Physics 212
Classical and Modern Physics
Spring 2008
Solutions for HandIn Set #5
Problem A28 — Falling Loops
As the loop falls out of the magnetic field region, the
flux through the loop decreases, and so an emf drives a
current through the loop. We can write the flux as
Φ
B
=
NBA
cos
θ
=
B
0
wx,
where
N
= 1, cos
θ
= 1, and
x
is the length of the loop
still in the magnetic field, so that
A
=
wx
. The emf is
then just the time derivative of the flux,
E
=
d
Φ
B
dt
=
B
0
wv,
since
dx
dt
=
v
, the speed at which the loop falls.
The
magnitude of the induced current is
I
=
E
R
=
B
0
wv
R
.
Two forces act on the loop: the force of gravity acts
down, and there is an upward directed magnetic force
acting on the counterclockwise current.
Newton’s second law gives
F
net
=
ma,
or
F
mag

F
grav
=
ma.
The magnitude of the magnetic force is given by
F
mag
=
IwB
=
B
0
wv
R
wB
0
=
B
2
0
w
2
v
R
which grows linearly with speed
v
.
After the loop is
dropped, the speed (and thus the magnetic force) grows
until the magnetic force and the gravitational force are
equal in magnitude. When this occurs the acceleration
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 Spring '08
 Ladd
 Physics, Magnetic Field, loop, BA cos

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