Goal: Distinguishing components of the Lorentz force
Source: 283 – effects of magnetic force
A charged particle moves into a region containing both an electric and
magnetic field. Which of the statements below are true?
- The particle cannot accelerate in the
direction of B.- The path of the particle must be a circle.
- Any change in the particle’s kinetic energy is caused by the E
field.
Goal: Reason regarding electric fields
Source: 283-10, E at origin due to charged rods
All of the configurations shown below consist of charged rods of the
same length L. The magnitude of the total charge is also the same for
each rod. The total charge in each rod is distributed uniformly.
For which
configuration(s) is the electric field vector at the origin in the
positive x direction?
(4) Because they are given lots of examples involving point charges,
spheres and rings, students often miss the fact that there are many
situations for which the direction of the field can be deduced even
though determining the value or formal expression for the field is way
beyond them.
Goal: Hone the concept of electric field
Source: UMPERG-283-365
Two
uniformly charged rods are positioned horizontally as shown. The top
rod is positively charged and the bottom rod is negatively charged. The
total electric field at the origin
(6) By symmetry the field must point along the y-axis. Students
who do not understand that the field points away from positive charges
and towards negative charges would select #1 thinking that the fields
cancel.
Goal: Reasoning with magnetic forces
Source: UMPERG-283-626
In the following situations a charge q moves in a uniform magnetic
field. The strength of the magnetic field is indicated by the density
of field lines. In all cases the speed of the charge is the same. For
which situation(s) will the charge q have the largest displacement in a
given time T.
(5) Since the speed cannot change, the greatest displacement will occur
when the path is a straight line. Some students may answer #10 thinking
that the time matters.
Goal: Reasoning with magnetic forces
Source: UMPERG-283-625
In the following situations a charge q moves in a uniform magnetic
field. The strength of the magnetic field is indicated by the density
of field lines. In all cases the speed of the charge is the same. For
which situation(s) will the charge q travel the greatest distance in a
given time T?
(9) The magnetic force can change the direction of the velocity but not
the speed. The distance traveled, therefore, cannot depend upon either
the strength or orientation of the magnetic field. It is important to
have students who pick one of the other choices verbalize their reasons.
Some students may interpret the question as asking for the
‘displacement’ and, thinking that the time is needed, respond #10.
Actually the result for displacement is #5.
Goal: Reasoning with thermodynamics
Source: UMPERG-ctqpe212
The
two curves shown are isotherms. A system is taken from state A to state
B along the T2 isotherm. State C has the volume of state A
and the pressure of state B.
Which of the following is true:
(5) This is the only answer that is definitely correct. Students
will likely assume that the system is an ideal gas. Many will choose #3
and this should engender a discussion of the sign of work. Physicists
use the convention that positive work is that done BY the system.
Chemists frequently use the opposite convention.
The one choice giving students the most reasoning difficulty is #4. A
reasoning path that eliminates that choice is as follows. Q(AB) is
positive: gas does positive work on surroundings, so heat has to be
added to keep energy the same. Q(CA) is positive: no work is done, so
heat has to be added to raise the temperature. Q (CB) is positive: heat
must be added to expand the gas and still maintain constant pressure. So
Q(CB) [constant pressure] > Q(CA) [constant volume]; Q(CA) – Q(CB) < 0
cannot equal Q(AB) > 0
Goal: Reasoning with thermodynamics
Source: ctqpe214
A
system consisting of a quantity of ideal gas has the two isotherms
shown. The system, initially at state C, can be taken along path CA to
final state A or along path CB to state B.
Which of the following is true:
(1) Since the internal energy of an ideal gas depends only on
temperature, states A and B have the same internal energy. Along path CB
the system does work requiring more heat to be added than along path CA.
Goal: Problem solving in thermodynamics
Source: UMPERG-ctqpe198
An
amount of an ideal gas is taken around the process shown.The amount of
heat extracted during process BC is
(4) Since no work is done the change in internal energy must be
due to heat extraction. Some students may think that the answer cannot
be determined because they do not know the number of moles. These are
likely thinking that they need to find the temperature at each state to
answer the question.
Goal: Reason with internal energy
Source: UMPERG-ctqpe194var
An
amount of an ideal gas is taken around the process shown. Which of the
following statements about the internal energy of the states is true?
(1) Students need to know only that the internal energy depends
upon the product of p and V. Alternatively, they can reason that,
according to the Ideal Gas Law, this product is proportional to the
temperature and the temperature determines the internal energy
Goal: Link representations
Source: UMPERG-ctqpe196
A vertical cylinder with a movable cap is cooled. The process
corresponding to this is
(4) Interpreting process diagrams is a very important skill for
students. good followup questions include; Is work done during this
process? … by or on the gas? How does the temperature at A compare to
that at C? How much heat was extracted during this process?
Commentary:
Answer
(3) The only cases that most students see is the one having E and B
perpendicular. As a result they discount the case of E and B parallel
and think statement A is also true.