Exam 1 solutions Physics 194 Name:
Instructions:
• The responsibility is yours to show me that you understand what you are doing so please show all your work. If you feel you are stuck on a particular part explain as much of your thinking as you can so I can give you the maximum amount of partial credit (explain the strategy you are using, draw pictures/graphs, etc.)
• Use the backs of the exam pages if you need extra space.
• Each problem is worth 10 points. The problems are not all of the same difficulty though, so budget your time wisely.
• The last pages of the exam contain useful physical constants, conversion factors, and other data. Please check there first if you feel you need more information than is given in the problem.
Good luck!
1. Two speakers (resistance 10.0 Ω each) wired in series are connected to a power supply (EMF 12.0 𝑉).
a. Determine the power output of each speaker.
b. One of the speakers is removed and wires reconnected so that only a single speaker remains in the circuit with the power supply. Determine the power output of the remaining speaker.
c. If your answers for a. and b. are the same, explain why that makes sense. If they are different, explain why that makes sense.
d. Describe how your answers to parts a. through c. would change if the speakers were originally wired in parallel.
Since the speakers are in series the potential difference each of them is equal to half the EMF of the power supply. The power output of one of them is then:
With only one speaker in the circuit, the potential difference across the speaker equals the EMF of the power supply. Its power output is then:
This is 4 times greater than when there are two speakers in the circuit. This makes sense since each electron that passes through the single speaker deposits all of its electric potential energy there, rather than just half of it. In addition, the equivalent resistance of the circuit is half as much when only one speaker is present. As a result, the current through the circuit is twice as
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great. So, not only do the electrons individually deposit all of their electric potential energy in the speaker, there are more of them passing through the speaker each second.
If the two speakers had been in parallel, the power output of each would have been 14.4 W since the potential difference across each would be 12.0 V. Removing one of them does not change the potential difference across the one that remains.
2. Two glass spheres are mounted on a table and are 20 cm apart. The left sphere has a charge of 1.0 μC and the right sphere has a charge of -2.0 μC. You place a cart with low-friction metal wheels halfway between the spheres and hold it at rest with your hand. The cart’s mass is 2.5 kg and its charge stays constant at +5.0 μC.
a. In physics we use visual representations (diagrams, sketches, graphs, etc.) to help us understand situations. While you are answering the following parts, include whatever representations are most useful.
b. Determine the magnitude and direction of the total E-field produced by the spheres at the location of the cart.
c. You let go of the cart. Determine the magnitude and direction of the net force exerted on the cart the instant after.
d. Decide if the following statement is true or not true. Explain your reasoning.
The cart moves with constant acceleration once you let it go.
I’ll draw a sketch of the situation, and a force diagram for the cart. I’m assuming the static friction force exerted by the table on the cart is zero.
I’ll use the principle of superposition to add the E-field contributions produced by each sphere to get the total E-field at the location of the cart. Both E-field contributions point to the right, so both are positive, so the total E-field will point to the right as well.
𝐸 = 𝐸*+ 𝐸,= -+𝑘|𝑞*|
𝑟*2 3 + -+𝑘
|𝑞,|
𝑟,2 3 = 𝑘
-|𝑞*|
𝑟*2 +
|𝑞,|
𝑟,2 3
= -9.0 × 106𝑁 ∙ 𝑚2
𝐶2 3
-|+1.0 × 10;<𝐶|
(0.10 𝑚)2 +
|−2.0 × 10;<𝐶|
To determine the net force exerted on the cart when it is released I just need to multiply the E-field at that location by the cart’s charge (x-component here since the force diagram is balanced vertically).
𝐹C DDD⃗FG H,J = 𝑞H𝐸J
= (+5.0 × 10;<𝐶)(+2.7 × 10<𝑁 𝐶⁄ )
= +13.5 𝑁
Since this ended up positive, the net force exerted on the cart is to the right.
For part d., the statement is not true. The E-field produced by the two spheres is not uniform, so the net force exerted on the cart will not be constant, which means the acceleration of the cart will not be constant.
3. Give detailed explanations for why the following happens. Draw charge diagrams to illustrate your explanations.
a. A metal spoon is placed on a Styrofoam plate. Several tiny bits of aluminum foil are placed into the bowl of the spoon. A glass rod is rubbed with silk making the rod
positively charged. Everything else is initially neutral. The rod is then touched to the end of the handle of the spoon. The instant this happens, the aluminum foil bits fly out of the spoon.
b. Another metal spoon is placed on another Styrofoam plate. Several new bits of aluminum foil are placed into the bowl of this second spoon. The glass rod is rubbed with silk again (to ensure the rod is still positively charged). Again, everything else is initially neutral. This time, the rod is brought towards the aluminum foil bits from directly above. Once the rod gets near the foil bits, the bits jump up, touch the rod, then fall back into the bowl of the spoon.
When the positive rod touches the handle of the spoon electrons are pulled out of the spoon (and therefore also from the aluminum foil bits) and into the rod. Since the
When the rod is brought above the bits the bits polarize and are attracted to the rod. Once they touch the rod they lose some electrons to the rod and become positive. Because of this they are now repelled by the rod, and
actually slightly attracted to the spoon since the spoon gets polarized by the bits as the bits fall downward.
4. A region of uniform 𝐵D⃗-field points into the page and has magnitude 5.0 𝑇. A proton is traveling to the left with speed 3.0 × 10<𝑚 𝑠⁄ as it enters this region. See the last page of
the exam for physical constants you might need.
a. Your goal is to add an 𝐸D⃗-field to this region so that the proton continues to move to the left with constant speed. Draw a labeled diagram showing both fields, the proton, and its direction of motion.
b. Determine the magnitude of the 𝐸D⃗-field that will accomplish this.
c. After the proton has traveled part of the way through the region the 𝐸D⃗-field fails leaving only the original 𝐵D⃗-field. Draw a labeled diagram showing the 𝐵D⃗-field, the proton, and the path it will now take. Determine the value(s) of any relevant physical quantity(ies) that describe this path.
d. Describe how your answers to parts a. through c. would change if the moving particle were an electron instead of a proton.
The B-field exerts a downward force on the proton (right hand rule) which means the E-field must point upward so that it exerts an upward force on the proton. If the E-field has the following magnitude the net force exerted on the proton will be zero and it will move to the left with constant speed.
P 𝐹FG Q,R = 0
𝐹SD⃗ FG Q,R + 𝐹CD⃗ FG Q,R= 0 −|𝑞|𝑣𝐵 sin 𝜃 + |𝑞|𝐸 = 0 (𝜃 = 90° so sin 𝜃 = 1)
𝐸 = 𝑣𝐵 = (3.0 × 10<𝑚 𝑠⁄ )(5.0 𝑇) = 1.5 × 10Z𝑁 𝐶⁄
When the E-field fails the proton will then move with uniform circular motion with radius 𝑟 = 𝑚𝑣
|𝑞|𝐵=
(1.67 × 10;2Z𝑘𝑔)(3.0 × 10<𝑚 𝑠⁄ )
If it were an electron instead of a proton the magnetic force exerted on it would be upwards and the electric force exerted on it would be downwards. Even though the electron is
negatively charged the same configuration of E and B-fields will cause it to travel with constant velocity. When the E-field fails the electron will move with uniform circular motion but it will be an upwards oriented circle with much smaller radius. 𝑟 = 𝑚𝑣
|𝑞|𝐵=
(9.11 × 10;^]𝑘𝑔)(3.0 × 10<𝑚 𝑠⁄ )
(1.60 × 10;]6𝐶)(5.0 𝑇)
= 3.4 × 10;<𝑚
5. You discovered in lab this semester that it is possible to use the magnetic field to induce an electric current in a circuit without needing a power supply. You have the following
equipment available to design an experiment to demonstrate this phenomenon: • A circular coil of wire. The coil has 50 windings and a radius of 5 cm. It has a
resistance of 0.50 Ω.
• An electromagnet. It acts like a bar magnet except that it can be turned on and off. It creates a magnetic field near its poles with magnitude 3.00 𝑇. It takes 0.50 𝑠 for the electromagnet to fully turn on, and 0.50 𝑠 to shut down.
• An ammeter to measure current.
a. Describe your experiment. Include a labeled diagram. Make sure your diagram includes any relevant magnetic fields and the direction of the induced current.
b. Determine the value of the induced current and the times during which it will be present.
I’ll take the electromagnet, place it close to the coil, and point its north pole at the center of the coil. I’ll connect the coil to the ammeter so I can measure the current that will be induced in it. The
electromagnet will be off to begin with. I’ll then turn the electromagnet on. This will cause a current to be induced in the coil. The direction of the induced current will be in whatever direction is needed so that the magnetic field it creates will attempt to restore the magnetic flux through the coil to its value the moment before. The direction of the induced current and magnetic field produced by it are shown in the diagram.
The coil is going to act like a circuit with a single power supply and a single resistor. I can determine the current flowing in the circuit using Kirchhoff’s loop rule and Ohm’s law. This current will only be
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present during the 0.50 s that the magnetic field of the electromagnet is changing.
The electromotive force (EMF) produced by the electromagnetic induction effect can be determined using Faraday’s law.
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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/kg2 Mass of Earth 5.97 : 1024 kgAverage 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
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KSpeed of sound in air (0°) 340 m/s
Coulomb’s law constant k 9.0 :910 N
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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
;Z
𝑇 𝑚 𝐴
⁄
=
𝜇
c2𝜋
Vacuum