Transcript Part 3 Set 1 - FacStaff Home Page for CBU

Magnetism
We are probably all familiar with the fact that
magnets repel and attract each other. This is
similar to electric charges. This suggests
that we can propose the following law
(like Newton’s Law of Gravity and Coulomb’s
Law for Electricity):
Fmagnetic = X p1 p2 / r122
where X would be the magnetic constant,
analogous to G and k, that describes the
strength of the magnetic force.
Magnetic Poles
Fmagnetic = X p1 p2 / r122 . The “p” in the
equation stands for pole. In magnetism, we
have two kinds of poles, and we call them
North and South. Like poles repel and
unlike poles attract, just as electric charges do.
However, unlike charges, we always have
two poles! If we break a magnet (which
has two poles) in half, we have not
separated the two poles, rather we have two
(smaller) magnets that both have two poles!
Magnetic Poles
N
N
S
S
N
S
Break one bar magnet in half, and you have
two smaller bar magnets!
Magnetism
Since we cannot seem to isolate one magnetic
pole like we could electric charges, the
force equation that is similar to Newton’s
and Coulombs Laws turns out to be not very
useful.
We do have a more useful alternative, though.
It turns out that charges experience a force
when moving near magnets. Note: the
charge must be moving near a magnet,
not just being near a magnet!
Magnetic Force
Just as charges set up an electric field, and other
charges in the vicinity feel an electric force due to
that electric field (Fel = qE), we can work with the
idea that magnets set up a magnetic field in space,
and charges moving through that field experience
a magnetic force. In addition to the necessity
of moving, there is one more difference:
the direction of the magnetic force is
perpendicular to the magnetic field
direction, and perpendicular to the
velocity of the moving charge!
Magnetic Force Law
magnitude:
Fmagnetic = q v B sin(qvB)
direction: right hand rule:
thumb = hand x fingers
Point your hand in the direction of v, curl you
fingers in the direction of B, and the force
will be in the direction of your thumb; if
the charge is negative, the force direction
is opposite that of your thumb.
Units
Fmagnetic = q v B sin(qvB)
This law effectively defines the magnetic
field, B (just like Felectric=qE defined E).
The MKS units of B are:
Tesla = Nt / [Coul-m/s] = Nt-sec / Coul-m .
This unit turns out to be a very large one. We
have a smaller unit, a Gauss, where:
10,000 Gauss = 1 Tesla .
Example
A proton is moving at a speed of 3 x 104 m/s
towards the West through a magnetic field
of strength 500 Gauss directed South. What
is the strength and direction of the magnetic
force on the proton at this instant?
v = 3 x 104 m/s, West
B = 500 Gauss * 1 Tesla/10,000 Gauss = .05 T,
South;
q = +e = 1.6 x 10-19 Coul.
Example, cont.
magnitude:
Fmagnetic = q v B sin(qvB)
direction: right hand rule
magnitude: F = (1.6 x 10-19 Coul) * (3 x 104 m/s)
* (.05 T) * sin(90o) = 2.4 x 10-16 Nt.
direction: thumb = hand x fingers
= West x South = UP.
Note: although the force looks small,
consider the acceleration: a = F/m =
2.4 x 10-16 Nt / 1.67 x 10-27 kg = 1.44 x 1011 m/s2.
Magnetic Forces
We’ll play with magnetic forces on moving
charges in the Magnetic Deflection
experiment in lab. At that time we’ll also
discuss and experiment with the earth’s
magnetic field.
Magnetic Force and Motion
Since the magnetic field is perpendicular to
the velocity, and if the magnetic force is the
only force acting on a moving charge, the
force will cause the charge to go in a circle:
SF = ma,
Fmag = q v B, and acirc = w2r = v2/r
gives: q v B = mv2/r, or r = mv/qB .
Mass Spectrograph
We can design an instrument in which we can
control the magnetic field, B. If we ionize
(almost always singly) the material, we
know the charge, q. We can use known
voltages to get a known speed for the ions,
v. We can then have the beam circle in the
field and hit a target, and from that we can
measure the radius, r. Hence, we can then
calculate the mass using r = mv/qB .
Example
To see if this is really feasible, let’s try using
realistic numbers to see what the radius
should be for a proton moving in a B field:
q = 1.6 x 10-19 Coul;
B = .05 Teslas (500 G)
m = 1.67 x 10-27 kg;
Vacc = 500 volts gives:
(1/2)mv2 = qV, or v = [2qV/m]1/2 = 3.1 x 105 m/s
r = mv/qB = (1.67 x 10-27kg)*(3.1 x 105 m/s) / (1.6
x 10-19 Coul)*(0.05 T) = .065 m = 6.5 cm.
Mass Spectrometer
Heavier masses will give bigger radii, but we
can shrink the radii if they become too big
by using bigger magnetic fields. Note that
by measuring quantities that we can easily
measure (charge, radius, Voltage, magnetic
field), we can determine very tiny masses!
In one of our experiments in lab (Charge to
Mass of the Electron), we will essentially
determine the mass of an electron!
Velocity Selector
To better control the velocity of the ions in a
mass spectrometer, we can use a
combination of electric and magnetic fields
to select a particular velocity for the ions
entering the main magnetic field. To do
this, we recognize that a magnetic force
depends on qvB, while an electric force
depends on qE.
Velocity Selector
If we make the (positive) ion beam go through an
area with an Electric field directed up and a
Magnetic field directed out, then for the beam to
NOT be deflected, we need qvB = qE, which
gives v=E/B. If the incoming ions go too fast,
the magnetic force “wins” and they go too low; if
they go too slow, the electric force “wins” and
they go too high to enter the main field.
parallel plates
Incoming ions
B
-Q
E
+Q
to main
field
Velocity Selector
As we saw in Part 1, the Electric Field for
parallel plates is constant. As we saw in
Part 2, the Electric Field is related to the
voltage by: Ey = -DV/Dy . This means we
can control E by controlling the voltage
across the plates. This, then, means we can
control the velocity of the ions entering the
main magnetic field by controlling the
voltage across the selector plates:
v = V/dB
Other uses
A cyclotron is an instrument used to
accelerate charged particles to very high
speeds. It uses magnetic fields to bend the
charges around in circles to keep them in
one place while they’re being accelerated.
Magnetic bottles can be used to contain high
energy plasmas in fusion research.
Cyclotron
The idea behind a cyclotron is to have charges
move in a circle, so the motion does not
carry the charges too far away. We can
accelerate the charges using voltage, but we
do so not with one big voltage, but with
having the charges “fall” through a smaller
voltage several (many) times.
Cyclotron
To make the charges go in a circle, we use an
external magnetic field.
To make the charges accelerate (speed up), we
use a voltage difference.
To make the charges keep “falling” through
the voltage difference, we place the charges
inside a metal short hollow cylinder, cut the
cylinder in half.
We then place an alternating voltage across
the two halves of the cylinder.
Cyclotron
A charged particle (q) is injected into the
center of the cylinder with some initial
speed, v. Due to the magnetic field, it will
then go in a semicircle.
B
B
It will then be
V
accelerated
B
B
across the gap between
the two halves of the cylinder (each called a
“dee”) using a voltage, V.
Cyclotron
After it is accelerated across the gap, the
speed will be larger, and from (r = mv/qB),
the radius of the semicircle will be larger.
During the time it is inside the half cylinder
(one of the dees), the voltage across the two
dees will be switched in polarity. Thus,
when the charge again approaches the gap,
it will not slow down, but rather speed up
again!
The larger radii each time will prevent the
particle from traversing exactly the same
path each time.
Cyclotron
Thus we get continual increases in speed
reusing the same voltage source! The limit
to the maximum speed is the maximum
radius of the dees, not the maximum
voltage we can obtain without causing
lightning.
Let’s see how long it takes the charge to make
one half circle to see how long we have to
switch the polarity of the voltage source:
Cyclotron
Starting with: r = mv/qB, and using v = wr,
where w = 2pf = 2p/T, we have:
r = m(2pr/T) / qB , or solving for T:
T = 2pm / qB .
Note that the period does NOT depend on
either the radius or the speed. That means
the time to switch the polarity of the voltage
source is constant!
Cyclotron - example
r = mv/qB and T = 2pm/ qB
To accelerate a proton to 1/10 the speed of
light (3 x 107 m/s) using a magnetic field of
1 Tesla, we would need an r of:
r = [(1.67 x 10-27kg)*(3 x 107m/s)] /[ (1.6 x 10-19 C)
*(1 T)] = 0.313 m .
T = [(2 * 3.1415 * 1.67 x 10-27 kg)]/ [(1.6 x 10-19 C)
*(1 T)] = 6.6 x 10-8 sec, or
f = 1/T = 15 MHz.
Cyclotron - example cont.
If we were to accelerate a proton to 1/10 the
speed of light using a single voltage, we
would need a voltage of:
Conservation of Energy: (1/2)*m*v2 = q*V, or
V = (1/2)*m*v2/q = (1/2)*(1.67 x 10-27 kg) *
(3 x 107 m/s)2 / 1.6 x 10-19 C = 4.7 Million
volts. That voltage would be hard to keep
from creating lightning and discharging!
Magnetic Forces
The Computer Homework Vol. 4 #1,
Magnetic Deflection, deals with magnetic
forces on charged particles. We’ll discuss
strategies for completing this assignment in
class.
Forces, Fields and Currents
If moving charges are acted upon by magnetic
fields (as we just saw), then currents, which
consist of moving charges, should be acted
upon by fields. Another way of saying the
same thing is: fields act on moving charges
[F = qvB sin(qvB)], and currents consist of
moving charges [qv = IDL] , so we have:
Fon current = I L B sin(qIB)
Magnetic Force on Current
Fon current = I L B sin(qIB)
with direction: thumb(force) =
hand(current) x fingers(field).
Example: If the earth has a magnetic field of
1 gauss directed North (for simplicity - it really
is directed down and to the North), what is the
magnetic force on a current of 100 amps
flowing East in a wire 50 meters in length
between two telephone poles.
Example
We are given: B = 1 gauss = 1 x 10-4 T, North
I = 100 amps, East; L = 50 meters.
Fon current = I L B sin(qIB) =
(100 amps) (50 meters) (1 x 10-4 T) sin(90o) =
0.005 Nts directed up. (This is negligible
compared to the weight of the cable!)
Let’s now consider how this force will work
on complete current circuits, not just
individual lengths. Also, how can we make
this magnetic force significant?
Forces on
rectangular current loop
Consider the situation in the figure below:
A current loop (with current direction going
counter-clockwise) is situated in a Magnetic
Field going from North to South poles.
N
B
I 
S
Forces on
rectangular current loop
We need to consider the forces on each of the
four sides of the current loop.
The force on the top and bottom of the loop
are zero, since the field and current are either
parallel or anti-parallel (qIB = 180o or 0o).
 I Ftop=0
N
B
I  Fbottom=0
S
Forces on
rectangular current loop
The current on the left side is going down while
the field is directed to the right, so that means
the force is directed out of the screen, and the
magnitude is: Fleft = I L B sin(90o) = I L B.
F
N I
B
S
L
Forces on
rectangular current loop
The current on the right side is going up while
the field is directed to the right, so that means
the force is directed into the screen, and the
magnitude is: Fright = I L B sin(90o) = I L B.
N
B
F
 I S
L
Net Force
Ftop = 0
Fbottom = 0
Fleft = ILB out
Fright = ILB in
As we can see, the NET FORCE (S F) is
zero.
However, since the force is pushing out on the
left and in on the right, there is a Torque!
The loop will tend to rotate about an axis
through the center.
Torque on
rectangular current loop
Recall that torque is: t = r F sin(qrF). In the
figure below we can see that r = w/2. Thus
the Fleft gives a torque of (w/2)ILB, and the
Fright also gives a torque of (w/2)ILB.
r
F
F
N I B
I S
L
w
Torque on
rectangular current loop
Since both torques are trying to rotate the loop
in the same direction, the total torque is:
S t = wILB. We note that wL = A (width times
length = Area). Also, we can have several
loops (N) that will each give a torque.
F
F
N I B
 I S
L
r
w
Torque on a loop
The final result for this loop is:
t = N A I B sin(qIB) sin(qrF) .
In this orientation, qIB = 90o and qrF = 90o .
If the loop does rotate, we see that qIB
remains at 90o (the current still goes up and
down, the field still goes to the right), but qrF
changes as the loop rotates!
t = N A I B sin(qrF) .
Electric Meter
One of the early types of current meters is one where
we have such a loop of current in a magnetic field,
and we have a restraining spring. As the current
increases, the torque increases, but we have a
restraining spring to keep it from rotating
completely around. The bigger the current, the
bigger the torque, and the loop will turn through a
bigger angle. Attach a pointer onto the loop, and
we have the (analogue) meter.
Electric Motor
To create an electric motor that will continue
to spin when a current is applied, we need
to keep the current going up the right side,
even when the side originally on the left
becomes the right side due to the spinning.
We can accomplish this by using a set of
brushes as indicated in the next slide.
Electric Motor
+
N
-
S
This diagram will be explained in class, since it
involves three dimensions. The two green “C’s”
are actually one ring that is split. The ring is in
and out of the screen.
Electric Motor
t = N A I B sin(qrF) .
Since we switch the current to make it always
run up whatever wire is on the left side, we
make sin(qrF) always positive.
To find the average torque, we need to
determine the average of sin(q) from 0o
through 180o.
Average of sin(q)
To find an average, we simply add up the
values and divide by the number of items
we have:
sin(q)avg = S sin(q) / S 1
With computers this is fairly easy to do. We
can also do it with the calculus (replace S
with  ) sin(q)avg = op sin(q)dq / op 1 dq
= [-cos(p) - -cos(0)] / [p - 0] = 2/p .
Average Torque
Thus we get for the average torque:
taverage = (2/p) N A I B .
But we usually describe motors by their
power, not by their torque. Recall that
Power is energy per time, and energy is
force through a distance. For rotations, this
becomes: energy is torque through an
angle, and power is torque through an angle
per time (but recall w = Dq/Dt).
Average Power
Putting all of this together gives:
Pavg = tavg w = w N A I B (2/p),
and with w = 2pf, we have:
Pavg = 4 f N A I B.
Note that the power depends not only on the
details of the motor (N, A, B) and the
current (I), but also on how fast the motor
spins (f)!
Example
Design an electric motor that has a power of
1/2 hp when it rotates at a frequency of 120
Hz (120 cycles/sec * 60 sec/min = 7200
rpm).
Pavg = 0.5 hp* (746 Watts/1 hp) = 373 Watts.
f = 120 Hz.
Design = ? (This means we specify N, A, I
and B such that P = 373 Watts when f=120
Hz. We have some “free” choices!)
Example, cont.
Pavg = 4 f N A I B.
Pavg = 373 Watts, f = 120 Hz.
Let’s start by choosing some reasonable
values: choose an area (A) of 10 cm x 10
cm = .01 m2; choose a magnetic field (B)
of 1000 Gauss (0.1 T). We can also choose
a current (I) of 5 amps. This means that we
can now solve for the number of turns, N.
Example, cont.
Pavg = 4 f N A I B.
Pavg = 373 Watts, f = 120 Hz.
A = .01 m2 B = 0.1 T I = 5 amps.
N = P / [4 f A I B]
= 373 Watts / [4 * 120 Hz * .01 m2 * 5 Amps * .1 T]
= 155 turns.
Hall Effect
Although we consider a current to be a flow
of positive charges, what actually flows in
normal electric currents: positive charges
(protons or positive ions) from high voltage
to low, or negative charges (electrons) from
low voltage to high?
How do we know this?
Hall Effect
Lets see if a magnetic field can help show
what happens.
We will consider a conventional current going
to the right in a magnetic field directed into
the page.

B

I

Hall Effect
If the current is carried by positive charges, the
positive charges are moving to the right and will
experience a magnetic force up. This will cause
some positive charges to accumulate on the top
creating a voltage difference between the top and
bottom, with the top being at a higher voltage.

F
B
+

v
I

Hall Effect
If the current is carried by negative charges, the
negative charges are moving to the left and will
experience a magnetic force up. This will cause
some negative charges to accumulate on the top
creating a voltage difference between the top and
bottom, with the top being at a lower voltage.


F
v
B
I

Metals and Semiconductors
Metals show a Hall voltage that indicates that
the negative charges are what carry the
current.
Semiconductors, however, show that either
negative (electrons) or positive (holes)
charges seem to carry the current. This
topic is explored in Solid State Physics.
Hall Effect and Gaussmeters
Once we have determined the Hall Effect for a
particular metal or semiconductor, we can then use
the Hall Effect voltage to measure the magnetic
field (both magnitude and direction). We merely
create a standard current through the material and
have a voltmeter measure any voltage difference
across opposite sides of the material. This voltage
difference is then proportional to the magnetic
field perpendicular to the current and
perpendicular to the sides of the material on which
the voltage is being measured.