Hall Effect
z
Electrons moving in a
wire. In this case the wire
is a rectangular slab with
width, d, and thickness, l.
z
The total cross sectional
area of the wire is A=l d.
z
B field points into the
screen.
Hall Effect
z
z
z
Electrons moving in a
wire (= current) can be
deflected by a B field
called the Hall effect
Creates a Hall potential
difference, V, across the
wire
Can measure the wire’s
charge density when at
equilibrium FE = FB
Hall Effect
z
z
z
z
Electrons have drift
velocity, vd in direction
opposite the current, i
B field into page causes
force, FB to right
Electrons pile up on
right hand side of strip
Leaves + charges on left
and produce an E field
inside the strip pointing
to right
Hall Effect
z
E field on electron
produces a FE to the left
z
Quickly have equilibrium
where FE = FB
E field gives a V across
the strip
z
V = Ed
z
Left side is at a higher
potential
Hall Effect
z
Can measure the number
of charge carriers per unit
volume, n, at equilibrium
FE = FB
FE = qE
r r
FB =| qv × B |
eE = evd B sin(90)
E = vd B
Hall Effect
z
Remember from Chpt. 27
that drift speed is
J
i
=
vd =
ne neA
iB
E = vd B =
neA
iB
n =
EeA
Hall Effect
z
Replacing E by
V = Ed
iB
iBd
n=
=
EeA
VeA
Hall Effect
z
If l is the thickness of the
strip
A
l =
d
z
Finally get
iB
n =
Vle
Magnetic Fields: Circular Motion
z
FB continually deflects path
of charged particles
z
If v and B are ⊥, FB
causes charged particles
to move in a circular path
z
If B points towards you
z
z
+ particles move clockwise.
– particles move counter
clockwise.
r
r r
FB = q v × B
Magnetic Fields: Circular Motion
z
Derive radius of circular path for particle of charge, q,
and mass, m, moving with velocity, v, which is ⊥ to B
field
r
r
FB =| qv × B |= qvB sin φ = qvB
z
z
Newton’s second law for circular motion is
Setting the forces equal and solving for r
z Faster particles move in larger circles
mv
r=
qB
v2
F = ma = m
r
v2
qvB = m
r
Exercise
A proton and an electron travel at
same v (in the plane of the page).
z
z
There is a B field into the page.
A) Which particle follows the
smaller circle?
r ∝ m/q , |q_e |=|q_p |=e, and mp > me ,
so the electron has the smaller circle
z
z
B) What direction does the electron move in?
Clockwise
Magnetic Fields: Circular Motion
Period, T, is the time for
one full revolution
2π r
2 π mv
2π m
=
=
T =
v
v qB
qB
z
z
Frequency, f, is the number of
revolutions per unit time
1
qB
f =
=
T
2π m
z
z
qB
ω = 2π f =
m
Angular frequency, ω, is
Only depend on q and m but not v
Cyclotron
z
Cyclotron
z
z
z
z
z
Particles starts at the center.
They circulate inside 2 hollow
metal D shaped objects
Alternate the electric sign of the
Dees so V across gap alternates
(the oscillator does this).
Whole thing immersed in
magnetic field B (green dots
pointing out of page) ⊥ to v
B approximately 1-10 T (tesla).
Cyclotron
z
Cyclotron
z
z
Proton starting in center
will move toward
negatively charged Dee
Inside Dee E field = 0
(inside conductor) but B
field causes proton to
move in circle with radius
which depends on v
mv
r=
qB
Cyclotron
z
Cyclotron
z
z
When proton enters gap
between Dees E field is
flipped so proton is again
attracted to negatively
charged Dee
Every time proton enters
gap the polarity of the
Dees is changed and the
proton is given another
kick (accelerated)
Cyclotron
z
Cyclotron
z
Key is that the frequency,
f, of the proton does not
depend on v and must
equal the fosc of the Dees
f = f osc
1
qB
f = =
T 2π m
qB = 2π mf osc