Physics C Electric Fields Name:_________________
AP Review
Charge Explain your reasoning
Net – charge: excess electrons
Net + charge: excess “holes”
SI unit: coulomb (C)
Quantum of charge:
the proton charge (e)
the electron charge (-e)
e = 6.02 10-19Coulombs
Conservation of Charge
The net charge in any process or reaction 40. If F is the magnitude of the force on the test
remains unchanged. charge due to only one of the other charges,
what is the magnitude of the net force acting on
Coulomb’s Law the test charge due to both of these charges?
Calculates magnitude of force between charges F
Force is repulsive if charges have the same sign. (A) Zero (B) (C) F
Force is attractive if charges have opposite signs.
2
(D) 2F (E) 2
F = kq1q2/r2
F: force (N) Show your work
k: constant 9.0 109 N m2/C2
q1, q2: charges (C)
r: distance between charge centers (m)
NOTE:
k = 1/(4o)
where o = 8.85 10-12 C2/ N m2
Multiple Forces on a Charges
Ftot = Fi
Problem: Electric Force (1998)
Problems: Electric Force (1998)
55. Suppose that an electron (charge -e) could
Questions 39-40 orbit a proton (charge +e) in a circular orbit of
constant radius R. Assuming that the proton is
stationary and only electrostatic forces act on the
particles, which of the following represents the
kinetic energy of the two-particle system?
1 e 1 e2
(A) (B)
As shown above, two particles, each of charge 4 0 R 8 0 R
+Q, are fixed at opposite corners of a square
that lies in the plane of the page. A positive 1 e2 1 e2
(C) (D) (E)
test charge +q is placed at a third corner. 8 0 R 4 0 R2
39. What is the direction of the force on the test 1 e2
charge due to the two other charges?
4 0 R 2
(A) (B)
Show your work
(C) (D)
(E)
11/17/2011 1
Problem: Electric Force (1988) touch them. With the rod held in place,
conductor 2 is moved to the right by pushing its
36 Two small spheres have equal charges q and stand, so that the conductors are separated.
are separated by a distance d. The force Which of the following is now true of
exerted on each sphere by the other has conductor 2 ?
magnitude F. If the charge on each sphere is (A) It is uncharged.
doubled and d is halved, the force on each (B) It is positively charged.
sphere has magnitude (C) It is negatively charged.
(A) F (B) 2F (C) 4F (D) 8F (E) 16F (D) It is charged, but its sign cannot be predicted.
(E) It is at the same potential that it was before the
Show your work charged rod was brought near.
Explain your reasoning
Electric Field
Exists in space due to the presence of charge.
Predicts what will happen to a charged particle Problem: Electric Field (1993)
put in that location in space.
Field is directed outward from the positive
charges creating it.
Field is directed inward toward the negative
charge creating it. 55. Two metal spheres that are initially uncharged
This equation calculates magnitude of field at point are mounted on insulating stands, as shown above.
in space near one spherically symmetric charge: A negatively charged rubber rod is brought close
to, but does not make contact with, sphere X.
E = kq/r2 (magnitude calculation) Sphere Y is then brought close to X on the side
E: field (N/C) opposite to the rubber rod. Y is allowed to touch X
k: constant 9.0 109 N m2/C2 and then is removed some distance away. The
q: charge (C) rubber rod is then moved far away from X and Y.
r: distance between center of charge and point What are the final charges on the spheres?
in space (m) Sphere X Sphere Y
A) Zero Zero
The equation above works only for spherically B) Negative Negative
symmetric charge distributions. C) Negative Positive
D) Positive Negative
NOTE: E) Positive Positive
Positive charges experience a force in the same
direction as the electric field is pointing. Explain your reasoning
Negative charges experience a force in the
opposite direction as the electric field.
Problem: Electric Field (1998)
38. Two initially uncharged conductors, 1 and 2,
are mounted on insulating stands and are in
contact, as shown above. A negatively
charged rod is brought near but does not
11/17/2011 2
Problem: Electric Field (1993) Calculating Force from Field
F = qE
36. From the electric field vector at a point, one
can determine which of the following? Problem: Force from Field (1988)
I. The direction of the electrostatic force
on a test charge of known sign at that point 38 A charged particle traveling with a velocity v in
II. The magnitude of the electrostatic an electric field E experiences a force F that must
force exerted per unit charge on a test be
charge at that point (A) parallel to v (B) perpendicular to v
III. The electrostatic charge at that point (C) parallel to v x E (D) parallel to E
A) I only B) III only C) I and II only (E) perpendicular to E
D) II and III only E) I, II, and III
Show your work
Explain your reasoning
Principle of Superposition
Problem: Electric Field (1993) Etot = Ei
The electric field at a given point in space is the
vector sum of the electric fields due to all
charges affecting that point in space.
Note: Electric field lines are NOT VECTORS,
but may be used to derive the direction of
37. A circular ring made of an insulating electric field vectors at given points. The
material is cut in half. One half is given a electric field vector is tangent to the field line
charge -q uniformly distributed along its arc. at any point in space.
The other half is given a charge + q also
uniformly distributed along its arc. The two Problem: Superposition (1998)
halves are then rejoined with insulation at the
junctions J, as shown above. If there is no Questions 45-46 refer to two charges located on the
change in the charge distributions, what is the line shown in the figure below, in which the charge at
direction of the net electrostatic force on an point I is +3q and the charge at point III is +2q. Point
electron located at the center of the circle? II is halfway between points I and III.
A) Toward the top of the page
B) Toward the bottom of the page
C) To the right
D) To the left 45. Other than at infinity, the electric field strength
E) Into the page. is zero at a point on the line in which of the
following ranges?
Explain your reasoning (A) To the left of I B) Between I and II
(C) Between II and III D) To the right of III
(E) None; the field is zero only at infinity.
Explain your reasoning
11/17/2011 3
Problem: Superposition (1998) 47. In which configuration is the electric field at P
equal to zero? (A,B,C,D,E)
Explain your reasoning
61. Two charged particles, each with a charge of
+q, are located along the x-axis at x = 2 and x = 4,
as shown above. Which of the following shows
the graph of the magnitude of the electric field
along the x-axis from the origin to x = 6 ?
48. In which configuration is the electric field at P
pointed at the midpoint between two of the
charges? (A,B,C,D,E)
Explain your reasoning
Problem: Electric Field (1984)
Explain your reasoning
42. The figure above shows a spherical distribution of
charge of radius R and constant charge density .
Which of the following graphs best represents the
electric field strength E as a function of the distance r
from the center of the sphere?
Problems: Superposition (1988)
Questions 47-48 relate to the following
configurations of electric charges located at the Explain your reasoning
vertices of an equilateral triangle. Point P is
equidistant from the charges.
11/17/2011 4
Problem: Electric Field (1988) = Q/V = dQ/dV
Questions 66-67 For continuous charge distributions
E = E = dE
Positive charge Q is uniformly distributed over a Electric Flux through Surface
thin ring of radius a that lies in a plane Dot product of area and a vector field.
perpendicular to the x-axis. with its center at the Can be visualized by counting field lines crossing a
origin 0, as shown above. surface.
Flux thru a surface of area A
67. Which of the following graphs best E = E • A
represents the electric field along the E: electric flux (N m2/C)
positive x-axis? E: electric field vector (N/C)
A: area (m2)
Electric Flux through Closed Surface
E = E • dA
Gaussian surface
The surface of a closed 3-dimensional shape in
Explain your reasoning space.
Net flux over a Gaussian surface is zero if surface
contains no charge.
Net flux is positive if the Gaussian surface contains
+ charge.
Net flux is negative if the Gaussian surface contains
- charge.
A Gaussian surface can be any arbitrary shape.
Gaussian surfaces are typically chosen for
convenience and high symmetry with regard the
Problem: Electric Field (1993) electric distribution they might enclose.
48. A conducting sphere of radius R carries a Gauss’ Law of Electricity
charge Q. Another conducting sphere has a q = oE
radius R/2, but carries the same charge. The q: net charge (C)
spheres are far apart. The ratio of the o: electrical permittivity of free space
electric field near the surface of the smaller 8.85 10-12F/m
sphere to the field near the surface of the 8.85 10-12C2/Nm2
larger sphere is most nearly E: electrical flux
A) 1/4 B) 1/2 C) 1 D) 2 E) 4 The net charge contained within a closed “Gaussian
surface” is directly related to the electrical flux
Show your work through that surface.
Gauss’ Law (expanded)
q = o E dA
Note that Gauss’ Law works best for those
situations where the electric field is constant or
zero over the Gaussian surface
Charge density or distribution
Linear ()
= Q/L = dQ/dL
Surface ()
= Q/A = dQ/dA
Volume ()
11/17/2011 5
Problem: Gauss’ Law (1998) Problem: Gauss’ Law (1993)
41. Gauss's law provides a convenient way to Questions 51-52
calculate the electric field outside and near
each of the following isolated charged
conductors EXCEPT a
(A) large plate
(B) sphere Two concentric, spherical conducting shells have
(C) cube radii r1 and r2 and charges Q1 and Q2, as shown
(D) long, solid rod above. Let r be the distance from the center of the
(E) long, hollow cylinder spheres and consider the region r1 < r < r2.
Expain your reasoning 51. In this region the electric field is proportional to
A) Q1/r2 B) (Q1 + Q2)/r2
C) (Q1 + Q2)/r D) Q1/r1 + Q2/r
E) Q1/r + Q2/r2
Show your work
Problem: Gauss’ Law (1993)
38. The net electric flux through a closed
surface is
A) infinite only if there are no charges enclosed by the surface
B) infinite only if the net charge enclosed by the surface is zero
C) zero if only negative charges are enclosed by the surface Problem: Gauss’ Law (1984)
D) zero if only positive charges are enclosed by the surface
E) zero if the net charge enclosed by the surface is zero
Explain your reasoning
40. A closed surface, in the shape of a cube of side
a, is oriented as shown above in a region where
there is a constant electric field of magnitude E
parallel to the x-axis. The total electric flux
through the cubical surface is
(A) -Ea2 (B) zero (C) Ea2 (D) 2Ea2 (E) 6Ea2
Show your work
11/17/2011 6
Electrical Potential Energy A charge changes its energy when it changes its
Work done against an electrostatic force to potential
assemble a system of charges from an infinite UE = Wext = qV
separation. or
For two charges K = qV
UE = k q1 q2/r
UE: electrical potential energy (J) Problem: Work and Potential Change (1998)
k: 1/(4o )
q1, q2: charges (C) 48. The work that must be done by an external
r: separation of charges (m) agent to move a point charge of 2 mC from the
For more than two charges origin to a point 3 m away is 5 J. What is the
UE = UE,i potential difference between the two points?
(A) 4 x 10-4 V (B) 10-2 V (C) 2.5 x 103 V
6 6
Electrical Potential (D) 2 x 10 V (E) 6 x 10 V
Exists around any charge or charge
configuration. Show your work:
Describes what electrical potential energy
another charge would have when placed in a
certain location in the electric field due to a
charge configuration.
For spherical charge distributions
V = kq / r
V: potential (volts)
k: 1/(4o )
q: charge (C)
r: distance from charge (m) Uniform or Constant Electric Fields
V = - Ed
Problem: Potential (1998) V: change in potential
E: Electric field magnitude
Questions 45-46 refer to two charges located on d: change in position
the line shown in the figure below, in which
the charge at point I is +3q and the charge at Charge configurations
point III is +2q. Point II is halfway between Discrete point or spherical charges
points I and III. V = Vi
Continuous charge distribution
V = dV
V = -E•dx
46. The electric potential is negative at some Going from potential to field
points on the line in which of the Ex = -V/x
following ranges? Ey = -V/y
(A) To the left of I Ez = -V/z
(B) Between I and II
(C) Between II and III Explain your reasoning:
(D) To the right of III
(E) None; the potential is never negative
Show your work:
Electrical Potential and Energy
11/17/2011 7
Problem: Potential and Work (1993)
56. The potential of an isolated conducting
sphere of radius R is given as a function of the
charge q on the sphere by the equation V = kq/R.
If the sphere is initially uncharged, the work W
required to gradually increase the total charge on
the sphere from zero to Q is given by which of 47. The graph above shows the electric potential V
the following expressions? in a region of space as a function of position along
A) W = kQ/R the x-axis. At which point would a charged particle
B) W = kQ2/R experience the force of greatest magnitude?
Q (A) A (B) B (C) C (D) D (E) E
C) W = 0
( kq / R ) dq
Q Explain your reasoning:
D) W = ( kq 2 / R)dq
0
Q
E) W = ( kq / R 2 )dq
0
Show your work
Problem: Field and Potential (1993)
41. Four positive charges of magnitude q are
arranged at the corners of a square, as shown above.
Problem: Field from Potential (1998) At the center C of the square, the potential due to
one charge alone is Vo and the electric field due to
Questions 59-60 one charge alone has magnitude Eo. Which of the
In a region of space, a spherically symmetric following correctly gives the electric potential and
electric potential is given as a function of r, the magnitude of the electric field at the center of
the distance from the origin, by the equation the square due to all four charges?
V(r) = kr2, where k is a positive constant.
Electric Potential Electric Field
59. What is the magnitude of the electric A) Zero Zero
field at a point a distance r0 from the origin? B) Zero 2Eo
(A) Zero (B) kr0 (C) 2kr0 (D) C) 2 Vo 4Eo
kr02 (E) 2kr03/3 D) 4 Vo Zero
E) 4 Vo 2Eo
Show your work:
Explain your reasoning:
Problem: Potential to Field to Force (1998)
11/17/2011 8
Equipotential Surface Problem: Equipotential Surface (1998)
A line connecting points of equal electrical
potential. 62. A positive electric charge is moved at a
A charge does not change its energy when it constant speed between two locations in an electric
moves along an equipotential surface. field, with no work done by or against the field at
any time during the motion. This situation can occur
Problem: Equipotential Surfaces (1993) only if the
(A) charge is moved in the direction of the field
Questions 53-54 (B) charge is moved opposite to the direction of the
field
(C) charge is moved perpendicular to an equipotential
line
(D) charge is moved along an equipotential line
(E) electric field is uniform
A battery or batteries connected to two parallel
plates produce the equipotential lines Explain your reasoning:
between the plates shown above.
53 Which of the following configurations is
most likely to produce these equipotential lines?
Explain your reasoning:
Isolated conductors
Charge resides entirely on surface.
Charges spread out as far as possible.
Electric field in interior is zero
Electric potential in interior is constant
There is no effective difference between an isolated
hollow and and isolated solid charged conductor
54. The force on an electron located on the
0-volt potential line is Capacitor
A) 0 N Consists of two “plates” in close proximity.
B) I N, directed to the right When “charged”, there is a voltage across the plates,
C) I N, directed to the left and they bear equal and opposite charges.
D) directed to the right, but its magnitude Stores electrical energy.
cannot be determined without knowing the
distance between the lines Capacitance
E) directed to the left, but its magnitude cannot C=q/V
be determined without knowing the distance C: capacitance (Farads)
between the lines q: charge on positive plate (Coulombs)
V: potential difference between plates in Volts
Explain your reasoning:
Capacitors in Circuits
Circuit drawing
11/17/2011 9
A capacitor acts like a zero-resistance short Equivalent capacitance
circuit right after a switch is closed. After the If you have capacitors in series, you add the
capacitor is charged, it acts like an open reciprocal of the capacitances and then take the
circuit. reciprocal of the result
Problem: Capacitors in Circuit (1993) 1/Ceq = Ci)
If you have capacitors in parallel, you add the
capacitances
Ceq = Ci
In the circuit shown above, the battery supplies a This is the opposite of the way you calculate
constant voltage V when the switch S is equivalent resistance.
closed. The value of the capacitance is C,
and the value of the resistances are R1 and R2. Problem: Equivalent Capacitance (1998)
Questions 64-65
57. Immediately after the switch is closed, the
current supplied by the battery is
A) V/(R1 + R2) B) V/R1 C) V/R2
D) V(R1 + R2)/R1R2 E) zero
Show your work
Three identical capacitors, each of capacitance 3.0 F,
are connected in a circuit with a 12 V battery as
shown above.
64. The equivalent capacitance between points
X and Z is
(A) 1.0 F (B) 2.0 F (C) 4.5 F
(D) 6.0 F (E) 9.0 F
Show your work
58. A long time after the switch has been closed,
the current supplied by the battery is
A) V/(R1 + R2) B) V/R1 C) V/R2
D) V(R1 + R2)/R1R2 E) zero
Show your work
65. The potential difference between points Y
and Z is
(A) zero (B) 3 V (C) 4 V (D) 8 V
(E) 9 V
Show your work
11/17/2011 10
Deriving capacitance Problems: Capacitors and Dielectrics (1998)
The goal is to derive an equation for capacitance Questions 69-70
of a capacitor given its general shape. A capacitor is constructed of two identical
Steps conducting plates parallel to each other and
a) Draw the capacitor separated by a distance d. The capacitor is
b) Identify symmetry charged to a potential difference of Vo by a
c) Draw Gaussian surface battery, which is then disconnected.
d) Write Gauss’ Law
e) Solve Gauss’ Law for E 69. If any edge effects are negligible, what is the
f) Develop function for V from E magnitude of the electric field between the plates?
g) Develop function for C from V (A) Vod (B) Vo/d (C) d/Vo (D)
Vo/d2 (E) Vo2/d
Capacitance of parallel plate capacitor Show your work
So often used in problems, this is a good one to
memorize
C = e0A/d
C: capacitance (F)
e: dielectric constant of filling
0 : permittivity (8.85 x 10-12 F/m)
A: plate area (m2) 70. A sheet of insulating plastic material is inserted
d: distance between plates(m) between the plates without otherwise disturbing
the system. What effect does this have on the
Problem: Capacitor and Voltage and Energy (1998) capacitance?
51. A parallel-plate capacitor has charge +Q on (A) It causes the capacitance to increase.
one plate and charge -Q on the other. The (B) It causes the capacitance to decrease.
plates, each of area A, are a distance d apart (C) None; the capacitance does not change.
and are separated by a vacuum. A single (D) Nothing can be said about the effect without
proton of charge +e, released from rest at the knowing the dielectric constant of the plastic.
surface of the positively charged plate, will (E) Nothing can be said about the effect without
arrive at the other plate with kinetic energy knowing the thickness of the sheet.
proportional to Show your work
edQ Q2 AeQ
(A) (B) (C) (D)
A eAd d
Q eQ 2
(E)
ed Ad
Problem: Capacitors and Dielectrics (1993)
Show your work
69. Which of the following capacitors, each of
which has plates of area A, would store the
most charge on the top plate for a given
potential difference V ?
Energy in Capacitors
Energy is stored in electric field
UE = ½ CV2
UE = electrical potential energy
C: capacitance
V: voltage
Dielectrics in Capacitors
Increase capacitance
= C/Co
Decrease field strength Show your work
= Eo /E
11/17/2011 11
Current Problem: Resistance (1998)
I = dq/dt
I: current in Amperes (A)
q: charge in Coulombs (C)
t: time in seconds (s)
Unit: Amperes
39. Two resistors of the same length, both
The direction of current flow is defined by the
direction the positive charge moves.
made of the same material, are connected
In most circuits the charge carriers are negative in a series to a battery as shown above.
and current flows in a direction opposite the Resistor II has a greater cross. sectional
direction taken by the charge carriers. area than resistor I. Which of the following
quantities has the same value for each
Problem: Current (1998) resistor?
Question 43 (A) Potential difference between the two ends (B)
A narrow beam of protons produces a current of Electric field strength within the resistor
1.6 x 10-3 A. There are 109 protons in each (C) Resistance
meter along the beam. (D) Current per unit area
43 Of the following, which is the best estimate (E) Current
of the average speed of the protons in the
beam? Show your work
(A) 10-15 m/s (B) 10-12 m/s (C) 10-7
7 12
m/s (D) 10 m/s (E) 10 m/s
Show your work
Conductors
High conductivity
Low resisitivity
Loose electrons (for most electrical circuits)
Insulators Resistor, R
High resistivity Represented by this symbol:
Low conductivity
Tightly held electrons (for most electrical
circuits) Reduces the amount of current flowing in a circuit
Resisitivity, Ohm’s Law
Property of a material which makes it resist the V = IR
flow of current through it. V : potential drop between two points (Volts)
Unit: Ohm-meters (m) I : current (Amps, A)
R : resistance (Ohms, )
Conductivity,
The inverse of resisitivity Electrical Power
= 1/ P = IV
P: Power in Watts
Resistance I: Current in Amperes
Depends on resistivity and on geometry V: Potential Drop in Volts
R = L/A Also, combined with Ohm’s Law
R : resistance (Ohms, ) P = iR2
: resistivity (Q m) P = V2/R
L : length (m)
A : cross-sectional area (m2)
11/17/2011 12
Resistors in series
Problem: Power (1998)
42. A wire of resistance R dissipates power P
Req = Ri
when a current I passes through it. The wire
is replaced by another wire with resistance
Resistors in parallel
3R. The power dissipated by the new wire
when the same current passes through it is
(A) P/9 (B) P/3 (C) P (D) 3P
(E) 6P
Show your work 1/Req = Ri)
Problem: Equivalent Resistance (1998)
37. Which of the following combinations of
4 resistors would dissipate 24 W when
connected to a 12 Volt battery?
Cell
What produces the current in a circuit.
Represented by this symbol:
Battery
Multiple cells in series.
Represented by this symbol (for a 2-cell battery)
Show your work
Electromotive Force
The maximum voltage a cell can supply.
Depends upon the chemistry of the cell.
Designated as EMF or as .
A misnomer: not a force at all!
Internal Resistance Problem: Equivalent Resistance (1993)
Resistance that is part of a cell.
Tends to increase as a cell ages.
Terminal Voltage
Voltage actually supplied by a cell when current 70 If the ammeter in the circuit above reads
is flowing. zero, what is the resistance R ?
Generally less than the EMF due to internal A) 1.5 B) 2 C) 4
resistance. D) 5 E) 6
VT = – IR
Show your work
11/17/2011 13
Kirchoff’s 1st Rule Problem: General Circuit (1993)
Junction rule.
The sum of the currents entering a junction Questions 45-47
equals the sum of the currents leaving the
junction.
Based on conservation of charge.
Kirchoff’s 2nd Rule In the circuit above, the emf's and the resistances
Loop rule. have the values shown. The current I in the
The net change in electrical potential in going circuit is 2 amperes.
around one complete loop in a circuit is equal
to zero. 45. The resistance R is
Based on conservation of energy. A) 1 B) 2 C) 3 D) 4
E) 6
Problem: Kirchoff (1984)
Show your work
41. In the circuit shown above, what is the
resistance R ?
(A) 3 (B) 4 (c) 6 (D) 12
(E) 18 46. The potential difference between points X
and Y is
State your reasoning A) 1.2 V B) 6.0 V C) 8.4 V D)
10.8 V E) 12.2 V
Show your work
Problem: Kirchoff (1998)
36. A resistor R and a capacitor C are connected 47. How much energy is dissipated by the 1.5-ohm
in series to a battery of terminal voltage V0. resistor in 60 seconds?
Which of the following equations relating the A) 6 J B) 180 J C) 360 J D) 720
current I in the circuit and the charge Q on J E) 1,440 J
the capacitor describes this circuit?
Show your work
(A) V0 + QC - I2R = 0 (B) V0 - Q/C - IR = 0
(C) V02 - Q2/2C - I2R = 0
(D) V0 - C(dQ/dt) - I2R = 0 (E) Q/C - IR = 0
Show your work
11/17/2011 14
RC Circuit Problem: RC Circuit (1984)
Resistor and Capacitor in series circuit.
= RC Questions 57-59 refer to the circuit shown below.
capacitive time constant
R: resistance in circuit
C: capacitance in circuit
The capacitive time constant tells at what time
Assume the capacitor C is initially uncharged. The
63 % of the capacitor is charged (or discharged).
following graphs may represent different
quantities related to the circuit as functions of
Charging a capacitor in an RC circuit
time t after the switch S is closed
q = qmax ( 1 – e-t/)
When switch is first closed, the capacitor acts
like a wire, or a short.
When switch is closed for a long time, the
capacitor acts like a gap in the wire.
Discharging a capacitor in an RC circuit
q = qmax (e-t/)
57. Which graph best represents the voltage versus
Problem: RC Circuit (1993) time across the resistor R ?
(A) (B) (C) (D) (E)
Questions 57-58
State your reasoning
In the circuit shown above, the battery supplies a
constant voltage V when the switch S is
closed. The value of the capacitance is C,
and the value of the resistances are R1 and R2.
57. Immediately after the switch is closed,
the current supplied by the battery is 58. Which graph best represents the current versus
A) V/(R1 + R2) B) V/R1 C) V/R2 time in the circuit?
D) V(R1 + R2)/R1R2 E) zero (A) (B) (C) (D) (E)
State your reasoning
State your reasoning
59. Which graph best represents the voltage across
58. A long time after the switch has been the capacitor versus time?
closed, the current supplied by the (A) (B) (C) (D) (E)
battery is
A) V/(R1 + R2) B) V/R1 C) V/R2 State your reasoning
D) V(R1 + R2)/R1R2 E) zero
State your reasoning
11/17/2011 15
Magnetic Field (B) Paths of Charged Particles in Magnetic Field
Formed by moving charge Circle
Affects moving charge When velocity is perpendicular to field
SI units: Tesla Helix
When velocity has component parallel to field
Static Magnetic Fields CAN Straight
accelerate charged particles by changing their When velocity is entirely parallel to field
direction
cause charged particles to move in circular or Problem: Path of charged particle (1998)
helical paths
Static Magnetic Fields CANNOT
change the speed or kinetic energy of charged
particles
do work on charged particles
Magnetic Force on Charged Particle
50. A uniform magnetic field B is parallel to the
F = qv B xy-plane and in the +y-direction, as shown above.
q: charge in Coulomb A proton p initially moves with velocity v in the
v: speed in meters/second
xy-plane at an angle to the magnetic field and
B: magnetic field in Tesla
the y-axis. The proton will subsequently follow
: angle between v and B what kind of path?
Magnetic Force on Current-carrying Wire (A) A straight-line path in the direction of v
F = iL B (B) A circular path in the xy-plane
i: current in Amps (C) A circular path in the yz-plane
L: length in meters (direction of current) (D) A helical path with its axis parallel to the y-axis
B: magnetic field in Tesla (E) A helical path with its axis parallel to the z-axis
: angle between L and B
Explain your reasoning:
Problem: Force / Torque (1993)
63. A square loop of wire 0.3 meter on a side
carries a current of 2 amperes and is
located in a uniform 0.05-tesla magnetic
field. The left side of the loop is aligned
along and attached to a fixed axis. When
the plane of the loop is parallel to the
magnetic field in the position shown above,
what is the magnitude of the torque exerted
on the loop about the axis?
A) 0.00225 Nm B) 0.0090 Nm
C) 0.278 Nm D) 1.11 Nm E) 111 Nm
Show your work:
11/17/2011 16
Magnetic Force as Centripetal Force Crossed Electric and Magnetic Fields
qvBsin = mv2/r Can be used as velocity discriminator
qB = mv/r qvB = Eq
q/m = v/(rB) v = E/B
Problem: Centripetal Force (1998) Problem: Electric/Magnetic Fields (1998)
57. A negatively charged particle in a uniform
magnetic field B moves in a circular path of radius 53. A beam of protons moves parallel to the x-axis
r, as shown above. Which of the following graphs in the positive x-direction, as shown above,
best depicts how the frequency of revolution f of through a region of crossed electric and magnetic
the particle depends on the radius r? fields balanced for zero deflection of the beam. If
the magnetic field is pointed in the positive
y-direction, in what direction must the electric
field be pointed?
(A) Positive y-direction
(B) Positive z-direction
(C) Negative x-direction
(D) Negative y-direction
(E) Negative z-direction
Explain your reasoning:
Explain your reasoning:
11/17/2011 17
Ampere’s Law Hand rule summary (RHR for positive)
Used to calculate magnetic fields from current.
One off our Maxwell equations Hand Rule for magnetic force on moving positive
oi = B•ds charge--version I
o: magnetic permeability of free space (4 x 10-7 Place your fingers in direction of velocity. Then
H/m) rotate your wrist so that your fingers can bend into
i: current enclosed by Amperian loop the direction of the field. Your thumb will be
(Amperes) pointing in the direction of the force.
B: magnetic field (Tesla) Hand Rule for magnetic force on current in wire--
s: distance around Amperian loop (meters) version I
Can calculate magnetic fields inside and outside wires Place your fingers in direction of current. Then
Used to calculate magnitude of magnetic fields rotate your wrist so that your fingers can bend into
inside solenoids and toroids the direction of the field. Your thumb will be
Best for high symmetry situations pointing in the direction of the force.
Problem: Ampere’s Law (1998) Hand Rule for magnetic force on moving positive
charge--version II (a mnemonic)
52. In which of the following cases does Put your Fingers in the direction of the Field
there exist a nonzero magnetic field that Put your thumb in the direction of the velocity or
can be conveniently determined by current (like a hitchhker)
Then your Palm is the direction of the Push on a
using Ampere's law?
Positive charge
(A) Outside a point charge that is at rest
(B) Inside a stationary cylinder carrying a uniformly
Hand Rule for magnetic force on moving negative
distributed charge
charge
(C) Inside a very long current-carrying solenoid
Use the method described above, then flip your
(D) At the center of a current-carrying loop of wire
thumb 180o. Alternately, you may use your left
(E) Outside a square current-carrying loop of wire
hand.
Explain your reasoning: Hand Rule for fields where current is straight
(for the magnetic field caused by the current)
Curve your fingers.
Place your thumb in direction of current. Then your
curved fingers point in direction of curved magnetic
field.
Hand Rule for fields where current is circular
(for the magnetic field caused by the current)
Problem: Ampere’s Law (1993) Curve your fingers.
Place your curved fingers in direction of current.
Then your thumb points in direction of magnetic
field in center of circular current.
65. Two long parallel wires are a distance 2a apart,
as shown above. Point P is in the plane of the
wires and a distance a from wire X. When there
is a current I in wire X and no current in wire Y,
the magnitude of the magnetic field at P is Bo.
When there are equal currents I in the same
direction in both wires, the magnitude of the
magnetic field at P is
A) 2Bo/3 B) Bo C) 10Bo/9
D) 4Bo/3 E) 2 Bo
Show your work
11/17/2011 18
Problem: Hand Rule (1998) Currents Near Each Other
Parallel Currents Attract
Questions 44 Anti-parallel Currents Repel
A narrow beam of protons produces a current of
1.6 x 10-3 A. There are 109 protons in each Problem: Magnetic Force on Wires (1998)
meter along the beam.
44. Which of the following describes the
lines of magnetic field in the vicinity of
the beam due to the beam's current?
(A) Concentric circles around the beam 49. A rigid, rectangular wire loop ABCD carrying
(B) Parallel to the beam current I1 lies in the plane of the page above a
(C) Radial and toward the beam very long wire carrying current I2 as shown
(D) Radial and away from the beam above. The net force on the loop is
(E) There is no magnetic field. (A) toward the wire
(B) away from the wire
(C) toward the left
Explain your reasoning: (D) toward the right
(E) zero
Explain your reasoning:
Problem: Magnetic Field (1993)
43. A cross section of a long solenoid that
carries current I is shown above. All of the
following statements about the magnetic field B
inside the solenoid are correct EXCEPT:
A) B is directed to the left.
B) An approximate value for the magnitude of B may be
determined by using Ampere's law.
C) The magnitude of B is proportional to the current I.
D) The magnitude of B is proportional to the number of
turns of wire per unit length.
E) The magnitude of B is proportional to the distance
from the axis of the solenoid.
Explain your reasoning:
11/17/2011 19
Magnetic Flux Motional emf
B = BA =BLv
B = BAcos() : induced potential
B: magnetic flux (Wb) L: length of bar or wire
B: magnetic field (T) V: speed of bar or wire
A: area (m2)
: angle between the magnetic field and a Problem: Motional emf (1998)
vector which is normal to the area
SI Unit: Weber (Wb)
1 Wb = 1 T m2
Faraday’s Law of Induction
The fourth Maxwell Equation
Changing magnetic flux induces a potential 54. A vertical length of copper wire moves to
= -dB/dt the right with a steady velocity v in the direction of
: induced potential (V) a constant horizontal magnetic field B as shown
B: magnetic flux (Wb) above. Which of the following describes the induced
t: time (s) charges on the ends of the wire?
Sometimes written in this way Top End Bottom End
Eds = -dB/dt (A) Positive Negative
since (B) Negative Positive
= Eds (C) Negative Zero
(D) Zero Negative
Problem: Faraday (1998) (E) Zero Zero
Show your work
68. A wire loop of area A is placed in a time-
varying but spatially uniform magnetic field
that is perpendicular to the plane of the loop, as
shown above. The induced emf in the loop is
given by = bAt1/2, where b is a constant. The
Lenz’s Law
time varying magnetic field could be given by
Tells the direction the field will point, and therefore,
1 1 1 / 2
(A) bAt 1 / 2 (B) bt the direction the current will flow.
2 2 Current will flow so as to oppose the change in
1 1/ 2 2 magnetic flux.
(C) bt (D) bAt 3 / 2
2 3
2 3/ 2
(E) bt
3
Show your work
11/17/2011 20
Problem: Lenz (1993) Show your work
66. In the figure above, the north pole
of the magnet is first moved down toward
the loop of wire, then withdrawn upward.
As viewed from above, the induced current
in the loop is Induced E-Field
A) always clockwise with increasing magnitude Caused by changing magnetic flux
B) always clockwise with decreasing magnitude Does not arise from charge, like static fields do
C) always counterclockwise with increasing Circular; loopy
magnitude Non-conservative
D) always counterclockwise with decreasing
magnitude Inductor (L)
E) first counterclockwise, then clockwise A coil in a circuit.
Resists change in current with an induced potential.
Stores energy in a magnetic field.
Show your work
Inductance and Faraday’s Law
L = -L di/dt (inductance defn)
L = -NdB/dt (Faraday’s Law)
Ldi/dt = NdB/dt (combined)
Ldi = NdB
Li = NB (true for all inductors)
Inductance in solenoids
Li = NB (general form)
For a solenoid, certain substitutions can be made.
Fore example:
Problem: Faraday and Lenz (1998) B = o n i
B = o n i A
N=nl
So the equation becomes, for a solenoid:
Li = n l (n o i) A
L = n2 l oA (air core)
L = n2 l oA B (other type of core)
L: inductance (H)
56. A square wire loop with side L and i: current (A)
resistance R is held at rest in a uniform n: coils per meter
magnetic field of magnitude B l: length (m)
directed out of the page, as shown above. The o: magnetic permeability of free space
field decreases with time t according to the B: magnetic permeability of core
equation B = a - bt, where a and b are posi-
tive constants. The current I induced in the Self Inductance
loop is When current changes in a coil, the magnetic flux
(A) zero through the coil changes, and the coil resists this
(B) (B) aL2/R, clockwise change by creating an opposing voltage. Hence
(C) aL2/R, counterclockwise the term “self-inductance”.
(D) bL2/R, clockwise
(E) bL2/R, counterclockwise
11/17/2011 21
The LR Circuit Kirchoff’s analysis (opening the switch)
Contains a resistance and inductance. L - V R = 0
Inductor slows down the approach to final
Ldi/dt – iR = 0
current value (whether current is increasing
-t/
or decreasing) i = imax(e )
L = L/R (inductive time constant)
The LR Circuit
L
Graph (closing the switch)
L I (A)
R
VR
Kirchoff’s analysis (closing the switch)
L - VR = 0
Ldi/dt – iR = 0 t (s)
-t/
i = imax(1 - e )
L = L/R (inductive time constant)
Graph (closing the switch)
I (A)
t(s)
11/17/2011 22
Problems: LR Circuit (1993) 61. After the switch has been closed for a long
time, it is opened at time t = 0. Which of the
Questions 59-61 relate to the following circuit in following graphs best represents the subsequent
which the switch S has been open for a long current i at point X as a function of time t ?
time. (A) i
O t
59. What is the instantaneous current at point X
immediately after the switch is closed? (B) i
A) 0 B) /R C) /2R D) /RL
E) L/2R
Show your work O t
(C) i
O t
(D) i
60. When the switch has been closed for a
long time what is the energy stored in the
inductor?
A) L/2R B) L2/2R2
O t
C) L2 /4R2 D) LR2/22 E) 2R2/4L
(E) i
Show your work
O t
Show your work
11/17/2011 23