Physics 194 Exam #1 practice problems solutions 1. A small electrically charged aluminum sphere (mass 20.0 𝑔, charge −5.00 𝜇𝐶) is placed on a plastic tabletop. Your
goal is to set up an electric field that will result in the sphere accelerating upwards at 13.0 𝑚 𝑠⁄ . over a distance of 1.00 𝑚.
a. Draw a force diagram for the sphere while it is in motion.
b. Determine the magnitude and direction of the E-field required to make the sphere move in this particular way. c. What assumptions did you make in part b.?
In order for the sphere (the system of interest) to have a constant acceleration the net force exerted on it must be constant. This means a uniform electric field is needed. Using Newton’s 2nd
law I can relate the acceleration of the sphere to the forces exerted on it. The force exerted by the E-field on the sphere points upward. The force exerted by Earth on the sphere points downward. Using the standard coordinate system (positive y-direction upward),
𝑎1 = 1
𝑚3 𝐹56 7 = 1
𝑚8𝐹9:⃗ 56 7,1+ 𝐹9:⃗ 56 7,1> = 1
𝑚?𝑞𝐸1− 𝑚𝑔B
𝐸1=𝑚
𝑞 ?𝑎1+ 𝑔B =
0.02 𝑘𝑔
−5.00 × 10EF𝐶(13.0 𝑚 𝑠⁄ .+ 9.8 𝑚 𝑠⁄ ) . 𝐸1= −9.12 × 10K𝑁 𝐶⁄
In order to make the sphere accelerate upwards at 13.0 𝑚 𝑠⁄ . the E-field must be uniform over the 1.00 𝑚 that the sphere is accelerating and point downward with a magnitude of 9.12 × 10K𝑁 𝐶⁄ .
In part b. I assumed that the air did not exert a significant force on the sphere as it accelerated upward. I also assumed that the table was not polarized significantly by the sphere and did not as a result exert an attractive electric force on the sphere.
2. “High voltage” power lines are often seen suspended high above the ground. Two of these power lines are suspended above a physics professor who is out for a walk. The power line on the left carries a current out of the page; the power line on the right carries a current into the page.
a. What is the direction of the magnetic field produced by the two currents at the location of the physics professor? He is on the ground halfway between the two wires. Draw a diagram illustrating your reasoning. Hint: Determine the direction of the magnetic field produced by each current separately, then add them together.
b. The physics professor has been shuffling his feet on the ground as he walks to the right which has made him become negatively charged. Determine the direction of the magnetic force (if any) exerted by the magnetic field on him. Explain your reasoning.
c. Does the power line on the left exert a force on the power line on the right? Explain your reasoning. If you think it does, determine its direction.
The magnetic field where the physics professor is standing will point upward. See the diagram. The magnetic field lines of each wire are circles, counterclockwise around the left power line, and clockwise around the right power line. Each of them contribute to the magnetic field where the physics professor is. When the two B-field vectors are added their horizontal components cancel leaving just an upward vertical component.
Using the right hand rule for magnetic force I can determine the direction of the magnetic force exerted by the field on the physics professor. His velocity is to the right, the field points up, which means (keeping in mind that he is negatively charged) the magnetic force exerted on him is into the page.
Yes, the left power line does exert a magnetic force on the right power line. First, using the right hand rule for the B-field I learn that the B-field produced by the left power line, at the location of the right power line, points up. Using the right hand rule for magnetic force, I learn that the magnetic force exerted by the B-field on the right power line points to the right.
3. Your friend is heating some food in the microwave. The two lights in the kitchen are off. You join your friend in the kitchen and turn the lights on as you enter. Your friend says “Since you turned the lights on I’ll have to leave the pizza in the microwave longer.” You’re not so sure about this, but your friend tries to explain. They take out a piece of paper and draw the circuit shown. “When you turned on the lights you closed the switch in the circuit. This means the current from the power supply is being split between the microwave and the two lights. Since the power output of a device is 𝑃 = 𝐼∆𝑉 and the current through the microwave is now lower, the power output of the microwave will be lower and the pizza will take longer to cook.”
a. You’re pretty sure you’ve never noticed having to heat something longer in the microwave if the lights were on. Explain the problem with your friend’s reasoning.
b. Assuming the circuit drawn by your friend’s is the way the kitchen is wired, determine the power output of the power supply when just the microwave is on. Then, determine the power output of the power supply when the microwave and two lights are on.
The problem with your friend’s argument is that the current through the power supply will remain the same when the switch is closed. This is not true. When the switch closes the equivalent resistance of the circuit decreases which results in a larger current through the power supply. This makes it possible for the current through the microwave, and
therefore its power output to remain the same. Another way to think about it is that because the three devices are in parallel, the potential difference across each device is the same, and in this case equal to the potential difference across the power supply, 120 V, regardless of whether or not the lights are on.
When just the microwave is on the power output of the power supply is: 𝑃Q = 𝐼Q∆𝑉Q= R
𝜀
𝑅UVW (𝜀) =
(120 𝑉).
9.6 Ω = 1500 𝑊 When the lights are on too:
𝑃Q = 𝐼Q∆𝑉Q= R 𝜀
𝑅UVW (𝜀) =
𝜀.
8𝑅1 [ +
1 𝑅\+
1 𝑅\>
E]=
(120 𝑉).
89.6 Ω +1 192 Ω +1 192 Ω>1 E]
= 1650 𝑊
•
1´
I
I
21
B
!
2B
!
B
!
q
Microwave
Light Light Power supply
to kitchen 120
V V
D =
9.6
R= W
192 R= W
4. An advanced materials lab uses strong magnetic fields to investigate the properties of new compounds. During a particular experiment the magnetic field in the lab is supposed to do the following, in sequence:
• Stage 1: Remain at 0.030 T for 1.0 s.
• Stage 2: Increase to 0.600 T over the next 0.5 s. • Stage 3: Remain at 0.600 T for 1.0 s.
• Stage 4: Decrease back to 0.030 T over the next 0.5 s. • Repeat the above steps a second time.
A circular wire coil (100 windings, 0.20 m radius, 5.0 W resistance) connected to an ammeter is placed in the lab so the above magnetic field points perpendicularly through the coil. This coil will be used to determine if the magnetic field in the lab is what it’s supposed to be.
a. Is it possible to use the ammeter reading to decide if the magnetic field in the lab is consistent with the above sequence? Explain your reasoning.
b. Sketch a graph of the current in the coil as a function of time for the above 6 second sequence. The graph should numerically precise along the horizontal axis. It does not need to be numerically precise along the vertical axis. c. Determine the magnitude of the current in the coil during the first time interval that the current is present. d. For each time interval that a current is present in the coil draw a labeled diagram showing its direction and also
the direction of the induced magnetic field. In the diagram(s) you draw, the coil should be in the plane of the page and the magnetic field produced by the lab equipment should point out of the page.
The ammeter reading will give some information about the magnetic field in the lab, but not all. Because there will only be a current in the coil during stages 2 and 4 (when the magnetic flux through the coil is changing), the ammeter will only reveal if the rate of change of the magnetic field is correct. It won’t be able to decide if the magnetic field has the correct magnitude during stages 1 and 3.
The first time interval there is a current is stage 2. I’ll use Faraday’s law of electromagnetic induction to determine the induced EMF in the coil, and choosing the coil’s surface normal to parallel the external magnetic field:
𝐼 = 𝜀 𝑅=
1 𝑅𝑁 ^
ΔΦa Δ𝑡 ^ =
𝑁 𝑅^
Φa,c− Φa,d Δ𝑡 ^ =
𝑁 𝑅^
𝐵c𝐴ccos 𝜃c− 𝐵d𝐴dcos 𝜃d
Δ𝑡 ^ =
𝑁𝐴 cos 𝜃
𝑅 ^
𝐵c− 𝐵d Δ𝑡 ^
𝐼 =(100)𝜋(0.20 𝑚).cos 0°
5.0 Ω ^
0.600 𝑇 − 0.030 𝑇
5. For each of the statements below, if true, write a response to convince Mike that the statement is true. If not true, write a response to convince Mike that the statement is not true.
a. As you take your jacket off you hear cracking sounds coming from the long sleeves of the shirt you are wearing. You then bring your arm close to some dust on the floor. The dust leaps through the air and sticks to your shirt sleeve. This means the dust was electrically charged.
b. A toy car has a positively charged glass sphere attached to its roof. A second toy car has a negatively charged plastic sphere attached to its roof. The two cars are released from rest and roll towards each other. The electric potential energy of the two-car system increases with time because they are getting closer.
6. Two plastic spheres are hanging from plastic threads as shown. A glass rod that has been rubbed with silk (removing electrons from the rod) is brought towards sphere #1 from the left. Sphere #1 swings toward the rod as it gets close. Sphere #2 does the same, but swings less. The rod is then used to “paint” sphere #1, which means touching all parts of the sphere with the rod. Once this is done and the rod is moved far away, the two spheres swing towards each other and remain touching. a. Explain in detail what is happening microscopically during this
sequence of events and why the spheres behaved in the way they did. Use charge diagrams to support your explanation.
b. The two plastic spheres are replaced by metal spheres. Everything happens the same except for: • When the rod is brought near, the two spheres swing toward it more dramatically.
• After the rod is used to “paint” sphere #1 and the rod is moved far away, the spheres swing towards each other, touch, then quickly swing away from each other, eventually coming to rest angled away from each other.
Explain in detail what is happening microscopically that causes the metal spheres to behave differently in these two parts of the experiment. Use charge diagrams to illustrate your explanation.
When the glass rod is brought near sphere 1, sphere 1 polarizes with its electrons shifting slightly to the left and its positively charged nuclei shifting slightly to the right. This causes sphere 1 to be attracted to the rod. The same thing happens to sphere 2, just less so because it is further away from the rod. When sphere 1 is painted with the rod,
electrons transfer from the sphere to the rod. Since sphere 1 is now positive, it polarizes sphere 2 causing the spheres to be attracted to each other. This is why they hang angled towards another and remain touching.
For the metal spheres, the initial polarization effect is enhanced because the metal spheres are conductors. This means their electrons are free to move throughout the spheres which means the separation between their positive and
negative regions will be greater. This leads to the metal spheres being attracted to the rod more strongly than the plastic spheres were. Later, when the spheres touch, electrons from sphere 2 can flow freely over to sphere 1. They do so until there two spheres are equally positively charged. This is why they repel, which is why they come to rest angled away from each other.
7. Two parallel metal plates are charged oppositely creating a uniform electric field of magnitude 6.00 × 10p𝑁 𝐶⁄ as shown. A beam of charged particles (could be positively or negatively charged) is fired with speed 𝑣 = 2.00 × 10F𝑚 𝑠⁄ into the electric field region. You realize these particles will be deflected from traveling in a straight path by the electric field. But, you then realize that it is possible to add a magnetic field to this region so that the particles will travel in a straight path anyway.
a. Explain how this is possible. On the diagram show the magnetic field that will accomplish this.
b. Determine the magnitude of this magnetic field.
There are two possibilities, the particles are positively charged or negatively charged. If they are positively charged the electric field will exert a downward force on them. This means the magnetic field needs to exert an upward force on them. Using the right hand rule, we find that a magnetic field pointing into the page will exert an upward force. If the particles are negatively charged the electric field will exert an upward force on them. But, a magnetic field into the page will exert a downward force on them. So, in either case, a magnetic field pointing into the page will allow the particles to possibly pass through the region on a straight path. This will happen if the electric and magnetic forces have equal magnitude.
⟹ 𝐵 =
6.00 × 10
7
𝑁 𝐶
⁄2.00 × 10
6𝑚 𝑠
⁄ = 30.0 𝑇8. A pair of light bulbs is connected to a pair of batteries as shown. a. Predict the power output of each light bulb.
b. Light bulb 2 is removed from the circuit and replaced with an ordinary length of wire. Your friend says “Light bulb number 1 will remain just as bright because
and the potential difference hasn’t changed since it’s still 3 volts total, and the resistance of the bulb hasn’t changed.” You’re not so sure about this… Explain the problems with your friend’s reasoning, then make a numerical prediction for what the power output of bulb 1 will become.
To predict the power output of each bulb I need to first predict the current flowing through each of them. Since the whole circuit is in series there is just one current to determine. I’ll use Kirchhoff’s loop rule:
Now I can predict the power output of each bulb: E B F F qE qvB E B v = Þ = Þ = 2
P
=
D
V R
1 2 1 2
1 2 1 2
1 2 1 2
0
0
3.00
0.200
15.0
R RV
V
V
V
IR
IR
V
I
A
R
R
e ee e
e e
D
+
D
+
D
+
D
=
Þ
+ +
-
-
=
+
Þ
=
=
=
+
W
+ + + + + + + + + + + + + + + +-E
!
+ + + + + + + + + + + + + + + + -ÄB
!
E
!
Ä
Ä
ÄÄ
Ä
ÄÄ
Ä
ÄÄ
Ä
ÄÄ
Ä
1
1.50
V
e
=
e
2 =1.50V1 5.00 R = W
Your friend is confused about what the symbols in the equation he is using mean. The in that equation is the potential difference across the light bulb, not the combined potential difference across the batteries. In addition, the potential difference across the light bulb will not remain the same when the circuit is modified since it depends on the current flowing through it. When bulb 2 is removed the overall resistance of the circuit changes so the current will change. To accurately predict the new power output of bulb 1 I have to predict the new current in the circuit:
Now I can predict the power output of bulb 1:
The power output increases dramatically.
9. Two parallel wires are carrying currents as shown in the diagram. The electrons that make up the current move at a very slow average speed, in this case approximately .
a. Draw a B-field line diagram of the magnetic field generated by . b. What is the magnitude and direction of the magnetic force, if any,
that is exerted on one of the moving electrons in the wire on the right. If this force is zero, explain why.
The magnetic field at the position of wire 2 points into the page and has a magnitude of
𝐵 =𝜇s𝐼] 2𝜋𝑟
The magnetic force exerted by the B-field on an electron in wire 2 will be towards wire 1. You can use the right hand rule to figure that out, but remember that the electron is moving in the opposite direction of the current and is negatively charged. The magnitude of this magnetic force is
𝐹a:⃗ 56 V = |𝑞|𝑣𝐵 sin 𝜃
𝐹a:⃗ 56 V = |𝑞|𝑣 x𝜇2𝜋𝑟s𝐼]y sin 𝜃
𝐹a:⃗ 56 V = |−1.6 × 10E]z𝐶|(6.9 × 10E{𝑚 𝑠⁄ )(4𝜋 × 10Ep𝑇 ⋅ 𝑚 𝐴⁄ )(2.00 𝐴)
2𝜋(0.1 𝑚) sin 90° 𝐹a:⃗ 56 V = 4.42 × 10E.z𝑁
That may seem very small, but remember that this is the magnetic force exerted on a single electron.
10. Nuclear fusion is a highly sought-after alternate energy source because it is environmentally friendly and uses very abundant fuels. However, it is a technical challenge to get two atomic nuclei to fuse. One way to get two nuclei to fuse is to strip off their electrons then accelerate the nuclei to high speed and collide them together. Using this idea
(
) (
)
(
) (
)
1 1
2 2
2 2
1
2 2
2
0.200
5.00
0.200
0.200
10.0
0.400
R R
R R
P
I R
A
W
P
I R
A
W
=
=
W
=
=
=
W
=
V
D
1 2 1
1 2 1
1 2
1
0
0
3.00
0.600
5.00
R
V
V
V
IR
V
I
A
R
e e
e e
e e
D
+
D
+
D
=
Þ
+ +
-
=
+
Þ
=
=
=
W
(
) (
)
1 1
2 2
1
0.600
5.00
1.80
R RP
=
I R
=
A
W
=
W
2
I
5
6.9 10
´
-m s
1 I
10.0cm
12.00
it is possible to fuse two deuterium nuclei into helium which results in a release of energy. Deuterium is an isotope of hydrogen, having one proton and one neutron in its nucleus.
a. If the two deuterium nuclei need to get within of each other for them to fuse into helium, determine what speed they must be fired at from far away to get that close.
b. Describe any assumptions you made.
c. Determine the electric field at the halfway point between the two nuclei when they are apart. Show your work/explain your reasoning.
Use energy principles with the system being the two deuterium nuclei. The initial state of the system will be the two nuclei very far apart. The final state of the system will be after the helium nucleus has formed.
I assumed the two deuterium nuclei were far enough away that the electric potential energy between them was approximately zero. I also assumed the two nuclei were point-like. Lastly, I assumed the two nuclei were fired at each other exactly head on.
When the nuclei are apart the electric field at the point halfway between them is zero. Since both nuclei are positively charged the electric field produced by each of them at that point will be in opposite directions, and since the magnitudes of those fields are equal they will exactly cancel.
15
1.3 10
´
-m
9
2 10 m
´
-(
)
(
)
(
(
)(
)(
)
)
, ,
2 2
,
2 2
,
2
9 2 2 19
2
, 27 27 15
6 ,
0
1
2 0 0
2
8.99 10 1.602 10 1.673 10 1.675 10 1.3 10
7.28 10
i f
i E i f E f
D D D i
f
p n D i
f
D i
p n f
D i
E W E
K U K U
q
m v k
r
e
m m v k
r
N m C C
ke v
m m r kg kg m
v m s
-- -
-+ =
Þ + + = +
æ ö
Þ ç ÷+ = +
è ø
Þ + =
´ × ´
Þ = =
+ ´ + ´ ´
Þ = ´
-CONVERSION FACTORS
Length
1 in = 2.54 cm
1 m = 39.4 in = 3.28 ft
1 mi = 5280 ft = 1609 m
1 km = 0.621 mi
1 angstrom (Å) = 10-10 m
1 light year = 9.46 : 1015 m
Volume
1 liter = 1000 cm3
1 gallon = 3.79 liters
Speed
1 mi/h = 1.61 km/h = 0.447 m/s
Mass
1 atomic mass unit (u) = 1.660 : 10-27 kg
(Earth exerts a 2.205 lb force on an object with 1 kg mass)
Force
1 lb = 4.45 N
Work and Energy
1 ft•lb = 1.356 N
#
m = 1.356 J1 cal = 4.180 J
1 eV = 1.60 : 10-19 J
1 kWh = 3.60 : 106 J
Power
1 W = 1 J/s = 0.738 ft
#
lb/s1 hp (U.S.) = 746 W = 550 ft
#
lb/s1 hp (metric) = 750 W
Pressure
1 atm = 1.01 : 105 N/m2= 14.7 lb/in2
= 760 torr
1 Pa = 1 N/m2
PHYSICAL CONSTANTS
Gravitational constant on Earth g 9.81 N/kg
Universal gravitational constant G 6.67 : 10-11 N
#
m2/kg2Mass of Earth 5.97 : 1024 kg
Average radius of Earth 6.38 : 106 m
Density of dry air (STP) 1.3 kg/m3
Density of water (4°C) 1000 kg/m3
Avogadro’s number NA 6.02 : 1023 particles (g atom)
Boltzmann’s constant k 1.38 : 10-23 J/K
Gas constant R 8.3 J/mol
#
KSpeed of sound in air (0°) 340 m/s
Coulomb’s law constant k 9.0 :910 N
#
m2/C2Speed of light c 3.00 : 108 m/s
Fundamental electric charge e 1.60 : 10-19 C
Electron mass me 9.11 : 10-31 kg
= 5.4858 : 10-4 u
Proton mass mp 1.67 : 10-27 kg
= 1.00727 u
Neutron mass mn 1.67 : 10-27 kg
= 1.00866 u
Planck’s constant h 6.63 : 10-34 J
#
sMagnetic
Constant
𝑘
~= 2.00 × 10
Ep
𝑇 𝑚 𝐴
⁄
=
𝜇
s2𝜋
Vacuum