Physics Challenge for
Teachers and Students
Boris Korsunsky, Column Editor
Weston High School, Weston, MA 02493
korsunbo@post.harvard.edu
Solution May 2010 Challenge
w Be There and Be Square
A square loop with side b is made of a wire of mass
m and negligible electric resistance. The loop is pivoted along its top horizontal side and placed in the
weak vertical uniform magnetic field B as shown.
The loop is then pulled to a horizontal position and
released. Eventually, the loop comes to rest due to air
resistance. Find the angle q that the plane of the loop
makes with the vertical at the final position. The inductance of the loop is L.
(Adapted from Physics Olympiads by A. Slobodyanjuk,
L. Markovich, A. Lavrinenko. Aversev, Minsk, 2003.)
B
g
The solutions to the May Challenge, Be There and Be
Square, came from all over the world. They were not
quite as numerous as we had expected. One possible
reason is that the problem was poorly worded and
thus unduly confusing (the column editor accepts full
responsibility and humbly apologizes for the mishap).
A small number of attentive readers noted that, as
stated, the problem was of no particular interest and
they proposed an alternative version for the initial
conditions (which was, incidentally, the intended one).
Interestingly, the vast majority of the successful solvers
simply interpreted the problem the way it was intended
(but not actually stated) without explicitly explaining
why they chose to alter the initial conditions.
Also, a number of contributors stopped short of a full
solution by providing the equation but not solving it.
Using the condition of “weakness” of the magnetic
field, the approximate solution could be found rather
easily; if no attempt to arrive at the final answer was
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THE PHYSICS TEACHER ◆ Vol. 48, 2010
made, the solutions were considered incomplete.
Here is one of the solutions that explains why the problem
should be rephrased and then proceeds to solve the altered
version:
Solution: The key to the problem is to note that the total
emf e induced in the loop is the negative time derivative of
the total magnetic flux  linked by the loop,
dΦ
(1)
= IR
ε=−
dt
using Ohm’s law in the last step, where I is the current
induced in the loop. But since the resistance R of the loop
is zero, we conclude that  must be a constant. Now in
general that total flux can be written as a sum of the flux
ext due to external magnetic field B and the flux ind in
the loop due to the induced current I. Specifically,
(2)
Φ = Bb2 sin θ ′ + LI at an arbitrary instant in time when the loop hangs at angle
q with respect to the vertical. (We are told that q oscillates for a while and finally stops at value q.) I have chosen
the sign convention that upward flux is positive and hence
current is positive if it is counterclockwise as seen looking
down on the loop from above (in accord with the righthand rule). Since  is a constant, we can determine its
value from the initial conditions in the problem. We are
told the loop was initially hanging vertically at rest (before
being pulled up to a horizontal position) so that qi = 0
and presumably the initial current in the loop was zero, Ii
= 0. In that case,  is initially zero, as we see from Eq. (2),
and consequently must always be zero. But that means that
when we pull the loop up into a horizontal position
(q = 90°), a clockwise current equal to –Bb2/L is induced in
the loop, according to Eq. (2). Now consider the following
sketch of the loop just after it is released from this position
and begins to fall.
Only the current along the edge farthest from the axis
(indicated by the dashed line above) contributes to the
The last step is to solve this equation for q.
Since the problem statement says that B is weak, we can
assume that the dimensionless constant A in Eq. (5) is
much smaller than one. Then q (in radians) is small. So let’s
use the small-angle approximation tan q < sin q < q in Eq.
(5) to get
I
net torque on the loop, so I have only marked I along
that edge. But since B is upward, the magnetic force on
that edge is leftward (according to the right-hand rule)
and thus gives rise to a restoring torque, as does gravity.
Therefore the loop is quickly pulled back to its original
vertical position [at which point I has decreased back to
zero according to Eq. (2)] and the solution of the problem
as stated is q = 0. Presumably this is not what was intended
as a solution to the problem!
We can fix the problem by altering the initial conditions.
Instead of starting with zero current when the loop is
hanging vertically, we should start with zero current when
the loop is in the horizontal position. But that cannot
be easily arranged for a complete loop because the loop
must have originally been outside of the field region, and
it would then have been brought into the field and hung
vertically from the pivot. An easy way to fix this issue is
to start with a small gap or open switch in the loop, so
that no induced current arises when it is swung up to the
horizontal position. Once it is horizontal, the switch can
be closed (or the gap soldered shut). Now we get a nonzero
solution for q as follows.
The initial conditions are now qi = 90° and Ii = 0 with the
loop in the horizontal position. Then Eq. (2) implies that
 = Bb2. Now as the loop drops, Lenz’s law implies that
the decreasing upward external flux must be countered by
an upward induced flux and thus by a counterclockwise
current, opposite in direction to the current in the
sketch above. In that case, the magnetic torque and the
gravitational torque are opposite in direction, and we can
find a nonzero final angle at which the two torques balance
each other,
(3)
( IbB )(b cos θ ) = ( mg )( 12 b sin θ ).
Here the left-hand side is the magnetic force on the
bottom edge multiplied by its moment arm, and the righthand side is the gravitational force on the loop times the
moment arm of the loop’s center of mass. At this final
angle, note that Eq. (2) implies that
(4)
Bb2 = Bb2 sin θ + LI . Eliminating I between Eqs. (3) and (4) leads to
tan θ = A(1 − sin θ ), where A ≡
2b 3 B 2
.
mgL
(5)
θ≈
A
≈ A, where
1+ A
A≡
2b 3 B 2
.
mgL
(Contributed by Carl E. Mungan, U. S. Naval Academy,
Annapolis, MD)
We are also pleased to recognize the following contributors:
Sharmila Balamurugan (undergraduate student, Chennai,
India)
Hratch Barsoumian (Haigazian University, Beirut, Lebanon)
Darío Castello (IES Front Marítim, Barcelona, Spain)
Matthew Cochran (Chaminade University of Honolulu,
Honolulu, HI)
R. C. Dhandhania (KalraShukla, Mumbai, India)
Don Easton (Lacombe, Alberta, Canada)
Fernando Ferreira (Universidade da Beira Interior, Covilhã,
Portugal)
Fredrick P. Gram (Cuyahoga Community College,
Cleveland, OH)
Art Hovey (retired, Milford, CT)
José Ignacio Íñiguez de la Torre (Universidad de Salamanca,
Salamanca, Spain)
Jack Kingston (Weston High School, Weston, CT)
Osvaldo D. Pavioni (Facultad de Ingeniería – U.N.C.P.B.A.,
Argentina)
Cássio dos Santos Sousa, student (Colégio Objetivo, São
Paulo, Brazil)
Adam Wyrzykowski, student (August Witkowski High
School, Cracov, Poland)
Many thanks to all contributors and we hope to hear from
you in the future!
Please send correspondence to:
Boris Korsunsky
korsunbo@post.harvard.edu
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