Chemistry Reference
In-Depth Information
Whether an object will float or not is determined by the balance between
the magnetic force and gravity. For a diamagnetic material, the induced
magnetic moment, m dia
M V , where M is the magnetisation (given by
eq. (6.6)) and V the volume, with m dia then given by
=
= χ
VB
µ
m dia
(6.44)
0
By integrating the work
B as the field is increased from zero to B
we can obtain the total magnetic energy of the object. Adding this to the
gravitational energy, mgz , where m is the mass and z the vertical position
coordinate, the total energy E is given by
d m dia
·
χ
V
B 2
E
=
mgz
(6.45)
2
µ
0
For the object to float, the total vertical force, F z
=−
E
/∂
z , must vanish so
that
V
χ
µ
B
B
mg
+
=
0
(6.46)
z
0
The equilibrium condition then becomes
B
B
= µ
ρ
g
0
(6.47)
z
χ
which we note involves only the density,
ρ
, of the levitated object, not its
10 3 kg
m 3 and
10 5 for adiamagnet or
10 3
mass. Ifwe take
ρ
/
χ ∼−
χ
1000 T 2 m 1
for a paramagnet, magnetic levitation then requires B
B
/∂
z
or 10 T 2 m 1 respectively. Taking l
0.1m as the typical size a of high field
magnet and assuming
l , we find that fields of order 1 or 10 T
are sufficient to cause levitation of para- and diamagnets.
We have not addressed here the equilibrium against horizontal displace-
ment - this ismore complex, but has been treated by Berry andGeim(1997),
who prove that a diamagnet can float in stable equilibriumabove amagnet,
whereas a paramagnet (which floats beneath the magnet) is always unsta-
ble to horizontal movements, accelerating away when displaced from the
vertical symmetry axis, as outlined in problems 6.6-6.8 below.
We conclude that magnetic forces are nevertheless generally small.
We think of them as significant because they can be comparable to or
even exceed gravitational forces. Yet if we assume that there are N
B
/∂
z
B
/
10 29 atoms
m 3 in a diamagnetic solid then from eq. (6.45) the energy stored
in a 10 Tfield is less than 1
/
eVper atom- orders ofmagnitude smaller than
the electrostatic interactions we shall be considering in the next chapter.
µ
 
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