Download PHYS 272 Key Points

PHYS 272 Key Points
David Blasing
Spring 2013
Contents
1 Disclaimer
4
2 Week of 1/7 to 1/13
2.1 Lab 0: VPython Review . . . . . . . . . . . . . . . . . . . . . . .
2.2 Rec 0: 1.X.66, and 1.X.80 . . . . . . . . . . . . . . . . . . . . . .
2.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . .
5
5
5
5
3 Week of 1/14 to 1/20
3.1 Lab 1: The Electric Field of Charged Particles . . . . . . . . . .
3.2 Rec 1: 14.P.24,14.P.52, 14.P.75, and 14.X.56 . . . . . . . . . . . .
3.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . .
7
7
7
7
4 Week of 1/21 to 1/27
4.1 Lab 2: The Electric Field of a Dipole . . . . . . . . . . . . . . . .
4.2 Rec 2: 15.X.30, 15.P.82, 15.P.83, and 15.X.58 . . . . . . . . . . .
4.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . .
8
8
8
8
5 Week of 01/28 to 02/03
5.1 Lab 3: Charge on Tape . . . . . . . . . . . . . . . . . . . . . . .
5.2 Rec 3: 16.X.12, 16.X.15 and 16.P.65 . . . . . . . . . . . . . . . .
5.3 Webassign Problems from this Week . . . . . . . . . . . . . . . .
9
9
9
9
6 Week of 02/04 to to 02/10
10
6.1 Lab 4: Electric Field of a Uniformly Charged Rod . . . . . . . . 10
6.2 Rec 4: 16.P.63, and 16.P.47 . . . . . . . . . . . . . . . . . . . . . 10
6.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 10
7 Week of 2/11 to 2/17
11
7.1 Lab 5: Potential Difference . . . . . . . . . . . . . . . . . . . . . 11
7.2 Rec 5: 15.X.39, 16.X.51, 16.X.52, 16.X.54, 17.X.39, and 17.X.53 . 11
7.3 Webassign Problems from This Week . . . . . . . . . . . . . . . . 11
8 Week of 02/18 to 2/24
13
8.1 Lab 6: The Magnetic Field of a Single Moving Charged Particle . 13
8.2 Rec 6: 17.P103, 17.P.99 . . . . . . . . . . . . . . . . . . . . . . . 13
8.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 13
9 Week of 2/25 to 3/3
15
9.1 Lab 7: Magnetic Field of Current-Carrying Wires . . . . . . . . . 15
9.2 Rec 7: 18.X.32, 18.X.40, 18.X.42, 18.X.47, and 18.X.50 . . . . . 15
9.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 15
2
10 Week of 3/4 to 3/10
16
10.1 Lab 8: Magnetic Dipoles . . . . . . . . . . . . . . . . . . . . . . . 16
10.2 Rec 8: 18.P.66, 19.X.26, 19.X.32, and 19.P.42 . . . . . . . . . . . 16
10.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 16
11 Week of 3/18 to 3/24
11.1 Lab 9: Energy Conservation in Circuits and Charge on a Capacitor
11.2 Rec 9: 19.P.71, and 20.P.81 . . . . . . . . . . . . . . . . . . . . .
11.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . .
17
17
17
17
12 Week of 3/25 to 3/31
18
12.1 Lab 10: Macroscopic View of RC Circuits . . . . . . . . . . . . . 18
12.2 Rec 17.P.71, 18.P.63, 18.X.69, 19.P.67 . . . . . . . . . . . . . . . 18
12.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 18
13 Week of 04/01 to 04/07
19
13.1 Lab 11: Motion of a Charged Particle in a Magnetic Field . . . . 19
13.2 Rec 11: 21.X.30, 21.P.37, 21.P.42, and 21.P.49 . . . . . . . . . . . 19
13.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 19
14 Week of 04/08 to 04/14
21
14.1 Lab 12: Faradays Effect and LC Circuits . . . . . . . . . . . . . . 21
14.2 Rec 12: 21.P.50, 21.P.79, 22.X.11, and 22.X.12 . . . . . . . . . . 21
14.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 21
15 Week of 04/15 to 04/21
22
15.1 Lab 13: Electromagnetic Radiation . . . . . . . . . . . . . . . . . 22
15.2 Rec 13: 22.P.23, 22.X.28, and 22.P.34 . . . . . . . . . . . . . . . 22
15.3 WebAssign Problems from this Week . . . . . . . . . . . . . . . . 22
1
Disclaimer
These are only the highlights that I think are important, not an exhaustive
(or official) list of the topics that might be on any homework/quiz/exam. These
will jog your memory, but you ought to memorize things from your textbook
or lecture notes. The topics covered here are only drawn from recitation, lab,
and the web assign homework’s. I cannot guarantee the following is error free.
Please inform me if there are any mistakes and I will fix them.
It is in your benefit to try and make a list of the most important topics on
your own. Please don’t let the fact that I made this one stop you from doing
that if you were going to already.
Aside: I will use some ”technical” notation but hopefully explain it sufficiently. Such notation is more compact and easier to use once you understand
it. Learning it now will save you time later. Some of the concepts below are
complicated, and will require some time to learn. Quickly reading will likely not
help much. I will try and send this out about every week, and I think it would
be beneficial for you to review periodically throughout the semester.
4
2
Week of 1/7 to 1/13
2.1
Lab 0: VPython Review
• Point charges produce electric fields that are completely in the radial direction. They point purely away or purely toward point charge. Electrons
produce radially in fields while protons produce radially out fields.
• Know how to do dimensional analysis - check if your equation actually
has appropriate units on both sides. This can be very useful when trouble
shooting.
2.2
Rec 0: 1.X.66, and 1.X.80
~ B
~ = (Ax ±Bx , Ay ±By , Az ±Bz ). It
• Know how to add/subtract vectors A±
doesn’t matter where a vector ”starts” for vector equality, it only matters
~=B
~ ↔ Ai = Bi for all i ∈ x, y, x. (Note
that every component is equal. A
that ≡ means ”is defined as”, so we can write ∈ ≡ in the set, and ↔ ≡ if
and only if.
~×B
~ = (Ay Bz − Az By , Az Bx − Ax Bz , Ax By −
• Vector Cross Product: A
~ and B
~ and has
Ay Bx ). The resultant vector is perpendicular to both A
~ B|sin(θ)
~
~ and B.
~
magnitude |A||
where θ is the smallest angle between A
2.3
WebAssign Problems from this Week
• Know the definition and basic arithmetic for vectors. Addition, subtraction, magnitude, dot product, multiplication by a scalar, definition of relative position vector, and how to get a unit vector from a regular vector.
q
~ = A2x + A2y + A2z
• The magnitude of any vector A is |A|
~ + B|
~ =
~ + |B|.
~
• Note: |A
6 |A|
~ is  = A~ . A unit vector is always dimensionless and
• A unit vector of A
~
|A|
only gives you information about the direction.
• Electric dipole: The Electric Field from a dipole is the just the sum of the
field from the two particles. If the observation point is much greater than
the separation distance for the two charged particles and happens to be on
~ k = k2qs
the parallel or perpendicular axis, the formula simplifies nicely:|E|
r3
kqs
~
and |E|⊥ = r3
• Spherically symmetric charge distributions produce electric fields outside
that look just like a point charge at the center of the sphere which has
charge equal to the amount in the whole spherical distribution.
5
• In a conductor, the electric field and drift velocity are related by the
~drif t = µE
~ where µ is the mobility (what units must the
mobility: V
mobility have?)
6
3
Week of 1/14 to 1/20
3.1
Lab 1: The Electric Field of Charged Particles
~ pointcharge =
• Coulomb’s law: E
kqr̂
|~
r |2 .
Note that k ≡
1
4π0 .
• Superposition: The electric or magnetic field at a point results from all of
the charges or currents around. Simply add their respective contributions.
• Relative position vector. The vector that points from r~i to r~f is r~f − r~i .
This is the vector that is used in coulomb’s law with the ~r ≡ as the
vector that points from the charge to point where the electric field is
being calculated.
3.2
Rec 1: 14.P.24,14.P.52, 14.P.75, and 14.X.56
• An electric field creates a force on anything with charge. Go between the
~ where F~ is force, q is charge, and E
~ is the electric field.
two with F~ = q E
~ as F~ = mA
~ where
You can also get information about the acceleration A
m is mass.
kqr̂
~ net = P
• Principal of Superposition: E
charges |~
r |2 . The sum goes over all
of the charges present in a given problem that create the electric field of
interest.
• Know what the electric field from an electric dipole looks like.
3.3
WebAssign Problems from this Week
• Neutral objects polarize in the presence of electric fields. They make tiny
little dipoles. The separation vector ~s in an electric dipole is the position
vector that points from the negative to the positive charge in the dipole.
This vector is parallel to the electric field.
~ = q~s where ~s is the separation vector of
• Atomic polarizability: P~ = αE
~ is the
the dipole, α is the atomic polarizability, q is the charge, and E
~
~
~
~
~
electric field. If you have E = 0, then P = 0. For determining P it only
~ is. It does not matter what charges are nearby.
matters if what E
• Shell Theorem: uniform spherical charge distributions create electric field’s
exactly like point charges (with charge equal to the total charge on the
spherical distribution) when outside the sphere and create zero electric
field on the inside of the sphere.
• Know what the electric field from an electric dipole looks like, and how to
calculate the magnitude and assign the direction when you are far from
the dipole on either the parallel or perpendicular axis.
7
4
Week of 1/21 to 1/27
4.1
Lab 2: The Electric Field of a Dipole
• Lab explored the electric field of a dipole in three dimensions. No ”new
physics”.
4.2
Rec 2: 15.X.30, 15.P.82, 15.P.83, and 15.X.58
• As a rule you can hold to, conductors cancel electric fields on the inside
of themselves. When placed in an external electric field, the charges in
the conductor redistribute only on the surface and cancel the electric field
on the inside. So conductors CAN make an electric field, but the only do
so to cancel the field that is already on the inside. The NET field on the
inside is zero after this redistribution process.
• Note that ”steady state” can mean different things in different situations.
Here it means that the charges have stopped flowing from the inside to the
surface. Later it will mean that the current has stopped changing, but in
contrast to the above, charges will still be moving around in this ’steady
state’ current. Steady state means that the parameters of interest in the
problem have stopped fluctuating. Applying the term to a given situation
takes a little bit of care.
• Sometimes electric fields are dynamic. One electric field redistributes
charge on something else, and then this redistribution creates another
electric field that redistributes charge elsewhere and so on. Recall the two
conducting spheres affecting each other in the recitation problem.
• Non-conductors will polarize in the presence of an electric field. The
charges do now flow like in a metal. They shift in their location a little
and make a bunch of small dipoles. This is why you can stick a ballon to a
wall after charging it by rubbing it on something. The ballon polarizes the
wall and then experiences an attraction force towards the induced dipoles.
4.3
WebAssign Problems from this Week
~ If you don’t have an electric field vector, then
• Polarization Vector P~ ∝ E.
you don’t have a polarization vector.
• The electric field of an infinite plane carrying a uniform charge surface
σ
~
density, σ ≡ charge
area is E = 20 â where â is a unit vector that points
perpendicularly in (out) for negative (positive) charge densities.
8
5
Week of 01/28 to 02/03
5.1
Lab 3: Charge on Tape
• Lab explored the charge on the tape to give practical experience dealing
with assumptions.
5.2
Rec 3: 16.X.12, 16.X.15 and 16.P.65
• You need to know the assumptions that allowed a formula to be derived.
~ =
• For a rod of length L carrying charge q, |E|
√
4π0
q
.
r 2 +(L/2)2
This
formula strictly valid only on the perpendicular axis at the mid point
of the wire. It is approximately valid when close to this. The formula
~ = 1 · 2q/L when L >> r. The direction is assigned
simplifies to |E|
4π0
r
by inspection through positive charges pointing out and negative charges
pointing in.
5.3
Webassign Problems from this Week
~ = 2πkσ[1 −
• The Electric field on the axis of a uniform disk of charge is E
√ 2r 2 ]r̂, σ is the charge density, q/area, r0 is the radius, and r̂ points
r +r0
perpendicularly in (out) for negative (positive) charge. As you zoom in
closer to the disk this simples twice. Keeping the leading term of rr0 is
~ = σ · [1 − r ]r̂, and neglecting any r dependence we have E
~ = σ r̂.
E
20
r0
r0
20
• The Electric field magnitude on the axis of a uniform ring of charge is
krq
where r is the perpendicular distance, r0 is the radius, and q is
(r 2 +r02 )3/2
the total charge.
9
6
Week of 02/04 to to 02/10
6.1
Lab 4: Electric Field of a Uniformly Charged Rod
• If you need to find the electric field of some random charge distribution,
this usually means coding. Split the distribution into point charges and
~ contributions using coulomb’s law to get E
~ net .
simply sum the E
6.2
Rec 4: 16.P.63, and 16.P.47
• See shell theorem already discussed above.
~ = ~0 on the inside of a conductor
• You used the fact that in a steady state, E
to get a relationship between the charge on the inside and outside of a
capacitor made of two parallel thin disks.
6.3
WebAssign Problems from this Week
• Kinetic energy is 12 m~v 2 , where m is the mass and ~v is the velocity vector,
and the speed is the magnitude of vecv.
• Know how to get potential from an electric field path integral: ∆V =
R ~r
~ This is typically three integrals (x,y,z), and dl
~ =
~ · dl.
Vf − Vi = − ~r f E
i
(dx, dy, dz) runs along any path from ~ri to r~f . Know how to do path
integrals, this is important. Ask me if you don’t! This integral simplifies
~ is constant over some portion (or all) of the total path. Along these
if E
segments (which might be the total path), the contribution to the change
R
~ is the total displacement only
~ ·L
~ where L
~ ≡ ~rf dl
in potential is −E
~
ri
along the part of the path where the electric field is constant.
• Positive charges go to regions of lower electric potential, while negative
charges go to regions of higher electric potential. Both go to regions of
lower potential energy.
• Electric potential change and potential energy change are related as V olts =
Joules
Coulomb . E = q · V where E is the energy, V is the voltage, and q is the
charge.
• A positive (negative) change in electric potential means the path between
the initial and final positions (~
ri and r~f ) was against (with) the electric
field.
10
7
Week of 2/11 to 2/17
7.1
Lab 5: Potential Difference
• This lab was about path integrals. Know how to set them up, particularly
~
the bounds. One mathematical note which applies to any vector field E
~
~
~
~
~
(not necessarily the electric field). If ∇ × E = 0 then E = ∇V where V is
RB
~ = V (B) − V (A). So
~ · dl
some scalar potential. This also means that A E
the path integral is independent of the path, and only depends on the end
points. There are other examples that you know of, gravitational potential
energy depends only on the height, not how you got to that height.
7.2
Rec 5: 15.X.39, 16.X.51, 16.X.52, 16.X.54, 17.X.39,
and 17.X.53
• Nothing ”new” here.
7.3
Webassign Problems from This Week
• Moving charges create magnetic fields. Superposition still applies and
you have to add (or integrate) over all of the moving charges (currents).
H ~
~
~ = µ0 I dl×r̂
r
B
4π
|~
r |2 where dl represents a path that runs over the currents. ~
points from infinitesimal current bit to the observation location.
• For an individual point charge, moving at non relativistic speeds, the
~ = µ0 q~v×r̂
integral collapses to just one term, B
4π |~
r |2 .
• Know Kirchhoff’s Voltage Law - Sum of the voltages around
H any loop is
~ = 0).
~ · dl
zero. Know the physics behind this, (think about this - E
~ ×E
~ = 0.
Kirchhoff’s Voltage Law only holds when the ∇
• Node Rule: All the current going into a node must also be flowing out.
• Kirchhoff’s voltage law and the current node rule come from two fundamental ideas in physics: conservation of energy and conversation of charge.
They can be used to get equations for current and voltages in a circuit
just like equivalent resistances can be used to get current through branches
and therefore voltage drop across different elements.
• In the presence of a dielectric, the electric field changes. With a dielectric
~ = σ where k is the dielectric constant. Still
inside a capacitor, |E|
k0
assign the direction by inspection from which plate is positive and which
is negative.
• The change in potential energy can be calculated
through two means:
R
~ or by defining a
directly through the force path integral, ∆U = F~ · dr,
system and calculating the change of total energy.
11
• Electric fields contain energy with density, u =
1 ~2
20 |E |
~2
(units are
joules
m3 ).
Magnetic fields have energy with density 2µ1 0 |B |.You need to integrate
over a volume if it varies, or multiply by the total volume if it is a constant
to recover the total energy.
12
8
Week of 02/18 to 2/24
8.1
Lab 6: The Magnetic Field of a Single Moving Charged
Particle
• Moving charges create magnetic fields. Superposition still applies and
you have to add (or integrate) over all of the moving charges (currents).
H ~
~
~ = µ0 I dl×r̂
r
B
4π
|~
r |2 where dl represents a path that runs over the currents. ~
points from infinitesimal current bit to the observation location.
• For an individual point charge, moving at non relativistic speeds, the
~ = µ0 q~v×r̂
integral collapses to just one term, B
4π |~
r |2 .
8.2
Rec 6: 17.P103, 17.P.99
• Capacitors and charge: q = cv where q is the charge, c is the capacitance
in Farads, and v is the voltage. A bit of culture. c is a geometric constant.
dq
Taking a time derivative of both sides gives c dV
dt = dt = I where I is the
current. So to experimentally measure capacitance, you put a time varying
voltage on something and measure the resulting current. Capacitance is
the constant of proportionality between the two.
• In the presence of a dielectric, the electric field changes inside a capacitor.
~ = σ where k is the dielectric conWith a dielectric inside a capacitor, |E|
k0
stant. Still assign the direction by inspection from which plate is positive
and which is negative. Each plate contributes half of the above value.
8.3
WebAssign Problems from this Week
• Currents make magnetic fields that circle around them. Look up a picture
the magnetic field from a wire.
• The magnetic field at a distance r away from the midpoint of a wire of
~ wire =
length L carrying current I, and along the perpendicular axis: B
µ0 √ LI
µ0 2I
θ̂
≈
θ̂
if
r
<<
L.
The
last
approximation
is
for
a rod
4π
4π r
2
2
r
r +(L/2)
of ”infinite” length. θ̂ is the unit vector in the direction of increasing theta
when a cylindrical coordinate system is used and the current is directed
along the positive z axis.
• Assign the direction using the right hand rule. Stick your thumb in the
direction of conventional current, and then your fingers will curl in the
direction of the magnetic field (θ̂).
~ Note the vector on the
• Force on a current carrying wire: F~ = LI~ × B.
~ is constant over a straight wire of length L.
current. This is valid if B
13
2
πR I
~ loop | = µ0 2N
• Magnetic fields from loops and wires with superposition:|B
4π (z 2 +R2 )3/2
where N is the number of loops (can be a fraction), I is the current, R
is the radius, and Z is the distance along the perpendicular axis. This is
valid only along the perpendicular axis.
µ0 2µ
• Magnetic field on the axis of a bar magnet or current loop goes as 4π
r3 .
µ is the magnetic dipole moment of the magnet or current loop. You have
to assign the direction based on the field pattern. The field goes away
from the magnet from the north end, and toward the magnetic from the
south end. µ for a current loop is defined as IA where I is the current in
the loop and A is the area of the loop, but for a magnet it is an intrinsic
property of the magnet.
• ”Magnetic Moment” (µ)often means the product of the area of a current
loop and the current running around it.
• Conventional current is the flow of of positive charges. What usually goes
on is actually electron current flowing in the other direction. Conventional
current flows from high potential to low potential, electron current flows
from low potential to high potential. (Both are examples of charges going
to places of lower potential energy.) Typically conventional current is
where electron current is ≡ Electrons
measured in Amps ≡ Coulombs
Sec
Sec
14
9
Week of 2/25 to 3/3
9.1
Lab 7: Magnetic Field of Current-Carrying Wires
• We measured the current and distance dependence of the magnetic field
from a long wire.
• The magnitude of the magnetic field at a distance r away from the midpoint of a wire carrying current I, of length L, and along the perpendicular
µ0 2I
~ wire | = µ0 √ LI
≈ 4π
axis: |B
4π
r with r << L
2
2
r
9.2
r +(L/2)
Rec 7: 18.X.32, 18.X.40, 18.X.42, 18.X.47, and 18.X.50
• No ”new physics here”. We explored the magnetic field from a few different
magnetic fields (point charge, current loop, and straight wire).
• One way to get the change in energy of a particle moving through a force
field (caused from the the force field) is through a work path integral,
R ~r
~ where F~ is the force acting on the particle. I
∆E = Ef − Ei = ~r f F~ · dl,
i
cannot think of an example where this integral is path dependent.
9.3
WebAssign Problems from this Week
• In amps, I = |e|nµEA where n is the electron density, e is the electron
charge, µ is the mobility, E is the electric field, and A is the cross sectional
area.
• In electron current, Ie = nµEA where n is the electron density, e is the
electron charge, µ is the mobility, E is the magnitude of the electric field,
and A is the cross sectional area.
• Connecting wires typically have very small resistance compared with circuit elements.
• The electromagnetic waves that set up the current in a circuit takes travels
at about the speed of light. So if you have a circuit of length L, the time
required to initialize the current is approximately L/C where C is the sped
do light. This is an under approximation because it also takes some time
for the charges to move from the inside of the wire to the surface and set
up the charge distribution that creates the electric field inside the wire.
15
10
10.1
Week of 3/4 to 3/10
Lab 8: Magnetic Dipoles
• Nothing ”new”.
10.2
Rec 8: 18.P.66, 19.X.26, 19.X.32, and 19.P.42
• The Drude model of electron flow adds a drag term to the equation of
motion for electrons. This is analogous to adding a drag term to a falling
object. In that example, the object reaches a terminal velocity. The
analogous result for the present case is that the wire reaches a steady
state current. The Drude model is still an approximation, but it is better
than saying that the electrons accelerate forever under the application of
an electric field.
10.3
WebAssign Problems from this Week
• If there is a bend in a wire, or a voltage drop across a circuit element, you
need to have a surface charge on the wire that makes the electric field that
either accelerates the electrons around the corner, or creates the electric
field that causes a drop in voltage across the circuit element.
Joules
) in circuits: IV or I 2 R. Energy dissipated is power
• Power (units ≡ Second
times time (as long as power is constant over the time interval). Note: the
brightness of a bulb is proportional to the power that it dissipates as a
resistive element.
• Current is conserved throughout a wire in steady state. You can use this
to relate the electric fields in different circuit elements.
• The conductivity, σ, is |q|nµ, where q is the magnitude of the charge on
the charge carriers, n is the charge carrier density, and µ is the mobility.
• Resistance in terms of conductivity: R =
ρ, the resistivity.
L
σA .
Sometimes
1
σ
is defined as,
• Conventional current flows out of the positive end of the battery, and into
the negative end. Electron current flows the opposite way.
• Charges on the surface of conductors are what make the electric fields on
the inside that drive currents.
• Capacitance of a parallel plate capacitor: C = kd0 A where k is the dielectric constant for whatever is in-between the plates of area A and separation
distance d.
• Time constant for an RC circuit is well...RC. Current is proportional to a
t
dying exponential, I ∝ e− RC .
16
11
11.1
Week of 3/18 to 3/24
Lab 9: Energy Conservation in Circuits and Charge
on a Capacitor
−t
• The charge on a charging capacitor in an RC circuit: q(t) = Qmax (1−e RC )
where Qmax = CVmax . R is the resistance of the circuit, and C is the
capacitance of the capacitor. The time constant, τ , is defined as RC.
• Resistors in parallel and series can bePmodeled by one resistor with
Pan
1
1
=
effective resistance. Series: Ref f =
R, and Parallel: Ref
R
f
where both sums go over all of the resistors you are attempting to make
an effective resistor model out of. You can replace multiple resistors with
single resistor with an ”effective resistance”.
11.2
Rec 9: 19.P.71, and 20.P.81
• Nothing new here this week.
11.3
WebAssign Problems from this Week
• Nothing new this week.
17
12
12.1
Week of 3/25 to 3/31
Lab 10: Macroscopic View of RC Circuits
• Resistive things have a slight temperature dependence. This meant that
a bulb is not ”ohmic” as the resistance changes as a function of how much
current you push through it (which heats it up).
12.2
Rec 17.P.71, 18.P.63, 18.X.69, 19.P.67
• Magnetic field from a current carrying air filled infinite length solenoid:
~ = µ0 N I where N ≡ turns and I is the current that each turn
Inside is |B|
length
~ = ~0.
carries. Assign the direction with the right hand rule. Outside, B
12.3
WebAssign Problems from this Week
~ × B.
~ This acts as a centripetal force (F~cent = − mV 2 r̂).
• F~mag = q V
r
This caused electrons and protons to go in circular motions. Get clockwise/counterclockwise through the right hand rule.
• When something acts as a centripetal force (causing something else to go
v2
in a circle) you can set it equal to F~cent = − m~
r r̂.
18
13
13.1
Week of 04/01 to 04/07
Lab 11: Motion of a Charged Particle in a Magnetic
Field
• We also saw that Electric and Magnetic Fields can couple to create unexpected motions such as the electron traveling in a perpendicular direction
to an applied electric field in a cycloid motion in a uniform magnetic field.
This means that if you had a uniform magnetic field parallel to the Earth’s
surface, and you dropped an electron, it WOULD NOT fall to the ground,
but go sideways in semi-circles. Thats weird, and physics is cool.
13.2
Rec 11: 21.X.30, 21.P.37, 21.P.42, and 21.P.49
• Nothing ”new” here again.
13.3
WebAssign Problems from this Week
• This is going to get mathematically gory. I will try to say in words afterwards what it all means.
R
~ where s is a
• Flux of any vector field: (say ~z) defined as F lux = s ~z · da
~
”mathematica surface” and da is an outward pointing surface area element.
For the problems that we do, this usually simplifies down to z⊥ · A where
A is the total area of the surface under consideration. This is valid to do
as long as the ~z is constant over the surface, and z⊥ is the perpendicular
component of ~z. You may have to break an object into multiple surfaces if
~z is different over different surfaces of the object. Flux can be conceptually
thought of as the amount of vector field poking through the surface.
~ ·E
~ = ρ , one can derive,
• Gauss’s Law and Gaussian Surfaces: from ∇
0
H
~ = qenc , usually you need to form a suitable Gaussian surface
~ · da
E
0
s
S where the electric field is a constant vector (maybe the zero vector)
~
over it. One then typically arrives at |E|A
= qenc
0 . This result can be
used to extract the magnitude of the electric field, and then you need to
apply the direction based the charge distribution. Look at this recitation
for examples or from the text for examples. Gaussian surfaces can be
confusing, but you can get quick results if you understand the method.
Take the time to understand this.
H
~ = ∂φ
~ ×E
~ = − ∂ B~ . from this, one can get that
~ · dl
• Faraday’s Law: ∇
E
∂t
∂t
P
H
~
~ For a suitable path P, you usually get |E|L
~ · da.
~ = ∂|B|A
where φ = s B
∂t .
This can be used to get the magnitude of the electric field. You still need
to assign the direction of the electric field. There are two ways to do this,
~
right hand rule and Lenz’s law: stick your finger in the direction of − ∂∂tB
~ or Lenz’s law, which states
and your fingers will curl in the direction of E,
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that the current that flows will create a magnetic field opposite to the
changing flux (if the magnetic field is getting stronger and points up, the
current will flow to make a magnetic field that points down).
• In both of the above two bullet points, you need to use symmetry. In the
first, try to find a gaussian surface S in which the electric field is constant
over, in the second, try to find a path P (which defines a surface S) in
which the electric field is constant along the path. You can use a right
~ Point your thumb in the direction
hand rule to find the direction of E.
~
~
opposite to ∂∂tB and your finger will point in the direction of E.
• Consider the two equations without any time varying electric fields. Am~ ×B
~ = µ0 J,
~ and ∇
~ ×E
~ = − ∂ B~ . (J~ is basically current for
pere’s Law: ∇
∂t
~ fields. If
~ fields that look just like ∂ B~ have E
here). So Currents have B
∂t
~
you have a ∂∂tB in a line, you will have a curling Electric field around it,
~ fields and what they look
etc. Use your intuition for how currents have B
~
∂B
~ fields.
like to make a first stab for how ∂t s have E
• from Ampere’s Law one can derive (same mathematical derivation) that
R
~ = µ0 Ienc . Where Ienc is the current enclosed by the surface with
~ · dl
B
a boundary curve of length L. This holds as long as the symmetry ensures
that the magnetic field is constant over the boundary (the perimeter/edge
~ = µ0 Ienc .
of the surface). In that case, we arrive at |B|L
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14
14.1
Week of 04/08 to 04/14
Lab 12: Faradays Effect and LC Circuits
Rb
• Average of a function: f (x) =
a
f (x)dx
b−a
• We also covered underdamped, critically damped, and over damped RCL
circuits. See the textbook for more information on this.
14.2
Rec 12: 21.P.50, 21.P.79, 22.X.11, and 22.X.12
• Nothing ”new” here again.
14.3
WebAssign Problems from this Week
~ ·B
~ =0
• One of Maxwell’s equations is ∇
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15
15.1
Week of 04/15 to 04/21
Lab 13: Electromagnetic Radiation
~ ×B
~ and the mag• Electro magnetic waves propagate in the direction of E
~ = c|B|
~ where c is the speed of light.
nitudes are related by |E|
~E/M = cÊ × B̂)
• E/M waves travel at the speed of light (V
• The final question was kind of interesting and seemed to be troublesome
to many groups. It might be worth going over again. The source of the
reradiated pulse is the accelerating electrons in the copper wire. The
a⊥
~ radiation = 1 −q~
formula you need is E
4π0 c2 r and is discussed on page 996
of your text.
15.2
Rec 13: 22.P.23, 22.X.28, and 22.P.34
• Nothing ”new” here.
15.3
WebAssign Problems from this Week
~ B
~ = µ0 J~+µ0 0 ∂ E~ So we can
• Ampere’s Law with Maxwell’s Correction ∇×
∂t
do the same game with pulling the magnetic field out of a path integral if
it is constant if we have a time varying electric field. Time varying electric
fields are B fields just like time varying B fields are E fields.
• For E/M Radiation, c = f λ where c is the speed of light, λ is the wavelength, and f is the frequency.
• The period: T = f1 is the time it takes for one complete oscillation to
occur. The angular frequency is: ω = 2πf = 2π
T
• Know how to geometrically get the perpendicular acceleration vector.
~ into a A
~ ⊥ +A
~ k and keep just the perpendicular component.
Break the A
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