Physics Challenge for
Teachers and Students
Solution to April 2008 Challenge
◗ A BLT Sandwich
Challenge: A square loop made of wire with negligible
resistance is placed on a horizontal frictionless table as
shown (top view). The mass of the loop is m and the
length of each side is b. A nonuniform vertical magnetic
field exists in the region; its magnitude is given by the
formula B = B0 (1 + kx), where B0 and k are known
constants. The loop is given a quick push with an initial
velocity v along the x-axis as shown. The loop stops after
a time interval t. Find the inductance L of the loop.
b
b
using Ohm’s law in the last step.1 But we are told
that the resistance R of the loop is zero. Therefore
b2dB / dt = LdI / dt. (That is, the back emf exactly
cancels the externally induced emf.) Integrating
both sides with respect to time then implies
that b2B = LI + c, where c is some constant that
depends on the initial conditions. Let’s assume
that the loop starts out with zero current before
it is placed on the table into the magnetic field.
Then c = 0 and the net flux, ext – back, linking
the loop is always zero. We conclude that the current in the loop is
I=
v
b2
B0 (1 + kx )
L
(2)
in the clockwise direction. But we now have
four wire segments (i.e., the four sides of the
Solution: Suppose that the external magnetic field loop) carrying current I in a magnetic field B(x).
Bext B = B0(1 + kx) is directed out of the page. Consequently there is a magnetic force on each
As the loop moves to the right, the externally
of them. The forces on the top and bottom sides
2
linked flux through the loop ext = Bextb increas- are equal and opposite. That on the right side is
es; its time derivative results in an emf of
–IbB(x), where the minus sign indicates that it is
ext = –d ext /dt, where the minus sign indicates directed leftward since I is down the page and B
that it drives a current I around the loop in the
is out of the page, and where we let x measure the
clockwise direction. Noting that dBext /dt = B0k
position of the right side of the loop. Thus the left
dx/dt is not constant (since the speed dx/dt of the side is located at position x – b and the force on
loop decreases monotonically from υ initially to
it is +IbB(x – b). Consequently the net magnetic
0 finally), the current I cannot be constant either. “braking” force on the loop is
We therefore have a time-varying clockwise curF = −Ib {B0[1 + k x ]− B0[1 + k( x − b )]}
rent in an inductor L and so we generate a back
emf back = –LdI / dt –d back/dt in the coun(3)
4 2
kb
B
0
ter-clockwise direction. Consequently the net
=−
(1 + kx )
L
clockwise emf around the loop is
after substituting Eq. (2).
dI
2 dB
(1)
ε =b
− L = IR
dt
dt
We can write this result in the form F =
–K (x – x0), where K = k2b4B20/L and x0 = –1/k
x
THE PHYSICS TEACHER ◆ Vol. 46, 2008
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are constants. This has the form of a Hookean
restoring force, where the effective spring constant
is K and the equilibrium position is located at x0.
Assume that the loop is originally placed at the
equilibrium position so that it doesn’t move before
it is given the impulsive push. The loop will subsequently oscillate about that position forever after
it is pushed, since there is no resistive dissipation
of energy. It first reaches the rightward turning
point in a time t equal to one-quarter of a period
T = 2π m / K so that
2
mL
π
t=
2
2 k b 4 B02
⇒
1 ⎛ 2kb 2 B0t ⎞⎟⎟
L = ⎜⎜⎜
⎟ .
m ⎜⎝ π ⎟⎠ (4)
By using the standard textbook expression for the
magnetic field of a straight wire segment, one can
fairly easily show that the inductance of a square
loop is
1 1
L=
2μ0b
π
∫∫
r /b 0
xdxdy
2
y x +y
2
,
(5)
where r is the radius of the wires and is assumed
to be small compared to b. This integral works out
to be approximately
L≈
2μ0b ⎛⎜ b ⎞⎟
ln ⎜⎜ ⎟⎟ ,
⎝r ⎠
π
(6)
which diverges if we let r go to zero. Instead, taking r = b/1000, this result gives L 4bμ0, where
μ0 = 0.4π μH/m is the permeability of free space.
For example, if b = 1 m then L 5 μH. The
important point is that Eq. (4) then implies that
t2 ∝
m
b(ΔB )2
,
(7)
assuming r << b, where ΔB B0kb is the variation in the external magnetic field strength across
the width of the loop. The stopping time increases
if the loop’s mass (inertia) is increased, while the
time decreases if the size of the loop or the magnetic field gradient is made larger.
Addendum
I would like to thank José Íñiguez for pointing out
that calculating the externally linked flux as ext =
Bextb 2 is not correct because the external magnetic
THE PHYSICS TEACHER ◆ Vol. 46, 2008
field varies across the length of the loop. Instead,
the external flux is
Φext = ∫ B dA =
x +b
∫
Bext ( x ′ )bdx ′
(8)
x
whose time derivative is
d Φext dx d
=
dt
dt dx
=υ
x +b
∫
Bext ( x ′ )bdx ′
(9)
x
ΔB 2
b .
Δx
But because B is a linear function of x, ΔB/Δx =
dB/dx = B0k, , and so Eq. (1) and the remainder
of the solution is nevertheless correct.
(Contributed by Carl E. Mungan, U. S. Naval
Academy, Annapolis, MD)
1.
In traditional textbook presentations of Faraday’s law
for the current induced in a loop placed in a varying
magnetic field, the loop’s inductance is assumed to
be negligible so that ext = IR. In the present problem, the loop’s resistance is instead negligible. More
generally, Eq. (1) can be viewed as the equation of
an LR series circuit with ext playing the role of the
usual battery.
We would also like to recognize the following
contributors:
Sanjeet Singh Adarsh (PTC, Pune, India)
Marianne Breinig (The University of Tennessee,
Knoxville, TN)
Alan J. DeWeerd (University of Redlands,
Redlands, CA)
F. Javier Doblas (Escuela Técnica Superior de
Ingenieros, Sevilla, Spain)
Don Easton (Lacombe, Alberta, Canada)
Bruce Gordon (Kimball Union Academy, Meriden,
NH)
Fredrick P. Gram (Cuyahoga Community College,
Cleveland, OH)
Art Hovey (Milford, CT)
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J. Iñiguez (Universidad de Salamanca, Salamanca,
Spain)
John Mallinckrodt (Cal Poly Pomona, Pomona,
CA)
Stephen McAndrew (Trinity Grammar School,
Summer Hill, NSW, Australia)
Bayani I. Ramirez (San Jacinto College South,
Houston, TX)
Finally, we would like to acknowledge Bartley
L. Cardon (MIT Lincoln Laboratory, South
Portland, ME), whose name was omitted from the
list of the March Challenge solvers, and apologize
to Dario Castello (Vasquez High School, Acton,
CA), whose last name was misspelled there.
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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THE PHYSICS TEACHER ◆ Vol. 46, 2008