Geology Reference
In-Depth Information
of that predicted by (
4.34
). The flux through any
cross-section
S
of the solenoid is:
Z
B
dS
D
BA
D
0
NA
ˆ.S/
D
L
I (4.35)
S
where
A
is the cross-section area,
N
is the total
number of turns, and
L
is the length of the
solenoid, so that
n
D
N
/
L
.Now,ifwevarythe
current
I
through the coil, we obtain a variation
of flux for each turn. Consequently, by Faraday's
law (
4.28
)anemf
appears, which
opposes
the
variations of current from which it was gener-
ated (remind the minus sign in Faraday's law).
The resulting additional current is said to be
self
-
induced
. To calculate the emf through the
solenoid, we multiply the emf of each turn by
N
:
E
@t
D
0
N
2
A
E
D
N
@ˆ
dI
dt
L
dI
dt
(4.36)
L
in (
4.36
), which depends from
the solenoid geometry, is called
self
-
inductance
of the solenoid and has units of Henry [H].
A very interesting phenomenon occurs if we
use a solenoid to generate the magnetic field
of Fig.
4.3
. It is not difficult to show that in
this instance the system becomes unstable. The
modified circuit is illustrated in Fig.
4.5
.Wehave
again a rotating disk and an external initial field
B
0
. As in the example of Fig.
4.3
, the induced
current flows toward the periphery of the disk.
However, in this new system, the contacts
C
and
A
and the disk shaft allow the current to
flow externally through a solenoid
S
.Thisflow
generates a new axial magnetic field
B
having a
magnitude determined by (
4.34
). At this point,
removing the initial field
B
0
and keeping the disk
in motion, we expect that the spontaneous field
B
continues to be sustained by the current that
it itself provides to generate. In this instance,
apart from a possibly weak initial magnetic field
B
0
seeding the system, no other external field is
needed to sustain the dynamo. Therefore, after
removal of the seed field
B
0
, the dynamo only
The quantity
L
Fig. 4.5
Self-exciting dynamo. A disk
D
rotates with
angular velocity
¨
in the axial magnetic field B, generated
by the current flowing through the solenoid
S
. Note the
direction of the current flow in the wire
requires a continuous supply of mechanical en-
ergy to drive the electrical conductor sufficiently
fast for self-excitation to be possible.
To understand what really happens, let us as-
sume that before the initial time
t
D
t
0
the circuit
be kept open, for example by disconnecting
C
or
A
,sothat
B
(0)
D
0 and the emf is only manifest
through a difference of potential
V
between the
contacts
C
and
A
.From(
4.30
) it results:
2
¨B
0
b
2
a
2
1
V
0
D
V.0/
D
(4.37)
At
t
D
t
0
the circuit is closed and a current will
flow through the solenoid, generating a magnetic
field
B
D
B
(
t
).
