Goal: Unspecified.
Source: Unspecified.
A person throws a ball straight up. The ball rises to a maximum height
and falls back down so that the person catches it. When is the
acceleration of the ball at its MAXIMUM?
Goal: Hone understanding of Faraday’s Law
Source: Hone understanding of Faraday’s Law
A long conducting bar moves with a constant velocity in a uniform magnetic
field. If the bar and the velocity of the bar are perpendicular to the
magnetic field as shown. Which of the following statements are true?
- At steady state there is an E field in
the bar- At steady state there is a current in the bar
- At steady state there is a mag. force on bar
(1) This question is often given as an example of Faraday’s law.
Relating Emf to flux change is difficult for some students to perceive
when there is no circuit. Creating an imaginary circuit helps, but many
students continue to get the direction of the field incorrect even
though the magnitude of the potential difference is now understood. It
is useful to view the process using the Lorentz force. This helps
students understand which charges migrate to which end and, therefore,
what the direction of the electric field in the bar is.
Goal: Reason regarding forces between current elements
Source: 283-660 Force between two wires
A very long wire lies in a plane with a short wire segment. The long
wire carries current I, while the short wire of length L carries current
i. The two wires are parallel to each other. Which of the following
statements are true?
- The direction of the magnetic force
exerted by the long wire on the short wire is directed away from the
long wire.- The magnitude of the force on the short wire is
μ0IiL/2πd.- The long wire experiences a force
of exactly the same magnitude as the force experienced by the short
wire.
(6) The force is attractive between the wires so statement A is false.
Students need to interpret the phrase ‘very long’ as implying that the
wire is infinite.
Goal: Reason regarding electrodynamics
Source: 283-635 Path of a charge in E&B fields.
A
charge has an initial velocity parallel to the y-axis in E and B fields.
Both fields point along the x axis. Which of the following statements
regarding the charge’s motion are correct?
(7) A common response is #4 because they forget that increasing pitch
implies that the speed changes.
Goal: Reason regarding electrodynamics
Source: 283-630 Path of a charge in E&B fields.
A
charge is released from rest in E and B fields. Both fields point along
the x axis. Which of the following statements regarding the charge’s
motion are correct?
(6) The different responses reveal the extent to which students
understand vector cross products and/or read the problem carefully. Some
students choose #8 because they do not like the way the motion is
expressed. They prefer descriptions such as, the charge first moves in a
straight line until it gets some speed then …
Goal: Reasoning regarding the Lorentz force
Source: 283-620 Moving bar magnet and charge
A bar
magnet moving with speed V passes below a stationary charge q. What can
be said about the magnitude of the magnetic force on the bar magnet and
the charge q.
Fbar and Fq are both zero.
Fbar is zero and Fq is not zero.
Fbar is not zero and Fq is zero.
Fbar and Fq are both non-zero.
(4) Many students have a lot of difficulty with this one. All of their
past experience has been with a moving charge in a magnetic field. They
may not think that it is equivalent to view the interaction from the
bar’s frame. Of course, they are correct, but the difference is
unimportant for purposes of recognizing that the force on the charge is
non-zero. They may invoke the third law by rote, without perceiving any
mechanism that could provide a force on the magnet. Discussing this in
some detail is a good idea.
Goal: Honing the right hand rule
Source: UMPERG-283 Mag Force
In a region of space there is a uniform magnetic field pointing in the
positive z direction. In what direction should a negative charge move
to experience a force in the positive x direction?
(6) Students will likely forget that the charge is negative.
Goal: Problem solving with dynamics
Source: UMPERG-ctqpe168
A
uniform disk with mass M and radius R rolls without slipping down an
incline 30° to the horizontal. The friction force acting through
the contact point is
(4) This problem requires students to use the 2nd law written in
terms of the CM acceleration and the rotational dynamic relation written
about the CM or the contact point. In either case they also need the
geometric constraint for rolling. This is a difficult problem for
students requiring a lot of additional knowledge, such as the moment of
inertia for a disk and, depending upon solution method, the Parallel
Axis Theorem.
Having gone to the trouble of solving the problem it is best to make
sure that the students glean as much as they can. A good followup
question is which would have a larger friction force, a hoop, a disk or
a sphere. They may try to reason from the acceleration of these objects
that the larger the acceleration, the smaller the friction force. The
friction force depends upon the mass, however, and the question cannot
be answered without knowledge of the masses.
Goal: Problem solving
Source: UMPERG-ctqpe160
A
uniform disk with R=0.2m rolls without slipping on a horizontal surface.
String is pulled in the horizontal direction with force 15N. Moment of
inertia of disk is 0.4 kg-m2. The acceleration of the center
of the disk is most nearly
(2) This problem can be done without knowing anything about the
friction force. To do so, though, requires knowing the Parallel Axis
Theorem for moments of inertia and the constraint between the linear and
rotational rates of motion for a rolling object. An alternate method is
to write the two equations for the linear motion of the center of mass
and the torque relation for rotation about the CM and then eliminate the
friction from the two equations.
Goal: Problem solve with rotational dynamics
Source: UMPERG-ctqpe156
A
uniform rod is hinged to a wall and held at a 30° angle by a thin
string that is attached to the ceiling and makes a 90° angle to rod.
The tension in the string is 10N. The weight of the rod is about
(4) Some students will use the wrong trigometric function and
conclude that the weight is 40N.
An interesting follow up question is to ask what is the hinge force.
Students often forget that both the sum of the forces and the sum of the
torques must be zero for static equilibrium.
Commentary:
None provided.