1.5 V
1.5 V
A. B. C. D.
1.5 VV1.5 V
V+
-
+
-V
+
-
BA
TT
ER
Y
+
BA
TT
ER
Y
+
ITOTAL
IA IB IC
IE
ID
BA
TT
ER
Y
+
BA
TT
ER
Y
+
BA
TT
ER
Y
+
BA
TT
ER
Y
+
BA
TT
ER
Y
+
3 V
Voltage
VDC
Measuring the voltage drop across a light bulb (DMM in parallel):
VR
BA
TT
ER
Y
+
Amperes
mA
Measuring the voltage drop across a light bulb (DMM in series):
A
R
BA
TT
ER
Y
+
Ohms ()
Measuring the resistance of a light bulb (component disconnected):
R
BA
TT
ER
Y
+
V
1.5
Circuit Position0
BA
TT
ER
Y
+
V
1.5
Circuit Position
Thumb pointsto North
Right-hand-wrap rule for finding direction of magnetic poles created by moving charges (current).
N
S
Wrap fingersin direction ofcurrent. q
If charge is negative,reverse poles.
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
N
S
N
S
N
S
S
S
SMAGNETIC
NMAGNETIC
N
S
++
+
+
++ +
+
+
+
+
+++
Excess positive charge on the surface of a spherical conductor.
-
-
-
Excess negative charge onthe surface of a cubic conductor.
-
-
--
-
-
--
-
-
- --
-
-
-
--
- ---
-
--
-
-
(Excess charges always repel each other to the surface.)
+
++
+
++
+ - Negatively charged attracts positive charge and repels negative charges on the surface of a conductor.
+---- --
-
---
---
+Positively charged object creates an electric field that rotates polar molecules in insulator.
+-MAGNIFY
insulator/dielectric material
rotate
+-
+-
Electrons deposited on the surface of a balloonby rubbing it against your hair do not spread out.
--
---
-- -
-
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