Download Exam 2 Review Slides

PHYSICS 222 EXAM 2 REVIEW
SI LEADER: ROSALIE DUBBERKE
TOPICS:
• Magnetic Fields and Forces
• Magnetic Flux
• LC/LR/LRC circuits
• AC power sources
• Mutual Inductance
•
An electron moving with a velocity = 5.0 × 107 m/s enters a region of space where perpendicular
electric and a magnetic fields are present. The electric field is E = 1j. What magnetic field will allow the
electron to go through the region without being deflected?
LETS SOLVE IT
•
An electron moving with a velocity = 5.0 × 107 m/s enters a region of space where perpendicular
electric and a magnetic fields are present. The electric field is E = 1j. What magnetic field will allow the
electron to go through the region without being deflected?
So what does this mean?
• The electric field will cause force on the electron (F = qE)
• The magnetic field will cause another force on the electron (F = qv x B)
• If the electron is to keep the same velocity without being deflected by wither field, these forces need to
balance each other.
LETS SOLVE IT
•
An electron moving with a velocity = 5.0 × 10^7 m/s enters a region of space where perpendicular
electric and a magnetic fields are present. The electric field is E = 1j. What magnetic field will allow the
electron to go through the region without being deflected?
• F = qE = qv x B
-1.6 x 10−19 N j = 8 x 10−12 i x B
-1.6 x 10−19 N j = (0 – 0)i + (0 – (8 x 10−12 )(𝐵𝑧 ))j + ((8 x 10−12) 𝐵𝑦 – 0)k
-1.6 x 10−19 N j = - (8 x 10−12 )(𝐵𝑧 )j
B = 2 x 10−8 T k
• Three particles travel through a region of space where the magnetic field is out of the page, as shown in
the figure. The electric charge of each of the three particles is, respectively,
LETS SOLVE IT
• F = qv x B
• Lets define the path particle 2 follows as positive x, and the magnetic field in positive z.
• F = (𝑞𝑣𝑦 𝐵𝑧 – 𝑞𝑣𝑧 𝐵𝑦 )x + (𝑞𝑣𝑧 𝐵𝑥 – 𝑞𝑣𝑥 𝐵𝑧 )y + (𝑞𝑣𝑥 𝐵𝑦 – 𝑞𝑣𝑦 𝐵𝑥 )z
• F = (– 𝑞𝑣𝑥 𝐵𝑧 )y
• From this we can see that if Q is positive, force is negative in the y direction, pushing the particle down. If Q is
negative, force is positive, pushing the particle up in the y direction. If q is 0, force is zero.
• So, Q1 is negative
• Q2 is neutral
• Q3 is positive
• The figure shows a velocity selector that can be used to measure the speed of a charged particle. A
beam of particles is directed along the axis of the instrument. A parallel plate capacitor sets up an
electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated
by 2.0 mm and the value of the magnetic field is 0.60 T, what voltage between the plates will allow
particles of speed 5.0 × 105 m/s to pass straight through without deflection?
LETS SOLVE IT
• We want to determine what voltage will make it so the force from the magnetic field is balanced by the
force of the electric field.
• Thus, 𝐹𝐸 =
𝑞𝑉
𝑑
= 𝐹𝑚 = 𝑞𝑣𝐵 →
𝑞𝑉
𝑑
= 𝑞𝑣𝐵 → 𝑉 = 𝑣𝐵𝑑 = 5 × 105
𝑚
𝑠
.6 𝑇 .002𝑚 = 600 𝑉
When an external magnetic field is applied along a positive z – direction to a diamagnetic material, the
magnetic moment induced inside the material
• A) points in a positive (+z) – direction
• B) points in a negative (–z) – direction
• C) iz zero
• D) points in a positive (+x) - direction
• E) is somewhere in the xy-plane (perpendicular to the z-direction)
LETS SOLVE IT
When an external magnetic field is applied along a positive z – direction to a diamagnetic material, the
magnetic moment induced inside the material
• A) points in a positive (+z) – direction
• B) points in a negative (–z) – direction
• C) iz zero
• D) points in a positive (+x) - direction
• E) is somewhere in the xy-plane (perpendicular to the z-direction)
By definition, magnetic moment in the diamagnetic material opposes the applied field, so it will
be in a negative (-z) – direction
A capacitor is charging in a simple RC circuit with a dc battery. Which one of the following statements about this
capacitor is accurate?
• A) The magnetic field between the capacitor plates is increasing with time because the charge on the plates is
increasing.
• B) There is a magnetic field between the capacitor plates because charge travels between the plates by
jumping from one plate to the other.
• C) There is no magnetic field between the capacitor plates because no charge travels between the plates.
• D) There is a magnetic field between the capacitor plates, even though no charge travels between them,
because the magnetic flux between the plates is changing.
• E) There is a magnetic field between the capacitor plates, even though no charge travels between them,
because the electric flux between the plates is changing.
LETS SOLVE IT!
• E is the correct answer!
• Why?
As the charge on the capacitor is changing, the electric field between the two plates is changing, leading to
a changing electric flux, which generates a magnetic field. However, it is important to realize that there is
no actual charge moving between the plates.
• The three loops of wire shown in the figure
are all subject to the same uniform
magnetic field that does not vary with time.
Loop 1 oscillates back and forth as the bob
in a pendulum, loop 2 rotates about a
vertical axis, and loop 3 oscillates up and
down at the end of a spring. Which loop, or
loops, will have an emf induced in them?
HINT:
𝑑∅
𝜀 = −𝑁
𝑑𝑡
LETS SOLVE IT
𝑑∅
• Using 𝜀 = −𝑁 𝑑𝑡 , we can see that if there is no change in flux, there is no induced emf (or current)
Loops 1 and 3 are moving, yes, but the don’t have any change in flux because the magnetic field isn’t
changing and neither is the perpendicular area of the loop, so loop 2 is the only possible answer.
• Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter
of 22.0 cm and carries a clockwise current of 16.0 A, as viewed from above, and the outer wire has a
diameter of 36.0 cm.
• What must be the direction (as viewed from above) of the current in the outer wire so that the net
magnetic field due to this combination of wires is zero at the common center of the wires?
LETS SOLVE IT
• Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter
of 22.0 cm and carries a clockwise current of 16.0 A, as viewed from above, and the outer wire has a
diameter of 36.0 cm.
• What must be the direction (as viewed from above) of the current in the outer wire so that the net
magnetic field due to this combination of wires is zero at the common center of the wires?
It should be counterclockwise. The first coils creates a magnetic field vector going into the table, so the second
wire needs to create a magnetic field vector coming out of the table to balance it. This means it should have a
counterclockwise current
•
The figure shows four different sets of insulated wires that cross each other at right angles without
actually making electrical contact. The magnitude of the current is the same in all the wires, and the
directions of current flow are as indicated. For which (if any) configurations will the magnetic field at
the center of the square formed by the wires be equal to zero?
LETS SOLVE IT
We know the current through each of the
wires is the same, which means they are
creating magnetic fields (following the
right hand rule) of equal magnitude.
Next, we need to balance the directions
C is the only possible correct answer,
because it is the only one where it has 2
magnetic field vectors going “into” it and 2
coming “out” of it.
• Two long parallel wires placed side-by-side on a horizontal table carry identical size currents in opposite
directions. The wire on your right carries current toward you, and the wire on your left carries current
away from you. From your point of view, the magnetic field at the point exactly midway between the
two wires
LETS SOLVE IT
• Two long parallel wires placed side-by-side on a horizontal table carry identical size currents in opposite
directions. The wire on your right carries current toward you, and the wire on your left carries current
away from you. From your point of view, the magnetic field at the point exactly midway between the
two wires
• We can use the right hand rule to determine that the magnetic field should be into the table.
A circular coil lies flat on a horizontal table. A bar magnet is held above its center with its North Pole
pointing down. The magnet is fixed and does not move. What kind of current does it induce in what
direction?
LETS SOLVE IT
A circular coil lies flat on a horizontal table. A bar magnet is held above its center with its North Pole
pointing down. The magnet is fixed and does not move. What kind of current does it induce in what
direction?
It induces no current! The magnet isn’t moving, the coil isn’t moving, so there is no change in flux,
meaning no current.
An L-shaped metal machine part is made of two equal-length segments that are perpendicular to each
other and carry a 4.50-A current as shown in the figure. This part has a total mass of 3.80 kg and a total
length of 3.00 m, and it is in an external 1.20-T magnetic field that is oriented perpendicular to the plane of
the part, as shown. What is the magnitude of the NET magnetic force that the field exerts on the part?
LETS SOLVE IT
𝐿
• Force on a length of wire is 𝐹 = 𝐼𝐿𝐵 for this problem, 𝐵 = 1.2 𝑇, 𝐼 = 4.5 𝐴, 𝐼 = 2 = 1.5𝑚
• We need to consider the two sections (top and left) separately, since their forces are in different
directions. Using the right hand rule, we can determine that the force on the top wire is down and the
force on the left wire is to the right.
• We calculate the magnitudes: 𝐹 = 𝐼𝐿𝐵 = 4.5 𝐴 1.5 m 1.2 T = 8.1 N
• Using Pythagoreans theorem, we can find the net flux magnitude: 𝐹𝑛𝑒𝑡 = 8.12 + 8.12 = 11.5
Two coils have mutual inductance of M= 3.25×10−4 H. The current 𝐼1 in the first coil increases at a uniform
rate of 860 A/s. What is the magnitude of the induced emf in the second coil?
LETS SOLVE IT
What about if we switch the current to the other coil?
𝑑𝑖
It makes no difference! They have the same value of M and 𝑑𝑡, so they have the same emf!
A 2.0-m long conducting wire is formed into a square and placed in the horizontal xy plane. A uniform
magnetic field is oriented 30 degrees above the horizontal with a strength of 9 T. What is the magnetic flux
through the square?
LETS SOLVE IT
A 2.0-m long conducting wire is formed into a square and placed in the horizontal xy plane. A uniform
magnetic field is oriented 30 degrees above the horizontal with a strength of 9 T. What is the magnetic flux
through the square?
Φ = Bacosθ
= (9T)(.25 𝑚2 )cos(90-30)
= 1.1 T𝑚2
30 °
30 °
• A series LR circuit consists of a 2.0-H inductor with negligible internal resistance, a 100-ohm resistor, an
open switch, and a 9.0-V ideal power source. After the switch is closed, what is the maximum power
delivered by the power supply?
HINT:
• How will the inductor affect the power, knowing we want the maximum power possible?
LETS SOLVE IT
𝑉2
𝑅
81
• 𝑃 = 𝐼𝑉 =
= 100 𝑊 = .81𝑊. Since the question asks about the maximum power, the current will be
maximal (and constant) and the inductor will just act like a piece of wire.
• There is an LC circuit with inductor L = 1 H, capacitor C = 1 F and an open switch. Initially capacitor is
fully charged. Now the switch is closed. How many times the voltage on the capacitor will be zero
during 341 seconds?
LETS SOLVE IT
• In a series LRC circuit, the frequency at which the circuit is at resonance is 𝑓0 . If you double the
resistance, the inductance, the capacitance, and the voltage amplitude of the ac source, what is the
new resonance frequency?
HINT
• At resonance, 𝜔 =
1
𝐿𝐶
LETS SOLVE IT
• The circuit is at resonance when 𝜔 =
1
𝐿𝐶

1
1

2𝐿2𝐶
2 𝐿𝐶
=
𝜔
2
SOME HELPFUL EQUATIONS FOR AC POWER SOURCES
(I would write these on
my equation sheet very
clearly so that they are
easy to understand)
• The 60-Hz ac source of a series circuit has a voltage amplitude of 120 V. The resistance, capacitive
reactance, and inductive reactance are as shown in the figure. What is the rms current in the circuit?
• (rms current is the root mean square current, so it’s 𝐼𝑟𝑚𝑠 =
𝐼𝑚𝑎𝑥
)
2
LETS SOLVE IT
• When an LRC series circuit is at resonance, which of the following statements about the circuit is
correct? (there may be more than one answer)
•
The reactance of the capacitor is zero
•
The reactance of the inductor is zero
•
The reactance due to the inductor and the capacitor is at its maximum value
•
The impedance is at its maximum value
•
The current amplitude is a maximum
LETS SOLVE IT
• When an LRC series circuit is at resonance, which of the following statements about the circuit is correct?
(There may be more than one)
•
The reactance of the capacitor is zero
•
The reactance of the inductor is zero
•
The reactance due to the inductor and the capacitor is at its maximum value
•
The impedance is at its maximum value
•
The current amplitude is a maximum
• E is the only correct answer.
• At resonance, 𝑋𝐿 = 𝑋𝐶 . This means that impedance is at a minimum, causing current to be at a maximum.
• Which of the following phasor diagrams shown below best represents a series LRC circuit at resonance?
LETS SOLVE IT
• Which of the following phasor diagrams shown below best represents a series LRC circuit at resonance?
In an LRC circuit, 𝑉𝐿 leads 𝑉𝑅 by 90 degrees, and 𝑉𝐶 by 180 degrees. Further, at resonance 𝑋𝐿 = 𝑋𝐶 so C
must be the correct choice.
When the current in a toroidal solenoid is changing at a rate of 0.03 A/s, the magnitude of the induced emf is 15
mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 2.8 mWb.
How many turns does the solenoid have?
LETS SOLVE IT
When the current in a toroidal solenoid is
changing at a rate of 0.03 A/s, the magnitude of
the induced emf is 15 mV. When the current
equals 1.40 A, the average flux through each turn
of the solenoid is 2.8 mWb.
How many turns does the solenoid have?
ε = 15 mV
𝑑𝑖
= .03 A/s
𝑑𝑡
I = 1.4 A
Φ = 2.5 mWb
∅
Φ = IA  A = 𝐼
dФ
𝑑𝑖
nΦ
ε = N dt  NA 𝑑𝑡  𝑑𝑖
𝐼
𝑑𝑡
Plug it in and solve!
250 turns
• A 4.0 mH coil carries a current of 5.0 A. How much magnetic field energy is stored in the coil's magnetic
field?
LETS SOLVE IT
• A 4.0 mH coil carries a current of 5.0 A. How much magnetic field energy is stored in the coil's magnetic
field?
• 𝑈=
1
2𝐿𝐼 2
= 50 mJ
QUESTIONS?