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PHYSICS 222 EXAM 2 REVIEW
SI LEADER: ROSALIE DUBBERKE
TOPICS:
• Kirchhoff’s laws
• Magnetic Fields and Forces
• Magnetic Flux
• LC/LR/LRC circuits
• AC power sources
• Mutual Inductance
WHAT ARE KIRCHHOFF’S LAWS?
• The sum of voltage drops through any loop are equal to zero
• The sum of currents going into any junction must equal the sum of the currents leaving it
• How can we use these?
QUESTION FROM FALL 2014 EXAM
𝑅1 = 5 𝑂ℎ𝑚𝑠
𝑅2 = 10 𝑂ℎ𝑚𝑠
r = 1 Ohm
ε= 9V
Find the voltage drop
across 𝑅2 ?
HINT
• Use the junction rule here:
• Use the Loop rule around the outer loop
• Use the loop rule again on the inner loop
LETS SOLVE IT
• 1) Draw your diagram
• 2) Set up your equations
𝐼1 = 𝐼2 + 𝐼3
𝜀 = 𝑟𝐼1 + 𝐼3 𝑅2
𝜀 = 𝑟𝐼1 + 𝐼3 𝑅2
(3 unknown variables, so 3 equations)
I1
I3
I2
Solve for I1:
LETS SOLVE IT
𝐼3 = 𝐼1 − 𝐼2
𝜀 − 𝐼1 𝑟
𝐼2 =
𝑅1
𝜀− 𝐼 𝑟
 ε = r𝐼1 + 𝑅2 [𝐼1 -( 𝑅 1 )]
1
3) Substitute and solve!!
𝐼1 = 𝐼2 + 𝐼3
𝜀 = 𝑟𝐼1 + 𝐼3 𝑅2
𝜀 = 𝑟𝐼1 + 𝐼3 𝑅2
𝐼1 = 2.076 A
Use 𝐼1 to find 𝐼3 𝑅2 :
9V – (1 Ω)(2.076 A) = 𝐼3 𝑅2
 𝐼3 𝑅2 = 6.9 V
MAGNETIC FIELDS
QUESTION FROM FALL 2014 EXAM
•
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
QUESTION FROM FALL 2013 EXAM
• 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
MAGNETISM AND CURRENT
QUESTION FROM FALL 2013 EXAM
• 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.
QUESTION FROM A PAST EXAM
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.
INDUCED CURRENT AND MUTUAL INDUCTANCE
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? Suppose that the current
described is in the second coil rather than the first. What is the magnitude of the induced emf in the first
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!
MAGNETIC FLUX
QUESTION FROM FALL 2013 EXAM
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
LR, LC, AND LRC CIRCUITS
QUESTION FROM THE 2013 FALL EXAM
• 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.
QUESTION FROM THE 2013 EXAM
• 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
AC POWER SOURCE
SOME HELPFUL EQUATIONS FOR AC POWER SOURCES
(I would write these on
my equation sheet very
clearly so that they are
easy to understand)
QUESTION FROM PAST EXAM
• 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
MORE QUESTIONS FOR PRACTICE
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
QUESTION FROM FALL 2013 EXAM
•
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.
QUESTION FROM A PAST EXAM
• 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
PROBLEM FROM A PAST EXAM
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.
• 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 clockwise. 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
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
• 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
QUESTIONS?