Physics Challenges

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Physics Challenges for 
Teachers and Students
◗ The Power of Two (A1) 
Challenge: Find the resistance between points A and B of 
an infinite circuit shown. Express your answer in terms of R. 
Solution 1. The problem is to find Rx, as illus-
trated in Fig. 1 below. 
My approach is to consider the first three resis-
tors plus “everything else” as a lumped resis-
tance R1. Then (see Fig. 2 below)
 
(1)R R R R a b ab
a bx
= + ≡
+
2 2 1|| ( || ).where
Solutions to April 2005 Challenges
We can then apply the same process to R1: 
consider it as three resistors plus a new “every-
thing else,” cleverly named R2. In general,
+
+= +
1
12 2 || ,
k k
k kr r (2)
where I have muddied the waters by divid-
ing everything by R: rwhatever = Rwhatever/R. 
The nature of the full solution is hinted at by 
rewriting Eq. (2) as
 
 (3)
+
+
= +
+
1
1
2
2
21
k
k
k k
k
r
r
.
 
We can write down a full expression for rx by 
starting with the expression for r0, and then 
repeatedly substituting into Eq. (3) the expres-
sions for r1, r2, etc. Provided only that we do it 
an infinite number of times, we have an exact 
expression for rx. The result turns out to be a 
continued fraction, which I have never, until 
this moment, had a use for:
 (4)
= +
+
+
+
+
+
+
+
4
1
2 11 24 21 48 41 816 81
xr
r
Equation (4) can be written in a more compact 
(but more obscure) notation as 
rx = + + + + + + + + + +
2
1
1
1
4
2
1
2
8
4
1
4
16
8
1
8
32
16
1
16
64
… .
Fig. 1. The infinite 2k ladder.
Fig. 2. The first two stages of the analysis 
showing how the ladder can be considered as 
three resistors of known value plus an unknown 
lumped resistor representing the rest of the cir-
cuit. This unknown lumped resistor can itself be 
considered a three knowns plus a new lumped 
resistor. The process is repeated ad infinitum to 
develop a complete expression for the resistance 
of the ladder. 
THE PHYSICS TEACHER ◆ Vol. 43, 2005 
I always need to write out too many terms 
before I’m sure I see the pattern. With the 
identification of the first term as b0, and the 
numerator and denominator of the jth fraction 
as aj and bj, respectively, we have
 
(5)
 
−
+
−
   = =    
1 2
2 1
1
1 odd 2 odd
even2 even
j
j jj
j
j j
b a
a jj
( ) /
/
.
The trouble with a continued fraction is that 
in order to evaluate it, you need to start at the 
end and work backward. Since the fraction in 
this case is infinite, that could get a bit tricky! 
Fortunately, there is a way to start at the begin-
ning and work forward recursively. The hope 
here is that the result will (rapidly, with luck) 
converge to some value. From a physical per-
spective (yes, there is a tiny flake of that in this 
problem), it is not unreasonable to expect con-
vergence. To a first approximation, rk ≅ 3 . 2k, 
which tends to infinity, and an infinite resis-
tance in parallel to a circuit makes no change 
to that circuit. 
The recursion relations for evaluating the 
continued fraction are as follows:1
 (6)− − − −= + = +1 2 1 2,j j j j j j j j j jA b A a A B b B a B
with starter values
 (7)1 1
0 0 0
1 0
.
1
A B
A b B
− −= =
= =
An approximation for rx is found by first cover-
ing the ears of any children present, and then 
recursing out to j = n to find 
 (8)nn
n
A
r
B
= .
An easy way to accomplish this recursion is 
with a spreadsheet program, and I find (to the 
maximum number of digits Excel will give me)
 Rx = 2.8507810593582R. (9)
For completeness, here is a table showing the 
convergence to this result:1
j bj aj Aj Bj fj
-1 1 0
0 2 2 1
1 1 1 3 1 3.00000000000000
2 4 1 14 5 2.80000000000000
3 1 2 20 7 2.85714285714286
4 8 2 188 66 2.84848484848485
5 1 4 268 94 2.85106382978723
6 16 4 5040 1768 2.85067873303167
7 1 8 7184 2520 2.85079365079365
8 32 8 270208 94784 2.85077650236327
9 1 16 385152 135104 2.85078162008527
10 64 16 28973056 10163200 2.85078085642317
11 1 32 41297920 14486528 2.85078108432883
12 128 32 6.12E+09 2.18E+09 2.85078105032099
13 1 64 8.86E+09 3.11E+09 2.85078106047022
14 256 64 2.66E+12 9.35E+11 2.85078105895576
15 1 128 3.8E+12 1.33E+12 2.85078105940773
16 512 128 2.29E+15 8.02E+14 2.85078105934029
17 1 256 3.26E+15 1.14E+15 2.85078105936042
18 1024 256 3.92E+18 1.38E+18 2.85078105935741
19 1 512 5.59E+18 1.96E+18 2.85078105935831
20 2048 512 1.35E+22 4.72E+21 2.85078105935818
21 1 1024 1.92E+22 6.73E+21 2.85078105935822
22 4096 1024 9.23E+25 3.24E+25 2.85078105935821
23 1 2048 1.32E+26 4.62E+25 2.85078105935821
Table I. Table showing the calculation of rx as a function 
of the number of terms in the recursion. Convergence is 
extremely rapid, although it takes a while until the 14th 
digit settles down.
1. W.H. Press, S.A. Teukolsky, W.T. Vetterling, and B.P. Flan-
nery, Numerical Recipes, 2nd ed. (Cambridge University 
Press, New York, 1992). 
(Submitted by Geoff Nunes, DuPont Central Research & 
Development, Swarthmore, PA)
Solution 2.
Fig. 1
Fig. 2
THE PHYSICS TEACHER ◆ Vol. 43, 2005
Let RT represent the total resistance between 
points A and B of the infinite circuit shown 
above in Fig. 1. If we isolate the portion of the 
circuit inside the box (see Fig. 2), we can see 
that this infinite circuit looks exactly like the 
original circuit except that the value of each 
resistor is now doubled. Therefore, the total 
resistance of this circuit is just twice that of the 
original, or 2RT. Now, the original circuit can 
be redrawn as shown in Fig. 3, where 2RT has 
replaced the entire circuit from the box. 
Since this circuit has a resistance RT, then RT 
= R+R+(R||2RT) (where || represents a parallel 
combination).
Mathematically:
−  = + + = +   + 
→ − − =
1
2 2
21 1
2 2
2 2
2 5 2 0.
T
T
T T
T T
RR
R R R
R R R R
R RR R
The only positive root of this quadratic equa-
tion is the answer: 
 +  =   
5 41
.
4T
R R
 
(Submitted by H. Scott Wiley, Science Academy of 
South Texas, Weslaco, TX)
◗ Don't Ask, Don't Tell (A2)
What resistor r must be connected between points 3 and 4 
so that the resistance between points 1 and 2 can be found 
without knowing the total number of resistors in the circuit 
shown below? Each resistor in this circuit is 10 Ω. 
THE PHYSICS TEACHER ◆ Vol. 43, 2005 
Fig. 3
Fig. 1
Both circuits must have the same equivalent 
resistance, X. In other words, X equals 10 in 
parallel with 20 + X.
X = (10)(20 + X) / (30 + X)
X2 +30X = 10X + 200
X2 +20X –200 = 0
X = 10 ( 3 – 1)
X = 7.321 Ω
So if we add a 7.321-Ω resistor at the end of 
the circuit given in the problem, the total resis-
tance of the circuit will be 7.321 Ω, regardless 
of the number of loops!
(Submitted by Asif Shakur, Salisbury University, 
Salisbury, MD)
◗ Out of the Loop (A3) 
In the circuit shown, when ideal voltmeters are connected 
to each emf source, the readings of the voltmeters are the 
same. In the circuit, resistance Rx is unknown, ε1 = 10 V; ε2 
= 5 V; r1 = 2 Ω; r2 = 1 Ω; R1 = 8 Ω; R2 = 9 Ω. Find the cur-
rent through Rx. 
Solution. There is no solution to the problem as 
stated for any possible choice of resistor Rx.
Solution. Let us consider the two INFINITE 
circuits of 10-Ω resistors shown in Fig. 1. 
Define current I1, I2, and Ix as the currents 
through resistors R1, R2, and Rx. We suppose 
that I1 is oriented upwards, I2 downwards, and 
Ix downwards. We apply Kirchhoff ’s loop rule to 
the outer loop of the diagram, which leads to
( ) ( )1 2 1 1 1 2 2 2 0.I r R I r Rε ε+ − + − + = 
Substituting the values of the known quantities 
leads us to the expression
I1 + I2 = 1.5 A. 
Experimentally, we are told that the voltmeters 
should give the same reading for the terminal 
voltages of the emf sources. Since we are not 
told the orientation of the voltmeters, we write 
that
ε ε− = −1 1 1 2 2 2 .I r I r
Again, substituting the valuesof known quanti-
ties leads to the requirement that
2I1 – I2 = 5 A or 2I1 + I2 = 15 A.
Combined with the result of the loop rule, this 
leads to two different solutions for I1 and I2, 
given by
 
= =−
= =−
1 2
1 2
13 2
 A and A
6 3
 OR
27
 A and 12 A.
2
I I
I I
Kirchhoff ’s junction rule informs us that Ix = 
I1 – I2, so the value of the current through the 
resistor Rx could be given by either
= =
17 51
A OR A.
6 2x x
I I
 
So far, so good. 
But there’s trouble in paradise. Note that 
both solutions require I2 to be negative—so we 
must have guessed wrong about its direction 
in the first place! When you look at the circuit, 
however, it’s sort of hard to see how that could 
be possible.
If we look a little deeper, we find the prob-
lem. Use Kirchhoff ’s loop rule on one of the 
other loops in the circuit—I’ll choose the left 
loop. This yields
( )1 1 1 1 0.x xI r R I Rε − + − =
Therefore, the necessary resistance of the ele-
ment Rx is given by
( )ε − +
= =− Ω − Ω1 1 1 1
70 70
 OR .
17 153x x
I r R
R
I
Since there’s no physical meaning to a nega-
tive resistance, we conclude that the conditions 
described in the problem are not attainable.
(Submitted by Craig Caylor, Westminster College, 
New Wilmington, PA)
Column Editor’s note: Many readers discov-
ered a glitch in this problem and yet provided 
a satisfactory analysis, instead of just taking the 
path of least (negative?) resistance and giving 
up….
◗ Ring, Ring, Ring...(A4)
A ring of mass m, diameter d, and resistance r is falling 
from a large height in a vertical magnetic field. The magni-
tude of the field changes with height: B = B0(1 + ky), where 
k is a known constant and y is the vertical coordinate. Find 
the terminal velocity of the ring. The plane of the ring 
remains horizontal as it falls. 
Solution 1. The ring falls under gravity in the 
nonuniform magnetic field. As the ring falls, 
eddy currents I arise in the ring because of 
the changing magnetic flux  and an emf  is 
induced in the ring. The direction of these cur-
rents is such as to repel the ring. The energy is 
dissipated in the form of heat. If the rings fall 
with constant velocity, then the rate at which 
the gravitational potential energy decreases 
must equal the rate at which heat is generated, 
and the magnetic force then is equal to the 
gravitational force.
Using Faraday’s law, the electromotive force is
THE PHYSICS TEACHER ◆ Vol. 43, 2005
y
B
 (1).
d
dt
ε
Φ= −
The magnetic flux at time t is given by
 
(2)
2
0 (1 ) ,4
d
BA B ky
π
Φ= = +
where A is the area of the ring. From Eq. (1)
 (3)0
2
.
4
B k d dy
dt
π
ε=−
Note that dy
dt
−
 
is the terminal velocity, Vt. 
From Ohm’s law. 
ε
= ,I
R 
where R is the resis-
tance of the ring.
The rate at which heat is generated is
 and the rate at which gravitational ε=
2
2I R
R 
potential decreases is 
 .dymg
dt
From energy conservation: 
 (4)ε =− =
2
.t
dy
mg mgV
R dt
Now by substituting  from Eq. (3) into Eq. 
(4) and using some algebra, we obtain the ter-
minal velocity
 .
π
=
2 2 2 4
0
16t
mgR
V
B k d
(Submitted by José Costa Leme, High School 
Lanheses, Viana do Castelo, Portugal)
Solution 2. Suppose for specificity that B0 and 
k are both positive. (The final solution only 
depends on these two quantities squared, so 
their actual signs are irrelevant.) I assume that y 
is positive upward. With these choices of signs, 
the (externally sourced) magnetic field B is 
directed upward along the axis of the ring and 
gets weaker as the ring falls downward (with 
terminal speed υ). This means the density of 
magnetic field lines must decrease as the ring 
falls, i.e., the field lines must diverge. I will 
assume this divergence from the central axis is 
THE PHYSICS TEACHER ◆ Vol. 43, 2005 
cylindrically symmetric, which implies that the 
off-axis field has a radially inward component 
Bρ. (I am using ρ to denote the radial coordi-
nate, to avoid confusion with the resistance r 
of the ring.) The situation is sketched below. 
To make it easier to visualize how this magnetic 
field could arise, I have imagined that a bar 
magnet is being held in the air above the ring, 
with its south (S) pole pointing downward.
To oppose the decrease in the external mag-
netic flux through the ring, there must be an 
induced magnetic field pointing upward, in 
accordance with Lenz’s law. This induced field 
is sourced by an induced current I in the ring. 
The directions of the induced field and current 
are related by the right-hand rule, implying that 
I is counter-clockwise as seen from above. But 
we now have a current I in a radially inward 
magnetic field Bρ, which gives rise to an upward 
magnetic force F on the ring, as can be seen in 
the following cross-sectional sketch of the ring. 
(This magnetic force is therefore a drag force 
retarding the fall of the ring, consistent with the 
fact that Lenz’s law is an expression of energy 
conservation. In particular, the induced current 
gives rise to joule heating. This thermal energy 
comes at the expense of kinetic energy, just as in 
the familiar demo where a magnet is dropped 
down an aluminum pipe. To see it another way, 
the magnetic dipole moment of the ring points 
upward, which implies the north pole of this 
electromagnet is on top and gets attracted to the 
south pole of the external bar magnet, thus slow-
ing the ring’s fall.)
The magnitude of the upward drag force is
ρ
ρπ µ= =
( )
(2 ) 2 ,
B R
F I R B
R
where R = d/2 is the radius of the ring and µ 
= IπR2 is the magnitude of its magnetic dipole 
moment. I have explicitly indicated in the last 
expression in Eq. (1) that Bρ is to be evaluated 
at the radius R of the ring.
Assume the ring has some infinitesimal height 
dy. Then it encloses a small cylindrical volume as 
sketched below, with the bottom surface at posi-
tion y and the upper surface at y + dy.
The arrows denote the perpendicular compo-
nent of the external magnetic field striking each 
face, where I assume the radius of the ring is 
small enough that I can neglect the radial varia-
tion of By across either end cap of the cylinder. 
Gauss’s law of magnetism applied to this cylin-
der becomes
 (2)ρ π π
π
= ⇒
− − +
+ =
∫
2
2
0
( )2 ( )
( ) 0.
y
y
d
B R Rdy B y R
B y dy R
B Ai
 
Noting that By(y + dy) – By(y) = DBy(y), Eq. 
(2) can be rewritten as
ρ = =
( ) ( ) ( )
2 .y
B R dB y dB y
R dy dy
I dropped the subscript y on B in the last step, 
since we will evaluate this derivative along the 
axis of the ring where the external magnetic 
field is purely vertical. Substituting Eq. (3) into 
(1) now gives
µ= .
dBF
dy
In the case of a small electric dipole ˆpy in a 
nonuniform electric field ˆEy , the electric force 
F on the dipole can be found by evaluating the 
virtual work dW the field does on the dipole 
during an infinitesimal displacement dy,
 (5)
,
EFdy dW dU pdE
dEF p
dy
= = − =
⇒ =
 
since iEU = −p E is the electrostatic potential 
energy. Equation (4) is the magnetic analog of 
Eq. (5), despite the fact that iBU = − Bµ is 
an orientational not a translational potential 
energy! (Note in our case that both µ and B 
point upward, and hence UB is negative, but it 
increases toward zero as B decreases during the 
fall, resulting in an upward force F = –dUB/dy.)
Next let’s find the induced current in the ring 
of cross-sectional area A,
 (6)
ε
υ
υ
Φ
= = = = =
+ =0 0
1 1 ( )
(1 ) ,
d d BA A dy dB
I
r r dt r dt r dt dy
A d A
B ky B k
r dy r 
where  is the induced emf around and 
 the induced flux linking the ring, using 
Faraday’s law to relate these two quantities. 
Consequently, the magnitude of the drag force 
in Eq. (4) is
 . (7)
υ υ
= + =
2 2 2
0 0
0(1 )
AB k A B kd
F A B ky
r dy r 
But in terminal velocity, this upward drag force 
mustexactly balance the downward gravita-
tional force mg on the ring. Equating these two 
 THE PHYSICS TEACHER ◆ Vol. 43, 2005
forces and substituting A = πd 2/4 finally gives
 . 
 (8)2 4 2 2
0
16
.
mgr
d B k
υ
π
=
(Submitted by Carl E. Mungan, U. S. Naval 
Academy, Annapolis, MD)
The following participants were recognized as 
the winners of the April contest: 
•Craig Caylor (Westminster College, New 
Wilmington, PA) – faculty
•Alan J. DeWeerd (University of Redlands, Red-
lands, CA) – faculty
•Scott Saltman (Phillips Exeter Academy, Exeter, 
NH) – faculty
•Asif Shakur (Salisbury University, Salisbury, 
MD) – faculty
•H. Scott Wiley (Science Academy of South 
Texas, Weslaco, TX) – faculty
•Yufei Zhao, (Don Mills Collegiate Institute, To-
ronto, Canada) – high school student
Congratulations!
We would also like to recognize the following 
contributors:
Erick Blomberg, student (Bradley University, 
Freeport, IL)
Phil Cahill (Lockheed Martin Corp., Rosemont, 
PA)
Anthony Charanis (East Brunswick HS, High-
land Park, NJ)
Jim DiCarlo (Phillips Exeter Academy, Exeter, 
NH)
Michael C. Faleski (Delta College, Midland, 
MI)
Fernando Ferreira (Universidade da Beira Inte-
rior, Covilhã, Portugal)
John F. Goehl, Jr. (Barry University, Miami 
Shores, FL)
THE PHYSICS TEACHER ◆ Vol. 43, 2005 
Matthew Chua Chin Heng, student (Catholic 
Junior College, Singapore)
Art Hovey (Amity Regional HS, Woodbridge, 
CT)
David Jones (Miami Palmetto Senior HS, Mi-
ami, FL)
Alexander Kuzma, student (Pechersk Interna-
tional School, Kiev, Ukraine)
Chor Hang Lam, student (Charter School of 
Wilmington, Wilmington, DE)
Pui K. Lam (University of Hawaii, Honolulu, 
HI)
Eldin W. C. Lim (National University of Singa-
pore, Singapore)
Demetris Nicolaides (Bloomfield College, 
Bloomfield, NJ)
Bayani I. Ramirez (San Jacinto College South, 
Houston, TX)
Gregory Ruffa (University of Minnesota, Min-
neapolis, MN)
Peter Sadowski, student (Archbishop Murphy 
HS, Everett, WA)
Gregory Stupp, student (Palmetto Senior HS, 
Miami, FL)
Leo H. van den Raadt (Heemstede, The Neth-
erlands)
John S. Wallingford (a hillside in the Western 
NC Mountains)
LeRoy Wenstrom (Mississippi School for Math 
& Science, Columbus, MS)
Ali Yazdi (Jefferson State Community College, 
Birmingham, AL)
We look forward to your contributions in the 
future!