Download PWE 19-3: Magnetic Levitation

Example 19-3 Magnetic Levitation
You set up a uniform horizontal magnetic field that points from south to north and has magnitude 2.00 * 1022 T. (This
is about 400 times stronger than Earth’s magnetic field, but easily achievable with common magnets.) You want to place
a straight copper wire of diameter 0.812 mm in this field, then run enough current through the wire so that the magnetic
force will make the wire “float” in midair. This is called magnetic levitation. What minimum current is required to make
this happen? The density of copper is 8.96 * 103 kg>m3.
Set Up
In order to make the wire “float,” there must be
an upward magnetic force on the wire that just
balances the downward gravitational force.
We know from Equation 19-5 that to maximize
the magnetic force, the current ­direction should
s.
be perpendicular to the ­magnetic field B
The right-hand rule then shows that the current should flow from west to east so that the
magnetic force is directed upward. We’re not
given the mass of the wire, but we can express
the mass (and hence the gravitational force on
the wire) in terms of its density. We’re also not
given the length of the wire; as we’ll see, this
will cancel out of the calculation.
Solve
Write expressions for the two forces (magnetic
and gravitational) that act on the wire.
Magnetic force on a
current-carrying wire:
F = i/B sin u
magnetic force
Definition of density:
m
r = V
(11-1)
Volume of a cylindrical
wire of cross-sectional
area A and length /:
B
i
(19-5)
up
West
East
South
i
North
gravitational force
down
V = A/
Assume the wire has length /. Since the current flows in a direction
perpendicular to the magnetic field, the angle u in Equation 19-5
is 90°. So the magnitude of the upward magnetic force is
F = i/B sin 90 = i/B
The magnitude of the downward gravitational force on the wire of
mass m is
w = mg
From Equation 11-1, the mass equals the density of copper multiplied
by the volume of the wire:
m = rV = rA/
The cross-sectional area A of the wire is that of a circle of radius r:
A = pr 2 so m = rA/ = r1pr 2 2/
So the magnitude of the gravitational force is
w = mg = r1pr 2 2/g
If the wire is floating in equilibrium, the
upward magnetic force must just balance the
downward gravitational force. Use this to solve
for the required current i.
In equilibrium the net vertical force on the wire is zero:
F 2 w = 0, so F = w and i/B = r1pr 2 2/g
The length / of the wire cancels out of this equation
(both the magnetic force and the gravitational force are
proportional to /):
iB = r(pr2)g
Solve for the current i:
i =
=
r1pr 2 2g
B
18.96 * 103 kg>m3 2 1p2 10.406 * 10-3 m2 2 19.80 m>s 2 2
= 2.27
Reflect
2.00 * 10-2 T
kg
T s 2
= 2.27 A
[Check on units: We know from Section 19-3 that
1 T = 1 N> 1A # m), and we also know that 1 N = 1 kg # m>s 2.
Therefore 1 T = 1 kg> 1A # s 2 2, and so 1 kg> 1T # s 2 2 = 1 A.]
A current of 2.27 A is relatively small, so this experiment in magnetic levitation is not too difficult to perform. Note that
the required current i is inversely proportional to the magnitude B of the magnetic field. You can see that if you tried to
make a wire “float” using Earth’s magnetic field, which is about 1>400 as strong as the field used here, you would need
to use an immense current of 400 * 2.27 A = 909 A. That’s not practical because a current of that magnitude would
cause the wire in this example to melt! (Recall from Equation 18-24 in Section 18-6 that the power into a resistor with
resistance R that carries current i is P = i2R. The wire in this example has a small cross-sectional area, so its resistance R
will be fairly large. The power delivered to the wire by a 909-A current will quickly increase its temperature to above the
melting point of copper, 1085°C.) So you needn’t worry about any of your electrical devices floating in midair when you
turn on the current.
A practical application of magnetic levitation is train design. By using magnetic forces to make a train float just
above the track, the rolling friction between the wheels and the track can be completely eliminated and very high speeds
achieved. (Magnetic forces are also used to propel the train forward.) A train of this type in commercial operation in
Shanghai, China reaches a top speed of 431 km>h (268 mi>h). Such train lines require special magnets and wires capable
of sustaining very high currents.