Download Magnetic Fields One goal of the course is to

Magnetic Fields
One goal of the course is to convey some
understanding of electromagnetic waves, which
were already mentioned in the context of radiative
heat transfer.
Electromagnetic waves involve both electric
fields, which we have been discussing, and
magnetic fields, which we discuss next.
Recall that electric fields describe the force
at each point in space which a test charge qt would
experience if it were placed at that point.
A magnetic field at a point in space can similarly be
detected by means of the force on a test charge
placed at that point in space. However the force
is different, both in direction and magnitude, than
it is in the case of electric fields.
Magnetic Fields
A magnetic field at a point in space can similarly be
detected by means of the force on a test charge
placed at that point in space. However the force
is different, both in direction and magnitude, than
it is in the case of electric fields. Unlike electric fields
, in a magnetic field a test charge ONLY experiences a
force if the test charge HAS A FINITE VELOCITY.
v
B
qt
F
In the situation
indicated here, the
force on the test
charge is pointing out
of the figure,
perpendicular to both
B and v and has
magnitude
qtvB .
Direction of the force on
The charges moving through the
magnetic field
N
S
Moving electrical
charges (whether in
a wire or not)
Magnetic field
from north to
south pole of a
magnet,
Right hand rule for determining
the direction of the force:
If v is not perpendicular to B then
the rule still works but the magnitude
of the force on the test charge
is qtvB sin α where α is the (smallest)
angle from v to B.
Units of magnetic field are
Newton-seconds/coulomb-meter
The unit is called a Tesla. A 1 Tesla field
is a very big field. A more commonly
used unit is the Gauss. There are 104 Gauss
in 1 Tesla. The earth's magnetic
field is 3.2 x 10-5 Tesla or .32 gauss. The
field of a typical refrigerator magnet
is about 100 gauss. Neodynium iron boride
magnets have fields of nearly a tesla.
In the experiment on the table, the
north pole of the magnet is above the
wire and the south pole is below the
wire. I am going to run a current through
the wire from left to right as you face the
apparatus. Which way will the wire swing?
A. Up
B. Toward the class
C. Down
D. Away from the class
An electron enters a region in a cathode ray
tube with initial velocity of magnitude 107 m/s
in the x direction. There is a magnetic field of
10-4 tesla in the z direction (into the plane of the drawing).
What is its position after its x coordinate has
increased by .1 m? e=1.6 x 10-19 coulombs, m=9.1 x 10-31 kg
(You can ignore changes in vx .)
y
B
.
v
x
20cm
a. (.1,-.018,0)
b. (.1,-.009,0)
c. (.1,.009,0)
d. (.1,0,-.009)
e. (.1,0,.009)
x
In the last problem, it wasn't really quite right to
ignore changes in vx . That is mainly because
y
B
.
v
x
x
20cm
a. The electron is also subject to electric fields
b. We ignored gravity.
c. The electron's path will be curved.
d. There was an initial (t=0) acceleration in
the x direction.
Path of an electron in a magnetic field:
With the field in the z direction the acceleration
is always in the direction perpendicular to the
direction of the velocity and in the xy plane
and always has the same magnitude (evB/m):
zB
y
v
a
What motion results?
a. a hyperbola
b. a parabola
c. a circle
d. an ellipse
x
Circle:
If it's a circle, the acceleration is constant in
magnitude and always perpendicular to the
velocity just as for a planet in a circular orbit.
Using the fact that for circular motion
the acceleration has magnitude v2/r , find
the radius of the circle traced out by
the electron in terms of e, v, B and m
a. r =v2 m/eB b. r =v m/eB c. r=eB/vm d. v2 m/2eB
Detecting magnetic fields.
In the lab, you will use a 'Hall probe'
to detect magnetic fields. Here
is how it works. There is a slab
of conducting material carrying charges,
as in a metal. Actually in some of the
materials used for Hall probes, the
moving charges have positive charge
and I will assume that they are positive
here.
B
q
V
v
Detecting magnetic fields.
In the lab, you will use a 'Hall probe'
to detect magnetic fields. Here
is how it works. There is a slab
of conducting material carrying charges,
as in a metal. Actually in some of the
materials used for Hall probes, the
moving charges have positive charge
and I will assume that they are positive
here.
B
q
V
v
Hall probe:
z
B
q
y
v
x
V
The charges move through the material
at constant speed (magnitude of v).
What is the direction of the magnetic
force on the charge q?
a. +x b. -x c. +y d. -y e. +z f. -z
z B y
q
v
x
F
V
OK the magnetic force is as indicated.
How will this force affect the motion of
the charge q?
a. move in a curved path up
b. move in a curved path down
c. acclerate in a straight line along x
d. decelerate in a straight line along x
e. move in a curved path toward the
back (y direction)
f. move in a curved path toward the
front.
z B y
-
q
A
v
x
+
V
That's what happens when you first connect up
the Hall probe. Very quickly, the charges hit the
front wall and get stuck there as indicated by
the + sign. Because the material is charge neutral,
negative charges then appear on the back wall as
indicated by the - sign. Suppose total charge Q
pile up on the front side and -Q on the back side.
What is the resulting y component of the electric
field inside the material? A is the area of the front
face.
a. (Q/(Aε0)) b. - (Q/(Aε0)) c. 0 d. very large
z B
--F q
E
----v
+ + +A + + + + +
w
x
V
Notice that the electric force due to the pile up
of charges on the sides opposes the magnetic
force indicated by F in the figure. So quite quickly
the path of the charges stops curving because the
magnetic force balances the electric force:
qEy =qvB so the electric field is Ey =vB
and a voltage between the front and back faces
of VHall =w|Ey| =wvB
z B
VHall
E
-------F
I, v
+ + +A + + + + +
A'
d
w
x
V
The voltage between the front and back faces
is VHall =w|Ey| =wvB .
If there are n charges per unit volume in the
material, the current in the x direction is
I = qnvA' so the Hall voltage is
VHall = (w/qnA')IB giving B= VHall( qnA'/Iw)
but A'=dw so B= VHall( qnd/I)
everything on the right is known or can be
measured, so the field B can be determined.
B= VHall( qnd/I)
If the magnetic field was .1 Tesla, the measured
current was one milliaamp, the thicknes of the
'Hall bar' was 1 millimeter, the carriers had
charge q= 1.6 x 10-19 C, and the measured
voltage was 10 millivolts then what was the number
of carriers per m3 in the material?
a.
b.
c.
d.
1.6 x 10-20 m-3
.626 x 10-20 m-3
.626 x 1020 m-3
1.6 x 1020 m-3
Why is I = qnvA' ?
Consider flow in the wire over a
short time Δt
the number in this volume pass through
the right face in time Δt.
That number is
n(A'vΔt)
A'
distance vΔt
and charge passing
through is
qn(A'vΔt)
The current I is
the charge/time=
qn(A'vΔt)/Δt=
qnvA' =I
How electric motors work.
The current moving through a wire is a collection
of moving charges so they will experience a force
if the wire is in a magnetic field. If we know the
number of charges n per unit volume in the wire,
the crossectional area A of the wire and the
speed v of the charges then we can figure out the
force on the wire due a magnetic field perpendicular
to the wire like this:
Force on one charge = qvB
number of charges in length L of wire =nLA
Force on a length L of wire =qvBnLA
current in the wire = number passing through A per
second x charge for each=qnAv =I
so force on a length L of wire = ILB
There is a coil sitting in a magnetic moment
on the bench. No current is flowing through the wire
N
S l
r
which way should the current flow to make the
wire jump out of the magnet?
a. l to r b. r to l
force
I
I
force
L
a
θ
If the dimensions and angle are as shown, what
is the torque on the rotor in terms a,L,B, I and θ?
a. ILBcosθ
b. IaLBcosθ
c. IaLBsinθ
d. IL2Bsinθ
B
coil
The quantity ILa is the current times the
area of the current loop and is called the
MAGNETIC MOMENT of the loop.
If you have some kind of misshapen loop,
you can imagine filling it with little rectangular loops.
All the torques from the little loops inside will cancel
and you are left with the currents on the edge
and the same expression for the torque with the
size of the magnetic moment given by IA where
A is the area of the (misshapen) loop.
This works for atoms too, and is the essential origin
of magnetism. You may think of the atoms as having
tiny currents flowing around in them giving rise
to magnetic moments. If all the magnetic moments
of the atoms in a material are aligned, the bulk
material has a magnetic moment, as in iron and
other materials from which magnets are made.
Notice that the units of magnetic moments are
force x distance/Tesla or joules/Tesla.
The magnetic moment of an iron atom in metallic iron
is roughly 2 x 10-23 joules/Tesla.
If the atoms are approximately 2 x 10-10 m apart, what
magnetic moment would you expect a disk of iron
.05m in diameter and 0.02 meters thick to have.
a. 31.25 joules/Tesla
b. 6.25x 10-9 joules/Tesla
c. 308.4 joules/Tesla
d. 3.17 Joules/Tesla
Answer: In this solution, I correct an error of a factor
of 4 which led to the intended correct answer (a.) being
4 times smaller than it should have been:
let the atomic moment be denoted matom , the distance
between atoms be a, the radius of the disc R and
the thickness of the disc d.
matom = iπ(a/2)2 , solve for i= 4matom /πa2
m, the moment of the entire disc is
m=iπ R2 x number of atomic layers
=iπ R2 (d/a) =4matom (R/a)2(d/a) =4(31.25)joules/Tesla
In this picture what are the commutator rings and
stationary brushes for?
A. To keep the connections to the battery clean.
B. To screen the magnetic field so it doesn’t interfere
with the battery
C. To switch the direction of of the current when the
armature turns ½ way around.
Fig. 11-7, p. 364
D. To conserve electricity.
What will happen if I replace the commutator by a
connection which keeps the current flowing
in the same direction in the loop all the time?
a. There will be no torque.
b. The direction of the torque will be reversed
all the time.
c. The direction of the torque will be reversed
1/2 the time.
d. The rotor will move to an upright position
and stop.
e. The rotor will move to an upright position,
oscillate for a while and stop.
A generator is a motor running backward.
crank
v
B
force
Resistive load
(appliance)
The direction of the current in the loop
is
a. clockwise looking down on loop
b. counterclockwise looking down on
loop
Fig. 11-7, p. 364
c. zero current d. alternating.
crank
v
Motion (up)
of charges in
wire makes force,
hence current this
way (for q>0)
Resistive load
(appliance)
I
B
I
force
Fig. 11-7, p. 364
crank
Resistive load
(appliance)
In this generator, if the crank is turned steadily, how will the
voltage at the device depend on the time?
A. It will be a constant (DC) voltage, not changing in time
B. It will be an AC voltage, varying from plus to minus and
back
C. It will be always of the same sign, but varying from zero to
a maximum value and back as the crank is turned
D. It will be a constant current but a time varying voltage.
Magnetic Fields are caused by currents.
We have seen how magnetic fields produce
forces on moving charges.
Moving charges (currents) also produce
magnetic fields.
Quantitatively, we confine attention to
the magnetic field produced by a current I
in a long wire. The field has magnitude
|B|= μ0I/2πr where r is the distance from the wire.
The direction is tangent to a circle at r around
the wire with direction given by a right hand rule.
μ0 = 1.26 x 10-6 Tm/A (a universal constant).
Forces between wires:
The force per unit length exerted by one wire on the other is
case 1.
case 2.
a.
b.
c.
d.
attractive
attractive
repulsive
repulsive
repulsive
attractive
attractive
repulsive
Let's also look at the details of the field distributions in these cases
Magnetic Fields are caused by
currents
As with electric fields, magnetic fields
from more than one current add
vectorially. You have been working
with that in homework 5. Some
important examples are illustrated in
these examples from your book
Solenoid
Summary on Magnetic Fields:
A magnetic field B at a point in space is a vector
such that a MOVING test charge qt with velocity v
at that point experiences a force perpendicular to both
B and v of magnitude qtvB sin α where α is the
angle between B and v and the direction is perpendicular
to the plane defined by B and v with sense (sign)
given by the right hand rule.
Currents are collections of moving charges and
the wires carrying currents in magnetic fields
experience forces which are the sum of the
forces on all those moving charges.
Currents I in wires are related to the charges q in them with
density (number per volume) n and velocity v
by I =Anvq where A is the cross section of the wire.
Magnetic fields summary (continued)
Currents also cause magnetic fields. The magnetic
field caused by a long straight wire has
magnitude B=μ0I/2πr where r is the distance from
the wire. The direction is tangent to a circle around
the wire of radius r and the sign is given by
a right hand rule.
Magnetic fields from a collection of wires add
vectorially to give the total magnetic field
at a point.
Magnetic moments are current loops. The magnitude
of the magnetic moment is the area of the loop
times the current flowing around the loop. Permanent
magnets can be regarded as having magnetic
moments which are the sum of the magnetic moments
of the atoms in the material of which the material is
made. Placed in an externally provided magnetic
field, magnetic moments experience torques.