Transcript magnetic field

Magnetism
magnetism
Magnetic fields are produced by moving electrical charges
– i.e., currents)
 macroscopic (e.g. currents in a wire)
 microscopic (electrons in atomic orbit and rotating
around their own axis)
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magnets
 the magnetic field produced by electrons tend to cancel each other,
so most materials are not magnetic
 in certain ‘ferromagnetic’ materials (iron) neighboring electrons can
couple and form domains (< 1mm) that are magnetic. Since there are
many domains that have different orientation, the material is overall
not magnetized
 when an external magnetic field is applied the fields in the different
domains align and the whole object becomes magnetic
 after the external field is removed, a material like iron becomes
unmagnetized quickly, but some remain magnetized and can be
used as ‘permanent’ magnets.
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para and ferro magnets
strawberry in a B-field
do not retain
any magnetism
in absence of
external field
retains domains
in which magnetic
field remain in the
absence of external
fields
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magnetic poles and fields
 magnets have ‘north’ and ‘south’ poles and field lines
point in the direction of force on a North magnetic pole.
 unlike the case of electrical fields, where positive charges
can exists separate from negative charges, north and
south poles always come together. There are no
monopoles discovered so far.
demo:
magnetic field lines (ohp)
broken magnet
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One big magnet!
demo:
compass needles
compass
Why is it higher here?
Note that the geographical North pole is in fact the
magnetic south pole
B=0.3-0.6 x 10-4 Tesla
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question
If you are standing exactly at the (magnetic) south
Pole (I.e. near the geographical north pole), and are
holding a compass parallel with the earth’s surface,
in which direction would the needle point?
a) It would point roughly to the geographical south
b) It could point anywhere
c) It would rotate with constant angular speed
The compass needle in fact wants to point into the earth
(along the direction of the field line). But if hold parallel
to earth, it can’t do that and will point wherever. There is
no reason for it to rotate though.
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charged particles moving in a magnetic field
 A charged particle q that is
moving with a velocity v in a
magnetic field B will feel a force
where q: charge of particle
v: velocity of paticle
B: magnetic field
: angle between velocity
vector and field direction
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direction of force on charged paricles in B-field
 magnitude of the force
demo: bending the beam I
 you can find the direction
of the force using the right
hand rule. It holds for
positive charges. For
negative charges switch
the direction of the force
In the 3pm lecture (Section 2), we will use the version
of Right Hand Rules given in the Textbook.
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example: electron in magnetic field
 an electron with v=1x106 m/s is
entering a area with B=1 T. The
field is directed into the screen.
a) in which direction will the
electron be bent, if at all?
b) how large is the force? what is
the acceleration?
x
x
x
x
x
x
x
x
a) use right hand rule:
x x x
thumb is velocity (initially to the right)
index finger is field (in the screen)
middle finger is force perpendicular to both
switch direction because negative charge
b) F=|q|vBsin=1.6x10-19 x 1x106 x 1=1.6x10-13 N
a=F/m=1.6x10-13 N/9.11x10-31 kg =1.76x1017 m/s2
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question
 A Magnesium ion (Z=12) with all
its electrons removed is moving
in a field of 0.1 T as shown. What
direction will the force act?
a) into the screen
b) out of the screen
c) parallel to the B field lines and
the screen
d) perpendicular to the B field lines
and parallel to the screen
e) in the direction of motion
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Mg
45o
v
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Charged particle in a magnetic field
 Let’s assume a charged particle is
moving in a uniform magnetic field so
that the velocity is perpendicular
to the field.
 The particle will follow a curved path
and is directed towards the center
 Use Newton’s second law and the
equation for centripetal acceleration
demo: bending the beam II
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Magnetic spectrometers
Beam from
cyclotrons
target chamber
S800 spectrometer
At the cyclotron
Bending angle ~ 150o
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question
In a nuclear reaction two types of fully ionized particles are created.
120Sn with Z=50 and v=12.8814x107 m/s (Tin)
120Sb with Z=51 and v=13.099x107 m/s (Antimony)
Both have a mass of 1.991x10-25 kg and pass through a 180o
magnetic spectrometer with B=1T. If the detector used to locate
the particles can separate events that are 2 mm away from each
other, are 120Sn and 120Sb separated?
r= mv/qB
For 120Sn:
M=1.991x10-25 kg v=12.8814x107 m/s
B=1T q=50x1.6x10-19 C.
RSn=3.2060 m
For 120Snb:
M=1.991x10-25 kg v=13.0990x107 m/s
B=1T q=51x1.6x10-19 C.
RSb=3.1961 m
RSn-RSb=3.206-3.1961=9.9x10-3 m = 9.9 mm thus separated
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What we did so far
 Moving charged particles make magnetic field
 North and South poles cannot exist independently
 The magnitude of a force on a charge particle in a
magnetic field: F=qvBsin where  is the angle
between v and B.
 The direction of the force is given by the (first)
right-hand rule
 for + particles: use directly
 for – particles: after using the right hand-rule,
reverse the direction of the force
 For a particle moving in a direction perpendicular
to a magnetic field
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Question
 a proton is moving from left to right into a field of which
the field lines point into the screen. As a result, the proton
will
a) continue along its original trajectory
b) bend upwards
c) bend downwards
d) bend into the screen
proton
e) bend out of the screen
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x
x
x
x
x
x
x
x
x
x
x
x
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magnetic force on a conducting wire
 consider positive charges moving through a
wire. Each particle feels a force, hence there is a
net force on the wire

N: total number of charges

n: charges per unit volume
 Use:
see earlier
 To get
I
 More general:
 where : angle between I and B vectors
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question: a floating wire
 a 1 m long copper wire of unknown mass is held horizontally with a
current of 1 A going through it. It is placed in a horizontal magnetic
field whose field lines are perpendicular to the wire. When the
magnetic field is 1 T, one can let go of the wire without if falling down.
top view
What is its mass?
I
B
electrons are moving left to right, so force due to B is up
(out of the screen). When floating Fgravity=FB-field
mg=Bil so m=Bil/g=1x1x1/9.81=0.102 kg
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question
x
 a rectangular looped copper
wire carrying a current is
placed horizontally in a B-field
pointing down. Disregarding
any other forces, it will move
a) in direction of vector A
b) in direction of vector B
c) in direction of vector C
d) in direction of vector D
e) none of the above
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A
x
x
x
x
x B
x
x
x
x
x
x
x
C
x
x
top view
x
D
x
I
x
x
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question
x
x
 a rectangular looped copper
x
x
wire carrying a current is
placed horizontally in a B-field
x B x
pointing down. Disregarding
any other forces, it will move
x
x
 a) in direction of vector A
 b) in direction of vector B
x
x
 c) in direction of vector C
 d) in direction of vector D
 e) none of the above  correct answer: e)
A
x
x
x
x
C
x
top view
x
D
x
I
x
x
x
It will not move at all. Forces on left
and right sides will cancel and
likewise for top and bottom
sides
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Torque on a current loop
Top view
Rotation axis
 Consider a current loop with
dimension a x b in a B-field parallel
b
I
to the loop.
B
 The force F on the right side (length
b): F=BIb (pointing into the screen in
the top view or downward in the
frontal view)
a
 The force F on the left side (length
F
b): F=BIb (pointing out of the screen
in the top view or upward in the
x
frontal view)
 force on up/down side (length a) is
F
frontal view
zero
 With the given rotation axis:
Torque: =Fd=(BIb x a/2) + (BIb x a/2) If there is a net torque,
=BIba=BIA with A=axb: surface of
the loop will rotate!
loop.
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Torque on a current loop
Top view
 Now the loop makes an angle  with
the B-field as shown right
 To calculate the torque we only need
the force perpendicular to the rotating
loop:
FL=Fsin
=FLd=(BIb x a/2)sin + (BIb x a/2) sin
=BIbasin=BIAsin
 If there would be N loops:
=BIANsin
F
frontal view  FL
sin=F/FL

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Rotation axis
b
I
B
a
F
frontal view

x
F
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So…
 The general equation for a torque on a loop of N windings of wire is:
 with
 B: magnetic field strength
 I: Current through the loop
 A: area of the loop (also holds for non-rectangular loops)
 N: number of windings
 : angle between B and line perpendicular to loop
 =IAN magnetic moment of the coil: it is a vector perpendicular to
the coil.  is also the angle between  and B. Note that  is
independent of B and , so it describes the properties of the coil

when placed in a field. Unit: Am2
B

I
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A
N
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note for loncapa: area of an ellipse: ab with a, b radii in the two directions
example

 A circular coil of 5 windings is
placed in a B-field of 2 T that
B

o
makes and angle =60 with
A
N
I
the line perpendicular to the
coil. The radius of the coil is 3
cm, and the current through
the coil is 0.5 A. What are:
A=r2= (0.03)2=2.82x10-3 m2
a) the area of the coil?
b) the magnetic moment of
=IAN=0.5 x 2.82x10-3 x 5=7.1x10-3Am2
the coil?
c)the torque one the coil?
=Bsin= 7.1x10-3 x 2 T x 0.866=1.23x10-2 Nm
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electric motor
By supplying electricity we can get some work done!
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creating magnetic field with current
 So far, we have seen that magnetic field can affect the motion of
charged particles.
 However, the reverse is also true: moving charge can create
magnetic fields.
 First seen by Hans Oersted who noted that a current through a wire
creates a magnetic field.
 A second right-hand rule can be used to find the direction of the
magnetic field
demo: Oersted experiment
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How to quantify the field
 0 = “permeability of free space” =
4 x 10-7 Tm/A
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an electron passing a wire

a)
b)
c)
d)
an electron with v=1x106 m/s is
moving parallel to a wire
carrying a current I=1A at a
distance of 2 cm, in the same
direction as the current
What is the direction of the
magnetic field near the electron
due to the wire?
what is the magnitude of the
magnetic field near the
electron?
what is the direction of the force
on the electron?
what is the magnitude of the
force on the electron?
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I=1A
q=-1.6x10-19C
2 cm
a) use 2nd right hand rule
B-field goes into the screen
b)
=4 x 10-7 x 1/(20.02)=1x10-5 T
c) use 1st right-hand rule and notice
that the electron is negative.
Force points to the right.
d) F=qvBsin=1.6E-19x1E6x1E-5x1=
= 1.6E-18 N (note sin(90)=1)
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question
 a proton is passing by a wire
carrying current and is moving
perpendicular to the wire, into the
screen
1) what is the direction of the B-field
near the proton?
 into the screen
 out of the screen
 to the left
 to the right
 up
2) what is the direction of the force on
the proton?
 to the left
 to the right
 up
 down
 no force at all
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I
x
proton moving
into the
screen
1) use 2nd right hand rule
(same as example on previous
slide)
2) use 1st right hand rule. velocity
is into the screen, B-field is into the
screen: no Force (sin(00)=0)
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magnetic force between two parallel wires
 if we place two parallel wires
next to each other, the current
in wire 2 creates a field near
wire 2, at distance d from wire 1:
d
 The force on wire 1 due to wire 2
is then:
 Note
 so that the force per unit length
is:
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attractive if same direction
repulsive if opposite direction
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question
 two wires are placed parallel, one carrying a current of 1A and the
other of 2A, in the same direction. The distance between the two wires
is 2 cm
 a) what is the magnitude of the B-field exactly in between the two
wires?
 b) if a proton moves parallel to the two wires with v=1x105 m/s,
exactly in between the two and in the same direction as the current,
what is the magnitude of the force on the proton?
 c) what is the force per unit length between the two wires?
a) B1=0I/(2r)=4x10-7x1/(20.01)=2x10-5 T
1A
2A
-7
-5
B2= 4x10 x2/(20.01)=4x10 T
B1: into the screen B2: out of the screen
2
Bnet=2x10-5 T out of the screen
1
b) F=qvBsin=1.6x10-19x105 x 2x10-5 x sin(90)=3.2x10-19 N
(directed to the right, use 1st right-hand rule)
c) F/l= 0I1I2/(2d)= 4x10-7x1x2/(20.02)=2x10-5 N
2cm
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note
 the procedure of the previous slide can be used for any
number of wires. In case of 4 wires (see lon-capa), one
can calculate the force of one on the wires by adding the
forces of each of the other three wires on that wire…
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other cases: the current loop
 magnetic field inside a current
loop
I
I
R
X
right-handed current
through loop:
B-field in the screen
left-handed current
through loop:
B-field out of the screen
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 example: A person
wants to find the current
in a superconducting
coil with diameter of 2
cm. She measures the
magnetic field at the
center to be 1x10-5 T.
What is the current?
I=2RBcenter/0=
2x0.01x10-5/4x10-7=
0.16 A
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other cases II: magnetic field of a solenoid
 a solenoid is a collection of coils
stacked on top of each other
 Inside a perfect solenoid, the field
lines are parallel and the field
uniform
 outside the solenoid, the field
pattern looks like that of a bar
magnet.
 For the field inside of a solenoid:
where I is the current and n is the
number of turns (n) per unit length l
of the solenoid
 note that the field at the center
does not depend on the radius of
the turns
B-field of solenoid
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example
 A perfect coil is 30 cm long and has 3000 windings. Its
radius is 2cm. What is the field strength along the central
line inside the coil if the current is 4 A?
B=0nI=4x10-7 x 3000/0.3 x 4 = 1x10-3 T use n=N/L
 The field strength along a line parallel to the central line
but 5mm away from the center is … along the central
line? a) lower than b) the same as c) higher than
inside the coil, the field is uniform
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