magnetic field B
at the electron path
r
path of
electrons
evacuated
doughnut
o
m
v
F
B
magnetic field B
directed up
E
o
B
directed
up and increasing
E
Figure 15
When the strong central field
B
0
in the
betatron is rapidly increased, it produces a
circular electric field that is used to
accelerate the electrons.
The electric field
E
is related to the flux
Φ
B
of the central field
B
0
by Faraday's law
E
⋅
d
=
–
d
Φ
B
/
dt
.
Figure 14b
Top view of the betatron showing the
evacuated doughnut, the path of the
electrons, and the magnetic fields
B
0
in the
center and
B
r
out at the electron path.
In
order to keep the electrons moving on a
circular path inside the doughnut, the
magnetic force
F
B
=
–
e
v
×
B
r
must have a
magnitude
F
B
=
m
v
2
r
where r is the radius of
the evacuated doughnut.

30-18
Faraday's Law
Let us consider an explicit example to get a feeling for
the kind of numbers involved.
In the 100 MeV betatron
built by General Electric, the electron orbital radius is
84 cm, and the magnetic field
B
0
is cycled
from 0 to
.8 tesla in about 4 milliseconds. (The field
B
0
is then
dropped back to 0 and a new batch of electrons are
accelerated. The cycle is repeated 60 times a second.)
The maximum flux
Φ
m
through the orbit is
Φ
m
=
B
0
max
π
r
2
= .8 tesla
× π ×
(.84m)
2
Φ
m
= 1.8 teslam
2
If this amount of flux is created in 4 milliseconds, then
the average value of the rate of change of magnetic flux
Φ
B
is
d
Φ
B
dt
=
Φ
m
.004 sec
=
1.8
.004
= 450 volts
Thus each electron gains 450 electron volts of kinetic
energy each time it goes once around its orbit.
Exercise 3
(a) How many times must the electron go around to
reach its final voltage of 100 MeV advertised by the
manufacturer?
(b) For a short while, until the electron’s kinetic energy
gets up to about the electron’s rest energy
m
0
c
2
, the
electron is traveling at speeds noticeably less than c.
After that the electron’s speed remains very close to c.
How many orbits does the electron have to make before
its kinetic energy equals its rest energy?
What fraction
of the total is this?
(c) How long does it take the electron to go from the point
that its kinetic energy equals its rest energy, up to the
maximum of 100 MeV?
Does this time fit within the 4
milliseconds that the magnetic flux is being increased?
TWO KINDS OF FIELDS
At the beginning of the chapter we showed that the line
integral
E
⋅
d
around a closed path was zero for any
electric field produced by static charges.
Now we see
that the line integral is not zero for the electric field
produced by a changing magnetic flux.
Instead it is
given by Faraday
’
s law
E
⋅
d
=
–
d
Φ
B
/dt
.
These
results are shown schematically in Figure (16) where
we are looking at the electric field of a charged rod in
(16a) and a betatron in (16b).
In Figure (17), we have sketched a wire loop with a
voltmeter, the arrangement we used in Figure (12) to
measure the
E
⋅
d
.
We will call this device an
“
E
⋅
d
meter
”
.
If you put the
E
⋅
d
meter over
the changing magnetic flux in Figure (16b), the voltme-
ter will show a reading of magnitude V
=
d
Φ
B
/dt
.
If
we put the
E
⋅
d
meter over the charged rod in

#### You've reached the end of your free preview.

Want to read all 28 pages?

- Fall '19
- Magnetic Field, Faraday