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screen menu,
calcs & info
menu buttons
# shows
0 volts for
each channel
cursor
control
autoset
measure
cursor
triggering
control
usual
channel
controls
Ideal Solenoid
B
Boutside=0
B
B
B
uniform
magnetic
field inside
Boutside=0
Use Ampere’s Law
to Find Magnetic Field
(Explain each step in your report.)
 B  ds   I
o total
enclosed
L
Amperian Loop
whole loop
 B  ds   NI
o
BIN
inside
solenoid

L
0
Bin dy  o NI
Bin  0 dy  o NI
L
Bin L  o NI
Bin 
N solenoid loops
enclosed, each
with current I.
o NI
L
Bin  o nI
where n is “loop density”
N/L of solenoid.
Cause and Effect in a Solenoid:
Ampere’s Law to Faraday’s Law
voltage
VR=RIR=RIL
The current in the solenoid creates a magnetic
field inside the solenoid due to Ampere’s Law.
Bampere
t
The changing magnetic field inside the solenoid
L causes a back EMF (voltage) due to Faraday’s
Law. Notice that dI/dt causes a phase shift.
Inside the solenoid:
dB
0
dt
B
S
N
V(velocity)

R
Direction of current
inside the resistor?
Transmitter
Oscillating transmitting
magnetic fields.
Oscillating transmitting
voltage.
Receiver
Transmitting magnetic fields
reach inside coils.
Oscillating voltage
received is measurable.
The LRC Circuit - AC Driven
voltage
VR
VL
VC
The LRC Circuit - AC Driven: Source from Addition
Vsource
voltage
VR
VL
VC
A
B
C
100 
50 mH
0.1 F
Iamplitude
Iamplitude
Large R
Small R
fdrive
fresonance
fdrive
fresonance
VR(t)
VR(t)
VS(t)
out of phase
45o
in phase
VS(t)
System: Charged hollow sphere
with inner radius a and outer radius b.
Charges: nonuniform charge distribution
in between (so not a conductor):
A
(r) 
r
Problem: The
electric field
is a radial vector
field due to the
symmetry of the
system. Find
the electric
field magnitude
in the radial
direction at
every distance
from the origin.
Required vector
calculus knowledge:

 dQ   dV
charge
volume

rI
r2
2

0
0
 d  sin d

r3
A 
 4    r 2 d
  r 
 4

 Ardr

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).
System: Charged infinite cylinder with radius a.
Charges: Nonuniform charge distribution
inside cylinder (so not a conductor):
Problem: The electric field
is a radial vector field due
to the symmetry of the
system. Find the
electric field magnitude
in the radial direction
at every distance
from the origin.
Required vector calculus knowledge:
2


0


 dQ   dV   d  dz  rdr 
(r)  Ar
charge
volume
Try solving over a finite height zo:

 dQ   dV

charge
volume

2
zo

0
0

 d  dz  rdr

 2zo  Arrdr
rI
r2


 2zo  Ar 2 dr


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.
Another view of drawing a Gaussian cylinder of radius r and
finite length zo around an infinite cylinder of charge (this one outside).
zo
r
System: Charged infinite slab of width w in x-y direction.
Required vector calculus knowledge:
Charges: Uniform slab of charge density :
 dQ   dV   dx  dy  dz 
charge
Problem: The electric field is a vector field
pointing perpendicular to the plane of the slab
due to the symmetry of the system. Find the
electric field magnitude in the perpendicular
direction at a given distance from the middle
of the slab.

xo
z1

z2




 dQ   dV

charge

volume

xo
yo
z
0
0
0
 dx  dy  d
z
 x o y o  dz

yo



Try solving over a finite box xo and yo:
xo
yo
volume


0
Set z=0 in middle of sla
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
Three representations of the same circuit:
V
V
+
BATTERY
V
BATTERY
0
BATTERY
(Note: bulb shape distorted.)
3.0
+
+
Circuit Position
bulb shape distorted.)
d
c
BATTERY
d
d
c
BATTERY
c
ATTERY
c
b
a
BATTERY
3V
+
b
a
+
3V
d
+
A.
B.
C.
1.5 V
1.5 V
1.5 V
1.5 V
.5 V
A.
B.
C.
1.5 V
.5 V
1.5 V
1.5 V
1.5 V
A.
.5 V
B.
+
V
-
1.5 V
D
C.
+
V
-
1.5 V
1.5 V
+
1.5 V
V
-
ITOTAL
BATTERY
+
ID
BATTERY
BATTERY
IE
IC
IB
IA
+
+
+
BATTERY
BATTERY
BATTERY
+
BATTERY
3V
+
+
Measuring the voltage drop across a light bulb (DMM in parallel)
Voltage
VDC
R
V
Measuring the voltage drop across a light bulb (DMM in series):
Amperes
mA
R
A
Measuring the resistance of a light bulb (component disconnected):
Ohms ()

R
V
1.5
+
BATTERY
0
Circuit Position
V
1.5
+
BATTERY
Circuit Position
Right-hand-wrap rule for finding direction of
magnetic poles created by moving charges (current)
Thumb points
to North
N
q
f charge is negative,
everse poles.
Wrap fingers
in direction of
current.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
N
S
N
S
N
S
S
S
SMAGNETIC
N
S
NMAGNETIC
Excess positive charge on
the surface of a sphere.
+
+
- -- -
+
+
+
+
+
+
+
+
+
Ex.1.
Excess negative charge on
the surface of a cube.
+
+
+
Ex.2.
- - - - -
(None of the excess charges rest inside the objects,
they always repel each other to the surface.)
Macroscopic charge separation across a neutral
conductor in the presence of an electric field.
-
+
-
+
++
+
+
+
+
- - -- - -

E
Negatively charged
object creates an
electric field.
-
Microscopic charge separation across a neutral
conductor in the presence of an electric field.
Positively charged object
creates an electric field
at surface of material.
+
E

+
+
rotate
+
insulator/dielectric material
Electrons deposited on the surface of a balloon
by rubbing it against your hair do not spread out.
- -- -