Massachusetts Institute of Technology
Spring Term 2005
8.02X Electricity and Magnetism
Problem Set 1 and Experiment Problem MI
Issued:
Due:
Wed, Feb 2
Problem set: Fri, Feb 11, 4PM in 4-337
MI experiment check-off: Fri, Feb 11, 3PM in 4-355
Please note that in general both students for each partnership must submit an
experiment write-up, answering the questions regarding the experiment. For experiment
MI however, no write-up is necessary. You should be able to demonstrate the answers to
the questions during experiment check-off!
Reading suggestions (from Young + Freedman, University Physics, 11th edition)
Fri: Electric Charge, Electric Induction 21-1, 21-2
Mon: Coulomb’s Law: 21-3
Wed: Electric Field: 21-4, 21-6
Fri: Electric Field cntd, Dipoles: 21-5, 21-7
Homework Problems (30 points)
Problem 1 (5 Points) : Consider an electron and a proton in a hydrogen atom.
(a) What is the ratio of the electric force to the gravitational force between them?
(b) If we move the electron and proton either closer together or farther apart, can we
find a configuration at which the gravitational force is bigger than the electric force?
Explain your answer in one or two sentences.
(c) Given your answer to (a) and (b), why does gravity dominate for astronomical
distances?
Problem 2 (5 Points): Suppose the charge of an electron was 10-9% smaller than it is in
nature (i.e. multiplied by a factor (1 – 1011)), while the charge of the proton was the same.
(a) Under these circumstances, what would be the ratio of the electrostatic force
between Earth and Moon to the gravitational force between them?
(b) Would they still form a stable system?
Problem 3 (5 points):
Consider the configuration shown below, with a positive charge Q0 at position x0=0 and
another positive charge Q1 at position x1 along the x-axis.
(a) At which position x2 could a third positive charge charge Q2 be added, such that
the total force on Q0 is 0? Give two examples of x2 and corresponding Q2, in
terms of x0, x1, Q0 and Q1 (or a subset of these variables).
(b)
Qualitatively, describe what would happen if Q0 was displaced by a small
distance Dx from x0=0 to x=Dx and then released (two sentences max.)?
Q0 > 0
Q1 > 0
x0 =0
x =x1
Problem 4 (5 Points): Young + Freedman, Exercise 21.9
Problem 5 (5 Points): Young + Freedman, Exercise 21.25
Problem 6 (5 Points): Young + Freedman, Exercise 21.31
Experiment Meter Introduction
Due Date: Check-off by Friday 3PM Feb 11 in lab 4-355. To satisfy the check-off for the Experiment
Meter Introduction you must do the following tasks. Please read the Check-off and Grading Policy
handout
Lab Hours: Mondays: 3-5 pm, Tuesday: 7:30-9:30 pm, Wednesday: 7:30-9:30 pm, Thursday: 3-5
pm, Fridays: 12-3 pm
Soldering, and Clip Leads:
•
•
•
•
Make Two Clip Leads
Solder Alligator Clips to your Multimeter Leads for each Multimeter
Solder Alligator Clips on your Transformer Leads
Put Battery in Multimeter
Measuring Voltage, Resistance and Current with the Multimeter:
•
•
•
Measure the resistance of the 20 1 resistor
Measure the voltage of the AA cell
Measure the current in a simple circuit
Measure the resistance of the 20 1 resistor: In order to zero the meter, short the test leads by
connecting them together. Then adjust the OHMS ADJUST knob (located to the left of center of the
MMM) so that the meter reads 0 ohms; the needle is then at its maximum deflection.
Set the range selector switch on the MMM to the RX1 range. Connect the test leads to the resistor. Measure
the resistance. You may want to make other resistance measurements. For example make a thick line with a
lead pencil and measure the resistance of the mark.
Measure the voltage of the AA cell: Set the range selector switch on the MMM to the 5 DCV range. Place
the AA cell in the battery holder. Connect the test leads to the leads from the holder. Measure the voltage.
Measure the current in a simple circuit: First set the range selector switch on the MMM to the 250m
DCA range. Make a simple circuit consisting of the 20 1 resistor, the AA cell, and the MMM. You can do
this by connecting the red lead of the MMM to the red lead (plus,+) of the AA cell holder. Use a clip lead
to connect the black lead (minus,-) of the AA cell holder to one end of the 20 1 resistor. Connect the black
lead of the MMM to the other end of the 20 1 resistor. Measure the current in the circuit. What effect do
you think the MMM has on the circuit?
8.02x – Problem Set 1 Solutions
Problem 1 (5 points)
a) The ratio of the electric force to the gravitational force for an electron
and a proton in a hydrogen atom is
Fe Fg =
=
k ree2
m m
G er2 p
=
k e2
=
G me mp
(1)
(1.6 × 10−19 C)2
9.0 × 109 N · m2 /C2
≈ 2 × 1039 .
2
6.67 × 10−11 N · m2 /kg (9.1 × 10−31 kg)(1.67 × 10−27 kg)
b) In a), we have seen that the ratio of the forces does not depend on the
distance, because both Newton’s gravitational law and Coulomb’s law have r−2
dependence. Therefore, the ratio Fe /Fg doesn’t change with distance.
c) The universe is charge neutral on large scales. Therefore, there is no
net electromagnetic force on astronomical distances. However, gravitational
mass adds up, and gravitation is only an attractive (and long-distance) force, it
dominates.
Problem 2 (5 points)
a) The ratio of the forces would be
Fe = k QE QM .
Fg G ME MM
(2)
We need to estimate the number of protons (and electrons) in Earth and in
the Moon. We assume protons contribute to roughly half the mass (neutrons
are the other part, and electrons are 1000 times lighter). Therefore, the Earth
has approximately NE ≈ (ME /mp )/2 protons. On the other hand, the Moon
has approximately NM ≈ (MM /mp )/2 protons. Assuming the electrons’ charge
is (1 − 10−11 ) times the proton charge, the Earth would carry a charge QE ≈
eNE × 10−11 , and the moon would have charge QM ≈ eNM × 10−11 . The ratio
of the forces would then be
ME
MM
−11
−11
e
×
10
e
×
10
Fe 2mp
k e2
= k 2mp
× 10−22 =
(3)
=
Fg G
M E MM
G 4m2p
=
9.0 × 109 N · m2 /C2
(1.6 × 10−19 C)2
× 10−22 ≈ 1 × 1014 .
2
6.67 × 10−11 N · m2 /kg (1.67 × 10−27 kg)2
1
b) This system would be highly unstable, the repulsive electrostatic force
would tear them apart. It would be 14 orders higher than the gravitational
attractive force!
Problem 3 (5 points)
a) The simplest example is placing another positive charge Q2 = Q1 symetrically at x2 = −x1 . Another example is placing a bigger second charge with
magnitude Q2 = 4Q1 at x2 = −2x1 .
We can also derive a general rule for finding Q2 and x2 . Let us assume the
net force on charge Q0 is zero. How do Q2 and x2 depend on Q1 and x1 ?
|F10 | = |F20 |,
Q0 Q1
Q0 Q2
= k 2 ,
k 2
x1
x2
2
Q1
x1
.
=
Q2
x2
(4)
b) When we displace Q0 by ∆x, its distance to charge Q1 decreases, and so
it is repelled more, also, it is repelled less from the other charge Q2 . Therefore it
is pushed back into the equilibrium position – the charge will start to oscillate.
Y&F 21.9 (5 points)
The force of gravity must equal the electric force
Fg = Fe ,
r=
s
ke2
=
gme
s
⇒
me g = k
e2
,
r2
⇒
(9.0 × 109 N · m2 /C2 )(1.602 × 10−19 C)2
≈ 5.1m.
(9.8m/s)(9.1 × 10−31 kg)
(5)
Y&F 21.25 (5 points)
~ = m~a. We also know that its accelerThe force on the α-particle is F~ = Eq
ation is such that in time t the particle stops. It was originally traveling to the
right with velocity ~v0 , and we want it to travel with velocity ~v = −~v0 in time t.
From ~v = ~v0 + ~a.t we have
~v = −~v0 = ~v0 + ~a.t
⇒
~a = −2
~v0
.
t
Putting ~a into the equation for the force, we obtain
~
Eq
~
E
2~v0
t
2(6.64 × 10−27 kg)(1.5 × 103 m/s))
2m~v0
=−
≈ 23.5N/C.
= −
qt
2(1.602 × 10−19 C)(2.65 × 10−6 s)
= m~a = −m
2
(6)
Y&F 21.31 (5 points)
a) The electric field doesn’t change the electron’s speed in the x-direction.
So the time of travel before he flies out of the field is
t = s/vx = 0.02m/1.6 × 106 m/s = 1.25 × 10−8 s.
Because it just misses the upper plate, its distance traveled in the y direction
is h/2 = 0.5cm. The motion in the y-direction is described by (the starting
y-velocity is zero)
y=
h
1
1 Fe 2
1 Ee 2
= ay t2 =
t =
t ,
2
2
2m
2 m
so the magnitude of the electric field is then
E=
h 2m
(0.01m)(9.1 × 10−31 kg)
=
≈ 364N/C.
2 et2
(1.602 × 10−19 C)(1.25 × 10−8 s)2
(7)
b) If the particle was a proton, its mass would be greater, so it would accelerate less – and NOT hit the plates. Its vertical displacement would be (note
that it will be downward, not upward, because the proton charge is positive!)
yp
1 Ee 2
1 2
ay t = −
t =
(8)
2
2 m
−19
1 (364N/C)(1.602 × 10 C)
= −
(1.25 × 10−8 s)2 ≈ −2.73 × 10−6 m.
2
1.67 × 10−27 kg
=
c) As in b), the proton will not hit the plates, because although it feels
the same electric force (in magnitude), its mass is greater, so it accelerates less.
Also, because its charge is opposite, its path will bend downwards, not upwards.
d) The acceleration of each particle in the electric field is a = Ee/m is of
2
the order 1010 m/s for the proton (and 1013 for the electron). We can see it is
much much greater than the gravitational acceleration g, and so it is reasonable
to neglect gravity here.
3
Massachusetts Institute of Technology
Spring Term 2005
8.02X Electricity and Magnetism
Problem Set 2 and Experiment Problem LVPS
Issued:
Due:
Fri, Feb 4
Fri, Feb 18, 4PM <- note Date + Time!
Please note that both students for each partnership must submit the LVPS experiment
write-up, answering the LVPS experiment questions found below and submit the
answers together with pset 2. This is in addition to the LVPS experiment check-off,
which is due Fri, 2/18, 3PM.
Reading suggestions (from Young & Freedman, University Physics, 11th edition)
Mon, 2/14: Electric Flux, Gaus’ Law: 22-1 to 22-5
Wed, 2/16: Electric Potential Energy, Electric Potential (1/2): 23-1, 23-2
Fri, 2/18: Electric Potential Energy, Electric Potential, Quiz Review: 23-3
Homework Problems (30 points)
Problem 1 (5 Points) Consider an infinite plane (non-conducting) with a uniform
negative surface charge density 1.
(a) Draw a sketch of the electric field (using field lines) close to the plane.
(b) Explain in words why the field has to look the way you have drawn (using
symmetry arguments). (Keep your explanation to 2-3 sentences)
Problem 2 (5 Points) Consider an ellipsoidal conducting object carrying a positive
charge Q (shown below).
(a) Draw a sketch of the elecric field (using field lines) very close to the surface of
the object. Your sketch should show where the magnitude of the electric field will
be biggest.
(b) Explain in words how the fact that we are looking at a conducting object
determines the direction of the field relative to the surface at very small distances
to the surface. (Keep your explanation to 2-3 sentences)
Problem 3 (5 points):
Consider an electric dipole consisting of point charges +q and –q, separated by a fixed
distance d.
(a) Determine the net force and net torque on the dipole in a uniform electric field E
as a function of the angle 2 between the axis of the dipole and the direction of the
field.
(b) Now the dipole is brought into the field of a fixed point charge Q, which is at
distance r from +q and distance (r+d) from –q. We observe that the dipole
accelerates towards Q. Is Q positive or negative?
(c) If the dipole has mass m, what will be the initial acceleration of the dipole in case
(b)?
Problem 4 (5 Points): Young + Freedman, Challenge Problem 22.65
Problem 5 (5 Points): Young + Freedman, Exercise 23.39
Problem 6 (5 Points): Young + Freedman, Problem 23.61
Experiment Low Voltage Power Supply (LVPS)
Due Date: Check-off by Friday, 2/18, 3PM in lab 4-355.
Lab Hours: Monday: 3-5 pm, Tuesday: 7:30-9:30pm, Wednesday: 7:30-9:30 pm,
Thursday: 3-5 pm, Fridays: 12-3 pm
There will be 2 bonus points for check-off on Monday, 2/14 and 1 bonus point on
Tuesday, 2/15!
Check-off Criteria:
•
•
•
•
•
Every partnership needs to build two LVPS’s.
One working LVPS must be demonstrated, with an output voltage range between
1.2 and 17V
All connections need to be properly soldered, with no shorts or bad contacts
You should be prepared to explain the testing procedure out lined below
You will NOT be responsible for a detailed explanation of how the LVPS
works (yet)!
•
LVPS Measurement:
To test the characteristics of the LVPS, we will see how the output voltage of the LVPS
changes when a load (the 8W filament of the 1157 lamp) is connected to the LVPS
output.
To perform the measurement, you will need to do the following:
1. Connect the LVPS to the wall transformer. Use a multimeter to determine
the output voltage of the LVPS. Verify that by turning the pot you can
vary the output voltage between approximately 1.2 and 17V.
2. Determine which filament of the 1157 lamp is the 8W filament and which
is the 26W filament.
3. For 8 different settings of the potentiometer between the minimum and
maximum, record the LVPS output voltage without the lamp connected
(no-load voltage) and with the 8W filament of the lamp connected (load
voltage) for the same pot setting.
LVPS Homwework tasks (20 points total):
(Data table, graph and answer need to be handed in by BOTH partners,
together with pset 2 on Fri, 2/20)
Task 1 (10 points): Record the no-load and load (Vno load and Vload) voltages in a table.
Plot Vload in a graph against Vno load.
Task 2 (5 points): Determine over which range of no-load voltages connecting the lamp
leaves the LVPS output voltage unchanged.
Task 3 (5 points): Describe how you distinguished the 8W and 27W filament.
MIT Department of Physics
Physics 8.02X
Spring 2005
Solution to Problem Set #2
Problem 1 (5 Points) Consider an infinite plane (non-conducting) with a uniform
negative surface charge density σ
(a) Draw a sketch of the electric field (using field lines) close to the plane.
(b) Explain in words why the field has to look the way you have drawn (using symmetry
arguments). (Keep your explanation to 2-3 sentences)
Field can only have component perpendicular to the plane: At each point, there is an
equal amount of charge to the right and to the left. The horizontal components always
cancel.
Problem 2 (5 Points) Consider an ellipsoidal conducting object carrying a positive
charge Q (shown below).
(a) Draw a sketch of the electric field (using field lines) very close to the surface of the
object. Your sketch should show where the magnitude of the electric field will be biggest.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
(b) Explain in words how the fact that we are looking at a conducting object determines
the direction of the field relative to the surface at very small distances to the surface.
(Keep your explanation to 2-3 sentences)
In electrostatics, E is perpendicular to the surface of the conductor. If not then, there
would be a tangential component results in movement and re-distribution of charges.
Problem 3 (5 points):
Consider an electric dipole consisting of point charges +q and –q, separated by a fixed
distance d.
(a) Determine the net force and net torque on the dipole in a uniform electric field E as a
function of the angle between the axis of the dipole and the direction of the field.
1
dθ
E -
+ F+
E
-
The force acting on + q is F+ = qE .
The force acting on −q is F− = − qE .
The net torque is τ = τ 1 + τ 2 = r1 × F1 + r2 × F2 = q ( r1 − r2 ) × E = qd × E
Therefore the magnitude of the net torque is
τ = qdE sin θ
and the direction of the net torque is perpendicular to this paper (outward), which is
determined by the right hand rule of cross product.
(b) Now the dipole is brought into the field of a fixed point charge Q, which is at distance
r from +q and distance (r+d) from –q. We observe that the dipole accelerates towards Q.
Is Q positive or negative?
r
Q
+
d
-
x
Since + q is closer to Q, force acting on + q is stronger than force acting on −q . This
indicates that Q is negative.
(c) If the dipole has mass m, what will be the initial acceleration of the dipole in case (b)?
⎛1
⎞
1
ma = F+ + F− = − kq Q ⎜ 2 −
⎟ xˆ
2
⎜ r (r + d ) ⎟
⎝
⎠
Since r
a=−
d , we have
−2
1
(r + d )
2
1 ⎛ d⎞
1 ⎛ 2d ⎞
= 2 ⎜1 + ⎟ ≈ 2 ⎜1 −
⎟⇒
r ⎝ r⎠
r ⎝
r ⎠
⎞
kq Q ⎛ 1
kq Q
1
xˆ = −
⎜ 2−
⎟
2
m ⎜⎝ r ( r + d ) ⎟⎠
mr 2
2kqd Q
⎛ ⎛ 2d ⎞ ⎞
⎜1 − ⎜ 1 − r ⎟ ⎟ xˆ = − mr 3 xˆ
⎠⎠
⎝ ⎝
Problem 4 (5 Points): Young + Freedman, Challenge Problem 22.65
∫ ρ dv = Q ⇒ ∫ 4π r ρ ( r ) dr = Q
⇒ ∫ 4π r α dr + ∫ 4π r 2α (1 − r R ) dr
R
a)
2
0
R/2
2
0
= 43 πα r 3
R
2
R/2
R/2
0
+ 43 π 2α r 3
R
R/2
− 44πR 2α r 4
R
R/2
= 85 πα R 3 = Q ⇒ α =
2
8Q
5π R 3
∫ Eids = ∫ ρ dv ε
b) Gauss’s law:
0
For r ≤ R 2 : 4π r 2 E = 43 π r 3α ε 0 ⇒ E =
When r = R / 2 , E =
For
R
2
4Q
15πε 0 R 2
≤r≤R:
4π r 2 E = ∫
R/2
0
4π r 2α dr + ∫
(r
r
R/2
4π r 2 2α (1 − r R ) dr
)=
=
4
3ε 0
π ( R2 ) α + 4πε 2α
=
4
3ε 0
π R 3 5π8QR 2 ( Rr ) − 23 ( Rr ) − 321 = 3ε4
E=
8Q r
αr
=
3ε 0 15π R 3 ε 0
3
0
3
Q
60πε 0 r
2
(
( 64 (
1
3
3 r
R/2
3
)
r 3
R
− 41R r 4
4
)
)
r
R/2
0
8Q
5
4
3ε 0
(
π R 3α 2 ( Rr ) − 23 ( Rr ) − 321
(2(
)
r 3
R
3
4
)
)
− 23 ( Rr ) − 321 ⇒
4
− 48 ( Rr ) − 1
4
4Q
15πε 0 R 2
Q
When r = R , E =
4πε 0 R 2
Q
For r ≥ R : 4π r 2 E = Q ε 0 ⇒ E =
4πε 0 r 2
Q
When r = R , E =
.
4πε 0 R 2
When r = R / 2 , E =
c)
4
3
π ( R2 ) α
d) E =
3
Q
=
4
3
π ( R2 )
3
Q
8Q
4
=
3
5π R 15
αr
8Q r
=
3ε 0 15π R 3 ε 0
15π mε 0 R 3
8Qe r
8Qe
mr +
=0⇒ω =
and T = 2π
8Qe
15π R 3 ε 0
15π mε 0 R 3
e) No, because the force acting on the electron is no longer proportional to the distance
the charge deviates from its original point.
Problem 5 (5 Points): Young + Freedman, Exercise 23.39
a) E =
q
4πε 0 r 2
when r ≥ R for both spherical shell and solid conductor.
∞
q
R
4πε 0 R
Then we have U ( R ) = ∫ E ( r ) dr =
.
3
b) −1200 = k
q
⇒ q = k −1 ×1.2 × 103 × 0.15 = −20nC
15cm
c) No, because the amount of net charge on the balloon is very small.
Problem 6 (5 Points): Young + Freedman, Problem 23.61
a) Using the result from 23.57, we have
Vab 1
50 × 103
1
E (r ) =
=
×
= 9.71× 104 V/m
−2
ln ( b / a ) r
⎛ 14 × 10 ⎞ 7 × 102
ln ⎜
−4 ⎟
⎝ 90 ×10 ⎠
b) 9.71× 104 V/m × q = 10 × 30 × 10−6 × 9.8 × 10−3 ⇒ q = 3 × 10−11 C
4
Massachusetts Institute of Technology
Spring Term 2005
8.02X Electricity and Magnetism
Problem Set 3
Issued:
Due:
Wed, Feb 16
Fri, Feb 25,, 4PM <- note Date + Time!
Reading suggestions (from Young & Freedman)
Fri, 2/18 : Quiz 1 Review
Tue, 2/22: Quiz
Wed, 2/23: Electric Potential cntd 23-3, 23-4
Fri, 2/25: Capacitance 24-1
Homework Problems (30 points total)
Problem 1 (10 points) Two pointlike test charges +q and –q are brought separately into
the vicinity of a pointlike charge +Q. First charge +q is brought to a distance r from +Q
and then +q is removed and –q is brought to the same distance r.
(a) At distance r from +Q, which of the two charges +q or –q has the greater
potential energy U, relative to U(infinity) = 0?
(b) How big is the difference in potential energy for +q and –q?
(c) What can you say about the electric potential in the two cases?
Problem 2 (10 points) A proton is shot towards a lead nucleus from very far away. The
proton reaches a closest distance to the nucleus of 10 fermi before being pushed back.
What is the initial speed of the proton?
Problem 3(10 points) Consider two spherical conductors with radii A and B (A>>B),
that are connected by a conducting rod. The spheres carry a combined net charge +Q.
(a) Explain why the electric potential of the surface of both spheres will be the
same
(b) Determine the ratio of the electric field strengths near the surfaces of the two
spheres.
EXPERIMENT HVPS Questions
Lab Hours: Mondays: 3-5 pm, Tuesday: 7:30-9:30 pm, Wednesday: 7:30-9:30 pm,
Thursday: 3-5 pm, Fridays: 12-3 pm
Due Date: Check-off by March 4 in lab 4-355. To satisfy the check-off for the
Building the HVPS you must do the following tasks.
You will be graded according to the following criteria:
1. Your HVPS works properly.
2. The HVPS is wired according to the wiring diagrams that are shown in the lab
write-up.
3. All connections are properly soldered without shorts.
4. You should be prepared to briefly explain the questions below.
5. You will not be responsible yet for a detailed explanation of how the HVPS
works.
Bonus Points Policy: Check-offs that are finished during the week of Feb 22-Feb 25
will receive 3 extra points. Check-offs on Mon 2/28 will receive 2 extra points, and
check-offs on Tue Mar 1 will receive 1 extra point.
Question 1: Testing The HVPS.
Measurements:
We will measure the voltage output of the HVPS with no-load and with the 1 M1
resistors as loads. Adjust the pot on your LVPS so that the output voltage of the LVPS is
between 1.5V and 12V . Use one of your multimeters to monitor the output voltage of the
LVPS throughout the following measurements. Set the other multimeter to the l K range
and connect the leads not at the output of the HVPS but just before the 1 M1 resistors
position (in the above diagram this connection is labeled output voltage). (Place your
positive red input into the appropriate input to the meter, labeled +DC 1000 V. The
black ground input goes in the usual place.)
1) Determine the range of output voltages for the HVPS.
Now make five measurements of your HVPS between100V to 1000V with three
loads, MMM only, the MMM and the two 1 M1 resistors, and the MMM and one 1M1
resistor . For each setting of the pot:
2) For each setting of the pot, measure the output voltage of the HVPS with your
MMM (note that the MMM is the only load).
3) Use a clip lead to connect the ends of the two 1 M1 resistors together so as to put
them in series across the output of the HVPS (in the above diagram this
connection is labeled (3)). For each setting of the pot, measure the new output
load voltage of the HVPS.
4) Use a clip lead to put only one of the 1 M1 resistors as a load across the output of
the HVPS (in the above diagram this connection is labeled (4)). For each setting
of the pot, measure the new output voltage of the HVPS.
Record your results in the accompanying data table at the end of this problem (which you
should hand in with your answers).
Question 2:
1) The multi-meter, on the 1K scale, acts like a resistor with resistance of
20,0001/ V DC . This means that to find the resistance you multiple
20,0001/ V DC by the full scale voltage reading which is 1000V . What is the
resistance RMMM of your multi-meter on the 1K scale?
2) Draw circuit diagrams of your HVPS, the multi-meter and the loads. Is your
multi-meter in parallel or in series with the load?
3) When the multi-meter is in parallel with the load the equivalent resistance Req is
given by
1
1
1
.
=
+
Req R MMM Rloa d
When the multi-meter is in series with the load the equivalent resistanceReq is
given by
Req = R MMM + Rloa d .
What is the equivalent resistance for loads of 2M1 and 1M1 loads?
4) Calculate the current output of your HVPS given by
V
I = o utput .
Req
Calculate the current output of your HVPS for each setting of the pot for both the
2M1 load and 1M1 load. Record your results in the last two columns of the
data table.
Massachusetts Institute of Technology
Spring Term 2005
8.02X Electricity and Magnetism
Problem Set 4
Issued:
Due:
Thu, Feb 24
Fri, Mar 4 4PM <- note Date + Time!
Reading suggestions (from Young & Freedman)
Fri, 2/25 : Electric Potential, Capacitance:23-4,24-1
Mon, 2/28: Energy Storage in Capacitors, Dielectrics:24-3, 24-4
Wed, 3/2: Capacitors in Circuits:24-2
Fri, 3/4: Conductors and Insulators, EF Experiment 25-1
Homework Problems (30 points total)
Problem 1 (8 points) Two point-like charges Q1 = 1C and Q2 = -1C are separated by a
distance of 1m. Suppose in an x-y coordinate system Q1 sits at (-0.5m, 0) and Q2 sits at
(+0.5m,0).
(a) What is the force on charge Q1 due to Q2?
(b) Find the minimum of the x-component of the field between –0.5 m < x < 0.5m.
What is the magnitude of the field in units of [V/m]?.
(c) Draw graphs of the x-component and y-component of the total electric field
Ex(x,y) and Ey(x,y) vs x between –0.5m < x < 0.5m for y=0, y=-10cm, y=+10cm
(the three curves for each component can be combined into one graph, if properly
labeled).
(d) Sketch the electric field of this charge configuration using fieldlines. Does this
sketch correspond to the graphs from (c)?
Problem 2 (8 points) Two point-like charges Q1 = 1C and Q2 = -2C sit at x1= -0.5m and
x2= +0.5m along the x-axis of some coordinate system.
a. Draw a graph the electric potential due to Q1, Q2 separately and the total electric
potential from x = -2m to x = +2m.
b. How could one approximate the total potential of Q1+Q2 for distances x >> 1m?
c. Draw a graph of the potential energy for a charge Q3 of –0.1C in the potential
created by Q1 andQ2 between –2m < x < 2m
Problem 3 (6 points) Young&Freedman, Problem 24-60
Problem 4 (8 points) Young&Freedman, Problem 24-71
Note that check-off and experiment write-up (FROM BOTH PARTNERS) for
experiment ‘HVPS’ are due on Fri, 3/4. HVPS questions have been provided in a
separate document. There will be 2 bonus points for HVPS on 2/28 and 1 bonus
point on 3/1.
8.02x – Problem Set 4 Solutions
Problem 1 (8 points)
a) The force on charge Q1 due to charge Q2 is attractive, therefore pointing
in the x+ direction. The distance between the charges is 1m. The magnitude
of the force is
F =k
Q1 Q2
≈ 9.0 × 109 N.
d2
(1)
b) The x-component of the electric field will be the smallest exactly in the
middle between the charges. If we moved away from x = 0, the field from one
of the charges would increase more than the amount the field from the other
charge would decrease. Therefore it will be the smallest in the middle. It will
point in the x+ direction, with magnitude
E=k
|Q1 |
|Q2 |
Q1
+k
= 8k 2 ≈ 72.0 × 109 V/m.
(d/2)2
(d/2)2
d
(2)
c) We will plot the components of the electric field, Ex and Ey for y = −0.1,
y = 0 and y = 0.1. The electric field has components
Ex
Ey
= Ex(1) + Ex(2) =
(3)
|Q2 |
Q1
x − x1
x − x2
p
p
+k
,
= k
(x − x1 )2 + y 2 (x − x1 )2 + y 2
(x − x2 )2 + y 2 (x − x2 )2 + y 2
= Ey(1) + Ey(2) =
= k
(4)
Q1
|Q2 |
y
y
p
p
−k
,
2
2
2
2
2
2
(x − x1 ) + y
(x − x2 ) + y
(x − x1 ) + y
(x − x2 )2 + y 2
First the Ex component plots (dashed line: y = −0.1 and y = 0.1, thick line:
y = 0):
1
And the Ey plots (dashed line: y = −0.1, thick line: y = 0, dotted line: y = 0.1):
Just as an extra, we also add a 3D plot of Ex and Ey . . .
d) For the plot of the field lines see the book, picture 21.26(b). From that
we can see that the x-component at y = 0 goes to infinity at the point charges,
and is smaller between them. For y = ±0.1, the x-component is almost zero at
x = ±0.5, because the field lines are almost vertical. When moving away from
the charges at small y, Ex quickly finds its maximum (dense field lines turning
towards x) and then decreases, because the field lines become much less dense.
On the other hand, the Ey -component is zero for y = 0, nonzero for y = ±10,
and changes sign at x = 0, which agrees with our plots.
Problem 2 (8 points)
a) We have two charges, Q1 = 1 C and Q2 = −2 C, positioned at x1 and x2 .
The potentials are
V1 = k
Q1
,
|x − x1 |
V2 = k
Q2
.
|x − x2 |
The plot of the potentials V1 , V2 and both of them combined:
2
(5)
b) For distances x ≫ 1m, the sum of the potentials can be roughly approximated by a potential from a single charge Q1 + Q2 = −1 C, located at x = 0.
We can support our claim by a simple computation:
V
= V1 + V2 = k
≈ k
Q1
Q2
+k
≈
|x − x1 |
|x − x2 |
Q1 + Q2
= V≈ .
|x|
(6)
Just to check where this rough approximation makes sense, we plot V and V≈
for large x.
c) The potential energy for Q3 = −0.1C in the potential of the two charges
is U3 = Q3 (V1 + V2 ).
Y&F 24.60 (6 points)
a) With the switch open, the charges on the capacitors distribute themselves
as on Fig.3. The capacitance of the top line is the same as the capacitance of
the bottom line (two capacitors in series) Ctop = 2µF. From the symmetry we
find that the charge will distribute itself evenly, Q1 = Q2 = Q/2. The total
capacitance of the net of capacitors is C = 4µF. From
Vab =
Q
2Q1
=
C
C
3
we get the total charge Q = 2Q1 = 210V4µF = 840µC. The potential differences
Vad and Vac will then be Vad = Q1 /3µF = 140V and Vac = Q1 /6µF = 70V. We
then obtain
Vcd = Vad − Vac = +70V.
(7)
b) When we now close the switch, the situation will look like in the second
picture. Because C12 = C34 = 9µF (C12 is 3µF and 6µF in parallel), we can
easily see from symmetry, that Vac = Vcb . So all the potential differences across
all of the capacitors will be the same, V = Vac ! How big are they?
Vab = Vac + Vcb = 2V
⇒
V = Vab /2 = 105V
(8)
across each of the four capacitors.
c) How much charge flowed through the line? We can compute the charges
on each of the conductors, because we know the potential differences across
them.
Q1 = Q4 = V × 3µF = 315µC,
Q2 = Q3 = V × 6µF = 630µC.
The total charge on the inner plates of the conductors is zero, −Q1 − Q2 + Q3 +
Q4 = 0. But before the switch was turned on, the total charge on the inside
plates of capacitors 1 and 3 was zero as well. However, now it is
Qtop = −Q1 + Q3 = 315µC,
(9)
so this amount of charge had to flow through the switch.
Y&F 24.71 (8 points)
The plane where the two slabs of dielectric meet, is an equipotential surface.
We can imagine a thin metallic plate in there – and nothing would change. The
capacitor now looks like a series of two capacitors with thicknesses d/2, with
capacitances
C1 = K1
ǫ0 A
ǫ0 A
= 2K1
,
d/2
d
C2 = K1
ǫ0 A
ǫ0 A
= 2K2
.
d/2
d
The total capacitance is then
¶
¶
µ
µ
ǫ0 A
2ǫ0 A
2K1 2K2
K1 K 2
C1 C2
C=
=
=
.
C1 + C2
d
2K1 + 2K2
d
K1 + K2
4
(10)
(11)
Massachusetts Institute of Technology
Spring Term 2005
8.02x Electricity and Magnetism
Problem Set 5
Issued:
Due:
Fri, Mar 4
Fri, Mar 11, 4PM <- note Date + Time!
Note that the EF write-up from both lab partners is also due on 3/11 4PM!
Reading suggestions (from Young & Freedman)
Mon: Electric Current, Resistivity, Circuits, chapters 25-1 to 25-5
Wed: DC Circuits, Kirchoffs Rules, chapters 26-1 to 26-3
Fri: Quiz 2 review
Homework Problems (30 points total)
Problem 1 (10 points) Consider a simple parallel plate capacitor of Area A for each
plate and separation d between the plates. The plates are given equal and opposite
charges Q and the capacitor is isolated from the rest of the world
(a) What is the potential difference between the plates in terms of the given
variables?
(b) What is the stored energy in the capacitor?
(c) Assume the plates are moved from a separation d to 2*d. How much does the
stored energy in the capacitor change?
(d) Show that the change in potential energy in question (c) is identical to the work
done on the plates when moving them from d to 2*d.
(e) Assume a dielectric with dielectric constant k=10 is inserted such that it fills the
gap of the capacitor. How much does the stored energy change? How is energy
conservation satisfied?
Problem 2 (5 points) For the HVPS experiment, you found that the voltage across the
output capacitor was lowest when the load had the lowest resistance. Explain this
observation.
Problem 4 (5 points) Young&Freedman, page 972, Question Q25.11
Problem 5 (5 points) Young&Freedman, page 975, 25.34
Problem 6 (5 points) Young&Freedman, page 975, 25.38
Note that check-off and experiment write-up for experiment ‘Electrostatic Force
(EF)’ are due on Fri, 3/11. EF questions were provided in a separate document.
There will be 2 bonus points for EF check-off on 3/7 and 1 bonus point on 3/8.
Massachusetts Institute of Technology
Spring Term 2005
8.02X Electricity and Magnetism
Problem Set 7
Issued:
Due:
Thu, March 24
Mon, April 4, 4PM <- note Date & Time!
Reading suggestions (from Young & Freedman)
Mon, 3/28 Magnetic Field, Lorentz Force
Wed,3/30 Source of Magnetic Fields, Law of Biot-Savart
Fri, 4/1 Ampere’s Law
Note that the next experiment is EB (Electric Breakdown). The EB questions will be
posted in a separate document. The EB experiment will be due on Monday, April 4.
Problem 1 (5 points):
A charged particle is moving through a uniform magnetic field. If an electric field that
points in the same direction as the magnetic field is turned on, describe the path the
charged particle will take (use a sketch).
Problem 2 (5 points):
Can you set a resting electron into motion with a constant magnetic field? Explain how
(or why not).
Problem 3 (10 points):
See drawing: An ion of mass M and charge q is initially at rest at the origin (x=0) at t=0.
A region of uniform electric field E pointing in the y direction extends from y=0 to y=s.
Above y=s, there is a region of magnetic field B, pointing in the z direction.
(a) When and where does the ion first cross the x-z plane located at y=s?
(b) What is the ions trajectory in the B-field region? Where does it cross the x-z plane
at y=s for the second time?
(c) Where does the ion come to rest?
y
B
s
E
0
X
Problem 4 (10 points):
Shown below is the cross section of a long coaxial cable consisting of an inner core with
radius r0 and an outer shell of radius r1, both centered at r=0. The inner core carries a
current I going into the paper plane, the outer shell carries the same current I in the
opposite direction. Using Ampere’s law, find the magnitude of the magnetic field B(r) in
the region r0 < r < r1 and r > r1.
r1
r0
Massachusetts Institute of Technology
Spring Term 2005
8.02X Electricity and Magnetism
Problem Set 8
Issued:
Due:
Sat, Apr 2
Fri, Apr 8 , 4PM <- note Date + Time!
Note that the exp MF write-up from both lab partners is also due on 4/8 4PM!
Reading suggestions (from Young & Freedman)
Mon: Electromagnetic Induction I
Wed: Electromagnetic Induction II
Fri: Quiz 3 Review
Homework Problems (30 points total)
Problem 1 (10 points) Two parallel wires carrying currents in the same direction are
found to attract each other.
(a) Give a step-by-step explanation of why the wires attract each other, due to the
field created by moving electric charges and the forces on moving charges. Use
pictures and words, but not equations
(b) Again using pictures and words show that you get repulsion if the currents in the
two wires run anti-parallel.
Problem 2 (10 points) You have a horizontal cathode ray tube (CRT) in a lab room
which has been adjusted such that the electron beam should produce a single spot in the
center of the front screen. However, you find that the spot is offset to the right, when
viewed from the front.
(a) You realize that the offset could be caused by an electric or magnetic field or by
a misalignment of the cathode. How can you determine the cause of the problem
without any instruments to detect electric or magnetic fields and without
removing the CRT from the room?
(b) If the cause was an magnetic field, what direction would the field have, relative to
the original direction of the electron beam?
Problem 3 (10 points) The Large Hadron Collider at CERN, Geneva, will accelerate
lead nuclei to an energy of 2.8 TeV/nucleon. The nuclei will follow a circular trajectory
of 27km circumference
(a) What is the momentum of each nucleus [note: The relativistic relationship
between energy and momentum is E = sqrt(p2c2 + m02 c4)]?
(b) How big is the magnetic field of the dipole magnets that hold the nuclei on their
trajectory?
(c) Would it be possible to build a linear accelerator of 100m length that achieves the
same final energy of the particles? What electric field would be necessary for this
and how does it compare to the typical fields inside atoms?
Experiment MF (Magnetic Force)
(20 points total)
Note that check-off and experiment write-up for experiment ‘Magnetic Force (MF)’
are due on Fri, 4/8. There will be 2 bonus points for MF check-off on 4/4 and 1
bonus point on 4/5.
The lab hours are Mon 3-5, Tue. 7:30-9:30 pm, Wed 7:30-9:30 pm, Thu 3-5, Fri 12-3.
You will be graded according to the following criteria:
1. Your Experiment: Magnetic Force apparatus works (you will need to demonstrate
a trial run during the check-off).
2. Your understanding of the underlining physical principles involved in the
experiment.
3. You may be asked a question during the check-off about how to calculate the
1
1Atgd
magnetic permeability constant µ 0 =
(
).
slope n1 n2 r
Problem 1 (10 points)
In your experiment, depending upon the winding direction of the coils, the current
through them will generate magnetic forces such that the coils will either attract or repel
each other causing them to move.
a) Is your apparatus set up to repel or attract? Briefly explain how you intend to
measure the current that flows through the coil that will produce a magnetic force
that will just balance the weight of the foil.
b) What is the radius r of your coils? What is the distance d between your coils?
c) You will measure the current that will produce a magnetic force that will just
balance one, two, and three squares of the foil. Suitable weights are 2 cm by 2 cm
squares of foil. For each number of weights, n, make several measurements of the
current that balances the weight. Average your values.
d) Make a table of your data with columns for weight (in terms of the number of
weights, n), the current (in A), and the current squared (in A2).
e) Make a plot of the current squared (in A2) vertically vs. the weight (in terms of
the number of weights, n) horizontally. If no weights were on your balance, and it
balanced at zero current, then origin is also a point.
f) Use a linear regression to find the slope and intercept of a straight-line fit of your
plot. (You may also find the best fit straight-line by eye.)
g) Estimate the error on the slope. This can be done by taking the extremal
(maximum and minimum) values of the slope and computing the difference with
the best fit slope.
h) Then from the best straight-line fit of your data, calculate the magnetic
permeability of space using
1
1Atgd
µ0 =
(
),
slope n1 n2 r
where g = 9.8m / s 2 , A is the area of your foil, 2 = 2.7 1 10 3 kg / m 3 is the density
of aluminum, t = 1.8 2 10 15 m is the thickness of the aluminum, n1 = 38 , n2 = 10 , r
is the radius of your coils, and d is the separation between your coils.
i) Consider the uncertainties in the measurements of slope, g, 1 , A, t, d, and r.
Which error contributes the most? Which errors can you safely ignore?
Problem 2 (10 points)
In the magnetic force experiment, a current I = 0.5A is passed in series through a 38 turn
coil taped to a table and a 10 turn coil which is taped to a balance directly above the 38
turn coil. The distance between the 10 turn coil and the 38 turn coil is d = 0.5cm . Each
coil has a radius of r =
cm .
a) Calculate the magnitude of the magnetic field originating from the 38 turn coil at
any point on the 10 turn coil. Clearly indicate any approximations you make.
b) Calculate the magnitude of the magnetic field originating from both coils along
the central axis at a height z = 0.5cm above the plane of the 38 turn coil. You
may take this point as the center of the 10 turn coil.
8.02x – Problem Set 8 Solutions
Problem 1 (10 points)
a) The currents run in the same direction x̂. The field created by the first
wire has direction ẑ at the place of the second wire. The electrons in the
second wire move to −x̂ (opposite the current), so the Lorentz force on them
has direction −ŷ, towards the first wire. The wires attract each other.
b) Now the currents run in opposite directions. Replace the velocity of
electrons in the second wire by −~v , so we get force −F~ (in the ŷ direction). The
wires thus repel each other.
Problem 2 (10 points)
The electrons in the CRT tube are deflected to the right, when looking from
the front. This could be caused by an electric field pointing horizontally to the
left, or a magnetic field pointing vertically upwards. Let’s try to distinguish
these cases by moving the apparatus around.
If we rotate the apparatus in the horizontal plane, and the beam is still
deflected to the right ⇒ the problem-causing field is a vertical magnetic field.
If we rotate the apparatus in the horizontal plane and find that the deflection
changes, and that there are two (opposite) orientations for which it is zero ⇒
the villain is the horizontal electric field.
1
Problem 3 (10 points)
a) Each nucleus has energy E = 208 × 2.8TeV = (208 × 2.8 × 1012 )(1.6 ×
10 )J = 9.3 × 10−5 J. Using the relativistic equation E 2 = p2 c2 + m20 c4 for
the energy and momentum, we get
r
E2
− m20 c2 =
(1)
p =
c2
sµ
¶2
9.3 × 10−5 J
2
=
− [208(1.67 × 10−27 kg)(3 × 108 m/s)] ≈
3 × 108 m/s
kg.m
≈ 3.1 × 10−13
.
s
−19
All of p comes just from the first term. We could have seen this much easier in the eV units. The rest mass (and energy) of the lead nucleus is E0 =
208(1.67 × 10−27 kg)(3 × 108 m/s)2 = 3.1 × 10−8 J = 0.2TeV. The total energy
when accelerated is 208×2.8TeV. Obviously, almost all of the energy comes from
momentum. Thus p ≈ 208 × 2.8TeV = (208 × 2.8)(1.6 × 10−19 )/(3 × 108 m/s) =
3.1 × 10−13 kg.m/s.
p
,
b) The radius of the path of a charged particle in a magnetic field is R = qB
so the magnetic field needed to keep the nuclei on their trajectory is
B=
3.1 × 10−13 kg.m/s
p
=
≈ 71T.
qR
(1.602 × 10−19 C)(27km)
(2)
c) A linear accelerator of length L = 100m would have to have a potential
difference of V = E/q ≈ 6 × 1014 between its ends. This would mean an electric
field of magnitude Ef ield = V /d ≈ 6 × 1012 V/m, which is ridiculously high. For
comparison, the fields inside atoms have strengths of order
Eatomic ≈
kq
(9.0 × 109 N · m2 /C2 )(1.602 × 10−19 C)
V
≈
≈ 1010 .
ra2
(10−10 m)2
m
2
(3)
Massachusetts Institute of Technology
Spring Term 2005
8.02X Electricity and Magnetism
Problem Set 9
Issued:
Due:
Wed, Apr 7
Fri, Apr 15 , 4PM <- note Date + Time!
Reading suggestions (from Young&Freedman)
Mon: Quiz 3
Wed: Faradays Law, Transformer Action, Inductance: 29-5, 31-6, 30-1 to 30-3
Fri: AC Circuits, 31-1 to 31-3 (Note: You can ignore phasor diagrams)
Homework Problems (30 points total)
Problem 1 (20 points) The 1200 dipole magnets of the LHC accelerator at CERN will
have a field of 8 Tesla. Approximate the magnet as two square loops of dimension 14m x
0.3 m, with a distance of 10cm between the two loops (similar to the arrangement in
question 4 of quiz 3).
(a) What direction should the field have to keep protons going clockwise on a
circular trajectory?
(b) What is the current in each loop needed to create a field of 8T in the midplane of
the two loops?
(c) Again using pictures and words show that you get repulsion if the currents in the
two wires run anti-parallel
(d) . Assume that the field is homogenous at 8T over a cross-section of 200 cm2 .
What is the total energy stored in the magnetic field of all dipoles combined?
Problem 2 (10 points) Explain why power is transformed to high voltages to be
transported over power lines
Experiment AMP (Amplifier)
(20 points total)
Note that check-off and experiment write-up for experiment ‘Amplifier (AMP)’ are
due on Fri, 4/29. AMP questions are provided below. There will be 4 bonus points
for MF check-off on or before 4/15, three bonus point for check-off on or before
4/22, two bonus points on 4/25 and 1 bonus point on 4/26.
The lab hours are Mon 3-5, Tues. 7:30-9:30 pm, Wed 7:30-9:30 pm, Thur 3-5, Fri 12-3.
You will be graded according to the following criteria:
1. Your Experiment: Amplifier apparatus works (you will need to demonstrate a trial
run during the check-off).
2. Your understanding of how to zero and calibrate your amplifier.
3. You may be asked a question during the check-off about how to calculate the gain
of your amplifier.
Problem 1: Data Analysis
After you have made and tabulated your measurements of input and output voltages for
the amplifier, plot the output voltage as a function of input voltage. Determine the gain
of the amplifier in the region where the response is linear and compare your value with
the theoretical expectation. Does the gain remain constant over the whole range of input
voltages?
Problem 2: Amplification
A negative feedback circuit in an amplifier circuit is shown in the figure below. In this
circuit, the resistors have the values R1 = 9.1k1 , R2 = 1001 , and the open loop
5
amplification is A 1 10 . Pin 3 is non-inverting and pin 2 is inverting.
a) What is the output voltage Vout of the op-amp in terms of the input voltage Vin to pin
3, and the voltage at pin 2, V2 , and the amplification A? (Note: all voltages are with
respect to common).
b) Derive an expression for the voltage at pin 2, in terms of R1 and R2 .
c) Derive an expression for the closed loop gain G = Vout Vin in terms of R1 , R2 , and A.
d) Calculate the value of the gain G, clearly indicating any approximations you have
made.
Problem 3: Calibration
The calibration circuit for the amplifier is shown in the figure below. When the multimeter is set on the 250 mV scale the multi-meter has a resistance of 5k1 .
a) When the 5k1 pot is turned 2/3 of the way in the direction of the +6V line from the
16V end, what is the voltage difference between common C and the 5k1 pot slider?
b) When the 5k1 pot is set as in part a), calculate the voltage difference between the
common C and point D which is connected via a clip lead to the input A?
c) What output voltage did you read when your pot was set as in part a) while you were
calibrating your amplifier?
MIT Department of Physics
Physics 8.02X
Spring 2005
Solution to Problem Set #9
Problem 1 (20 points) The 1200 dipole magnets of the LHC accelerator at
CERN will have a field of 8 Tesla. Approximate the magnet as two square loops
of dimension 14m x 0.3 m, with a distance of 10cm between the two loops
(similar to the arrangement in question 4 of quiz 3).
(a) What direction should the field have to keep protons going clockwise on a
circular trajectory?
0.1m
Magnet:
0.3m
14m
End view:
B
y
0.1m
x
0.3m
At the center, the four fields add up to for a net vertical field.
The accelerator, made by 1200 dipole magnets, is as the following:
B
B
B
v
proton
F
magnets
f
f
f f
F = qv × B . The direction of B is shown in figure above.
(b) What is the current in each loop needed to create a field of 8T in the midplane
of the two loops?
1
B
y
h=0.1m
x
w=0.3m
For each wire,
µI
µ0 I
B= 0 =
2π r 2π h 2 + w2 / 2
)
(
And the y component:
µ0 Iw
w
=
By = B
2
2
h 2 + w2 π ( h + w )
By symmetry, all wires contribute the same By , therefore in total
ByTotal = 4 By =
I=
4µ0 Iw
⇒
π ( h 2 + w2 )
Bπ ( h 2 + w2 )
4µ0 w
=
8π ( 0.12 + 0.32 )
4 × 4π ×10−7 × 0.3
= 1.7 ×106 A
(c) Again using pictures and words show that you get repulsion if the currents in
the two wires run anti-parallel.
The magnetic field created by wire 1 at wire 2
is shown in figure.
The force on wire 2 is given by
f
f f
F = IL × B
Which is in the +x direction: repulsion
y
x
I1
I2
(d) . Assume that the field is homogenous at 8T over a cross-section of 200 cm2 .
What is the total energy stored in the magnetic field of all dipoles combined?
The energy density of magnetic field is
B2
u=
2µ0
For each magnet, the energy stored is
U = uV
2
For the total 1200 magnet, the energy stored is
82
B2
× (14 × 0.3 × 0.1) = 1.3 × 1010 J
U tot = 1200uV = 1200
V = 1200 ×
−7
2 µ0
2 × 4π × 10
Problem 2 (10 points) Explain why power is transformed to high voltages to be
transported over power lines.
Energy loss (per unit time) in transmission line is due to the resistance of the
wires, i.e.
Ploss = I 2 Rloss
The power delivered is given by
Pdel = IV
For a given delivery power Pdel ,
Pdel2
Ploss =
Rloss
V
Therefore, the higher the voltage, the lower the loss.
3
Massachusetts Institute of Technology
Spring Term 2005
8.02x Electricity and Magnetism
Problem Set 10
Issued:
Due:
Sun, April 17
Fri, April 22, 4PM <- note Date & Time!
Reading suggestions from Young & Freedman
Mon, 4/18 Patriots day vacation
Wed, 4/20: AC Circuits, Inductors, RL Circuits 31.2, 30.2,30.4
Fri, 4/22: RLC circuits, Oscillations: 30.5, 30.6
Problem 1 (6 points):
Compare the oscillations of an LRC circuit to the vibration of a mass m on a spring. What
do L and C correspond to in the mechanical system? What is the mechanical analog to R?
Problem 2(6 points)
You have probably noticed that when a circuit carrying a large current is interrupted, a
spark occurs between the poles of the switch (or the poles of a plug that is pulled).
(a) Explain this phenomenon. Where does the energy for the spark come from?
(b) Assume an inductor L=1mH and a resistor R=101 are connected in series to a
battery providing V=100V. How much energy is stored in the inductor a long
time after the circuit is closed?.
Problem 3 (6 points) Young&Freedman, Problem 30.10
Problem 4 (6 points) Young&Freedman, Problem 30.14
Problem 5 (6 points) Young&Freedman, Problem 30.26
PROBLEM SET 10; 802x; SPRING 2005
1) The differential equation governing an RLC circuit is:
-Ldi/dt - Ri + q/C = 0. Using i= - dq/dt, we have,
Ld2q/dt2 + Rdq/dt +q/C = 0.
The differential equation governing a mass on a spring is (with velocity proportional viscous damping):
md2x/dt2 + bdx/dt + Kx = 0. Here the mass is m, K is the spring constant and b is the coefficient
of proportionality between velocity and the viscous retarding force.
Thus: M and L play the same roles; b and R play the same roles; and, K plays the same role as 1/C.
Mv is the momentum that will persist unless changed by a force, and Li is the flux in an inductor that
will persist unless changed by an external agent. The kinetic energy stored in motion is (1/2) mv2, while
energy is stored in the inductor as (1/2)Li2. The resistor is an agent for energy loss at the rate i2R. Energy
is lost to viscocity at the rate bv2. Energy is stored in a capacitor as (1/2)q2/C and energy is stored in the
spring as (1/2)Kx2.
2) The self-inductance of the circuit causes the current to persist until the voltage developed across the gap
acting as a capacitor causes it to stop. Now this gap usually has a very small capacitance and the current,
which we have assumed to be large, can charge the gap to a very large voltage. Thus the spark develops
when the air brakes down. The energy for the spark comes from the energy stored in the self-inductance of
the circuit, (1/2)Li2.
The equilibrium current is i = V/R = 100/10 = 10 amps. The energy stored in the inductor is (1/2Li2 =
(1/2)(1/1000)(100) = 1/20 joule.
3) a. Compare figure 30.18 and fig 30.6b. Note that points a and b are reversed. Thus, according to equation
30.8, dI/dt = (Vb – Va)/L = - 1.04V/0.260H = -4 A/s. Thus, the current is decreasing.
b. From a. we know that di = (-4A/s) dt. After integrating both sides of the expression with respect to t, we
obtain 1I = (-4A/s)1t and so I = (12.0A) – 4A/s * 2s = 4A.
4) a. U = P*t = (200W)(24h/dayx3600s/h) = 1.73x107J.
b. U = 1 L I2 and therefore L = 2U/I2 = 2 (1.73x107J)/(80A2) = 5406H.
5) When switch 1 is closed and switch 2 is open, the loop rule gives L dI/dt + IR = 0 and therefore
dI/dt = - I R/L. Integrating from I0 to I on the LHS and 0 to t on the RHS gives ln(I/I0) = - R/L t and
therefore I(t) = I0 exp(-t/(L/R))
Massachusetts Institute of Technology
Spring Term 2005
8.02x Electricity and Magnetism
Problem Set 11
Issued:
Due:
Sun, April 24
Fri, April 29, 4PM <- note Date & Time!
Note that Exp AMP check-off and write-up are also due 4/2\9
Reading suggestions (from Young&Freedman)
Mon, 4/25: RLC circuits, Resonance: 30-6, 31-3,31-5
Wed, 4/27: Displacement current, Maxwell’s Equations: 29-7
Fri, 4/29: Electromagnetic Waves: 32-1 to 32-3
\
Problem 1 (10 points)
How do you know that sound is not an electromagnetic wave? What kind of wave is it?
Problem 2(10 points) Young & Freedman 29.38
Problem 3(10 points) Young & Freedman 32.6
8.02x – Problem Set 11 Solutions
Problem 1 (10 points)
How to tell whether sound is an EM wave? Let us just list some of the
obvious differences. Sound waves do not propagate in vacuum, they need some
medium (gas, solid, or liquid) to support them. The speeds of propagation have
very different magnitudes for sound and EM waves.
However, to be sure to tell the two types of waves apart, we should measure
their effects. Simply said, you can see light and hear sound, but not the reverse.
An antenna (a metal rod) can “catch” EM waves, they would make the free
charges move in it. This effect can be measured. However, you will not get any
effect by trying to induce charge motion (or EM fields) by sound. On the other
hand, you can hear a sound wave (your ears measure pressure changes), but not
an EM wave.
Sound is a longitudial mechanical wave, molecules of air bumping and pushing each other, making the signal spread towards the receiever. Longitudial
means the displacements of air molecules are in the direction of the signal.
On the other hand, EM waves are transverse, meaning that the displacements
(electric and magnetic fields) change in the direction perpendicular to the signal
direction.
Y&F 29.38 (10 points)
a) Let us use the microscopical Ohm’s law. The electric field is proportional
to resistivity and current density.
E = ρj =
ρI
(2.0 × 10−8 Ωm)(16A)
V
=
= 0.15 .
A
2.1 × 10−6 m2
m
b) The field is changing at rate
µ ¶
ρ dI
d ρI
2.0 × 10−8 Ωm
V
dE
=
=
=
(4000A/s) = 38 .
−6
2
dt
dt A
A dt
2.1 × 10 m
ms
(1)
(2)
c) The displacement current density in the wire is (taking ǫ ≈ ǫ0 for copper)
jD = ǫ0
dE
A
= 3.4 × 10−10 2 .
dt
m
(3)
d) The magnitude of the displacement current is
2
ID = AjD = (2.1 × 10−6 m2 )(3.4 × 10−10 A/m ) = 7.14 × 10−16 A,
1
(4)
which is many orders of magnitude smaller than the conduction current. Therefore the magnetic field is very well approximated only by the magnetic field from
the conduction current,
µ0 Ic
µ0 16A
=
= 5.33 × 10−5 T.
(5)
2πr
2π 0.06m
Just for comparison, the magnitude of the magnetic field due to the displacement
current is totally negligible:
Bc =
BD =
µ0 7.14 × 10−16 A
µ0 ID
=
= 2.38 × 10−21 T.
2πr
2π
0.06m
(6)
Y&F 32.6 (10 points)
~
a) Our EM wave has B(x,
t) = ŷB0 sin(kx + ωt). Sketching sin(kx + ωt) for
time t = 0 and t = ∆t we realize it propagates in the −x̂ direction. In detail,
point A (x = 0) is where the amplitude is zero for time t = 0. At time t = ∆t,
point B where the amplitude is zero can be found from k∆x + ω∆t = 0, giving
us ∆x = −(ω/k)∆t. This means the wave has moved to the left.
b) We know the relationship between the wavevector and the wavelength,
k = 2π/λ. The wavelength is the distance the wave travels in one time period,
λ = cT = c/f . The frequency of the wave is then
kc
(1.38 × 104 rad/m)(3 × 108 m/s)
=
= 6.59 × 1011 Hz,
2π
2π
giving us also the angular frequency ω,
f=
ω = 2πf = 4.14 × 1011
rad
.
s
(7)
(8)
~ ×B
~ points in the direction
c) The direction of the electric field is such that E
~
~
of propagation. If B has direction ŷ, E must then point in the ẑ direction for
the wave to propagate in the −x̂ direction. The magnitude of E is simply
|E| = c|B|, so we get
~
B(x,
t)
~
E(x,
t)
= ŷB0 sin(kx + ωt),
(9)
= ẑ(cB0 ) sin(kx + ωt) =
¶
µ
¶ ¸
·µ
12 rad
4 rad
x + 4.14 × 10
t .
= ẑ(2.48V/m) sin 1.38 × 10
m
s
2