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PROFESSOR: The steady-state magnetic levitation
of current carrying conductors over a ground plane
demonstrates magnetic forces due to conduction currents.
This cylindrical conductor carries the current I.
The current is returned in the conducting plane just below.
It is alternating at a high enough frequency
that both the conductor and the ground plane
act as perfect conductors.
The current in the conductor and oppositely directed current
in the ground plane distribute themselves
into surfaces in just such a way that the magnetic flux
normal to the surfaces of the conductor and the ground plane
tends to be zero.
The magnetic field between the conductor and the ground plane
is the same as that of a line current
and its oppositely directed image.
These two currents are positioned
so that the magnetic field is tangential to the ground plane
and to the cylindrical conductor.
Parallel currents in opposite directions
cause a force of repulsion.
To better understand the repulsive force,
let's look at how the current and magnetic field interact
here, at the bottom surface of the cylindrical conductor.
At a given instant, the current density
is in the axial direction, while the magnetic field is phi
directed.
The current density is in the axial direction.
The magnetic field is phi directed.
The current density crossed into the magnetic flux density
is the force density, acting on the bottom of the conductor.
The force density acting on other parts of the conductor
also has horizontal components, but these cancel.
There is even some force density at the top of the conductor
acting downward, where J is the same,
but B has reversed direction.
But that force is smaller, because B
is smaller on the upper surface than on the lower surface.
This is because of the proximity of the lower
surface of the cylindrical conductor to the ground plane.
The net force f on the conductor is upward.
When the distance xi from the ground playing
to the conductor's center is much larger than the cylinder
radius R, the upper image current
is at about the cylinder's center,
while the image current below is the same distance
xi below the ground plane.
The Lorentz force is the length of the wire l
times the cross product of the current i in the conductor
with the magnetic field B from the image current, a distance 2
xi below the conductor.
We can equivalently determine the net force
per unit length acting on the cylindrical conductor
by using conservation of energy.
The vertical position of the conductor is xi.
The inductance for a cylindrical wire of length
l is a function of the vertical displacement xi.
R is the conductor radius.
When the conductor radius is very small
compared to the vertical displacement xi,
the inductance formula simplifies.
Conservation of energy tells us that the force in the wire
is one half i squared times the derivative of the inductance
with respect to xi.
When xi is much greater than R, the force
is identical to that derived using the Lorentz force.
In the experiment, the current-carrying wire
has been wound into a pancake-shaped coil.
Line current at 60 hertz is supplied to the coil
through flexible wires, which are connected to a variac.
Here is the actual experiment, with the pancake coil
excited by 60 hertz current from this variac.
The current is measured with this clip-on ammeter
and indicated on the oscilloscope.
The ground plane is this aluminum sheet.
Its thickness is 1.27 centimeters.
The coil has a thickness of about 1 and 1/2 centimeters.
[VIBRATIONS]
A typical current in the coil is 20 to 30 amperes,
so it's made of large-diameter wire to reduce its resistance.
The 250 turns are of number 10 aluminum wire.
Aluminum is chosen for its lightweight.
Its density is 2.7 grams per cubic centimeter.
So that, for a given current and thus magnetic force,
the coil will rise higher than if made of copper.
Although copper is a better conductor
by a factor of about 1.6, its density
is about 3.3 times greater than aluminum.
The scope trace indicates the coil current.
Here again, is what happens as the current is raised.
[VIBRATIONS]
The force on the conductor results
in vibrations, which we hear as telltale evidence
that the current is being increased.
The force is proportional to the square of the 60 hertz current.
These vibrations result from the 120 hertz second harmonic
component of the force.
When the RMS current is large enough
that the Lorentz force can overcome the coil weight,
the coil lifts off the ground plane.
Increasing the current further results
in an increase in the levitation force at a given altitude,
so the coil rises to make the upwards magnetic force
equal to that of the coil weight.
Let's see if we can estimate how high the pancake coil will
rise for a given coil current.
We are limited in height by the maximum current
before we blow a fuse.
Focusing attention on a single turn of the circular coil
with radius a, with a much larger than the wire diameter,
we can approximately use our inductance formula
for a straight wire of length l if we
take l to just be the coil circumference 2 pi a.
If we had N concentrated turns in our coil,
the inductance increases by N squared.
One factor N arises because there
are n turns, each carrying the current I. Thus
the net current that is effective in producing
the magnetic field is n I.
The second factor of N is because the flux
length by all N turns is n times that length by 1 turn.
So the inductance of the n turn coil
is roughly that of a one turn coil, multiplied by N squared.
In our experiment we have a pancake coil, rather than
a single concentrated circular coil with a single radius.
Our 250 turn coil has an average radius of about 9 centimeters.
Because of the large variation in radius from 2
and 1/2 to 16 centimeters, it is not clear
what effect of radius a to use in our approximate inductance
formula, so we choose to write it
in a generalized form in terms of the parameter--
L0.
We can estimate a value of L0 equal mu0
a N squared by using the average radius of 9 centimeters,
so that L0 equals 7 millihenries.
To estimate L0 for measurements, we
use this inductance meter to measure the coil inductance
as a function of the coil position xi,
where xi is measured from the ground plane
to the midpoint of the coil.
The coil inductance is measured at 1 kilohertz,
so that the ground plane acts as a perfect conductor.
At 1 kilohertz, the skin depth is
much less than the thickness of the conducting sheet.
The coil inductance increases as the coil height is increased.
Here, when the coil rests on the ground plane,
the coil inductance is about 1.7 millihenries
while, when the center is at about 5 centimeters
above the ground plane, the coil inductance
is about 5 millihenries.
Our measurements are shown as dots on the plot.
L0 has been estimated to give a perfect fit
to this mid-range data point, at xi equals 7.25 centimeters.
The value of L0 fitted this way is 1.3 millihenries, about
one-fifth what we estimated.
The discrepancy is, in part, due to the large size
of the coil with the winding distributed
over radial and axial positions.
Our model assumed all the n turns to be concentrated.
When we use L0 equal 1.3 millihenries,
our formula approximates the position dependence
of inductance over most of the range of xi.
Now that we know from empirical measurements
how the coil's self-inductance depends on coil position,
we find the magnetic force as a function of coil position
by differentiating the coil inductance with respect
to position xi.
The gravitational force that balances the magnetic force
is constant.
For a given RMS coil current i, the coil height xi
will adjust itself so that the magnetic force just
equals the coil weight--
Mg.
We solve this equation for i.
The current required to just lift the coil off the ground
plane can be estimated using as the position xi half the coil
thickness.
Then, with the pancake coil mass of 1.8 kilograms in the one g
gravity field of 9.8 meters per second squared,
with xi equals 0.75 centimeters--
half the coil height--
and with L0 equal 1.3 millihenries,
we calculate the lift-off coil current
to be about 14.3 amperes RMS.
We reconnect the pancake coil to the variac.
The clip-on ammeter measures the current,
which is displayed on the oscilloscope.
We raise the current slowly and find the coil lifts off--
[VIBRATIONS]
--at about 15.5 amperes RMS.
What we predicted was 14.3 amperes RMS.
As we raise the current further, the coil height
above the ground plane increases.
Here, the coil center rises by 2.25 centimeters
at 20 amperes RMS.
We take other data points by increasing the current further.
Here is our 20 ampere data point.
The solid line shows the estimated mid coil position
as a function of RMS coil current.
Discrepancies between theory and experiment
are due to our many approximations.
In particular, errors are amplified because our force
computation depends on the derivative of the inductance
with position xi.
Also, how good an approximation is it to assume that,
at 60 hertz, the coil and ground plane are perfect conductors?
The skin depth in aluminum at 60 hertz is 1.1 centimeters.
At first, one may think that, for the assumption
of a perfectly conducting ground plane to be accurate,
the frequency must be high enough
that the ground plane thickness is greater than the skin depth.
For our experiments thus far, the aluminum ground plane
thickness has been 1.27 centimeters.
This thickness is about a skin depth.
To demonstrate the dependence on ground plane thickness,
we can try a ground plane with half the thickness.
This plate has a thickness of 6.35 millimeters.
This is about half the skin depth.
Remember that with 1.27 centimeter thickness,
lift off occurred at 15.5 amperes.
With this half skin depth thick ground plane, lift off occurs--
[VIBRATIONS]
--at 16.3 rather than 15.5 amperes RMS.
This current is only slightly higher
than with the 1.27 centimeter thick ground plane.
Even though the ground plane thickness is now
about half the skin depth, it still lacks essentially
as a perfect conductor.
With the 1 millimeter thick plane,
the coil does not lift up even at 30 amperes RMS.
[VIBRATIONS]
The ground plane is now so thin that most of the magnetic field
penetrates through the ground plane.
We just blew a fuse again, but this time because of an arc.
The enamel insulation on the wires
may not be enough to prevent arcing to the sheet.
We have just seen an example of this.
Let's try it again now, with some gaffer's tape for better
insulation on the bottom.
[VIBRATIONS]
Here it is again, with a current of 30 to 40 amperes.
The coil doesn't lift off, but the induced current
does make the ground plane very hot.
Our coil has the making of a burner for a stove.
Our thin sheet could be the bottom of a pan.
Our experiment has been turned upside down.
[VIBRATIONS]
The protective border prevents spillage.
Let's turn the current up and let it heat for a while.
[VIBRATIONS CONTINUE]
[TAPPING]
[VIBRATIONS CONTINUE]
It takes a while, but it does cook.
The experiment demonstrates the magnetic induction
of current in conductors and both the resulting Lorenz
magnetic force of repulsion and the heating.