Download Chapter 27:

Magnetic Field and Magnetic
Forces
Young and Freedman
Chapter 27
Introduction
Review - electric fields
1) A charge (or collection of charges) produces an electric field in the
space around it.
2) The electric field exerts a force on any other charge q that is present
in the field.
What is coming up for magnetic fields
1. A MOVING charge (or charges) produce a magnetic field in the
space around it.
2. The magnetic field exerts a force on any other MOVING charge or
current that is present in the field.
Some Simple Phenomenology
The world is a big magnet
Magnetic and Electric Forces
Just Replace + & - with N & S ????
This is an extremely good idea that is, unfortunately
WRONG
The basic problem is that there are no “Free” N & S
poles.
Basic relationship between electric fields
and magnetic currents
Demonstrated in
Oersted’s Experiment
Place a compass near a wire
Compass deflects when an electric
current flows in the wire
N
Motion of a charged particle in a magnetic field
1. A moving charge or a electric current produces a magnetic field in
the surrounding space (It also produces an electric field)
2. The magnetic field exerts a force on any other moving charge or
current that is in the field.
Strategy:
We will begin with a discussion of the force
on a moving charge (part 2.)
Then we will discuss how a moving charge
makes the field (part 1.)
Some examples of the force on a moving
charge in a magnetic field
observation #1- The magnetic force is always perpendicular to the
magnetic field
observation #2- The magnetic force is always perpendicular to the
particle velocity
Magnetic Forces
• Four observations about a charge q moving in a
magnetic field B
• The force is:
– proportional to the charge q
– proportional to the velocity v
– perpendicular to both v and B
– proportional to sinφ, where φ is the angle between v and B
• This can be summarized as:
r r
F = qv " B
• The symbol represents a cross product (not a
multiplication!) of the velocity vector of the charged
particle and the magnetic field vector.
!
More on magnetic forces
• The magnetic force is zero if the velocity is
either parallel or anti-parallel to the magnetic
field.
sin(0) = sin(180) = 0
• The force has its maximum value when the
velocity and magnetic field are perpendicular
sin(90) = 1
• The force on a negative charge is in the
opposite direction
The vector or cross product
v r r
If C = A ! B then the magnitude
r of
where θ is the angle between A and
is given by the “Right Hand Rule”:
r r
v
Cr = A B sin ! ,
r
B . The direction of C
Advice on using the Right Hand Rule:
1) First determine the plane that contains A and B. The cross product will
point perpendicular to that plane. There are only two choices.
2) Use the Right Hand
r Rule
r tor pick which choice is correct.
3) If you are using F = qv ! B , Remember that a negative charge will
reverse the direction of the cross product!
Magnetic Force and Magnetic Field
r
r r
F = qv " B
F = q vB!
Units of Magnetic Field:
1 Tesla = 1 T = 1 Newton/(Ampere·meter)
10-4T =1 Gauss ~ Magnetic field of the earth
A steady 50 T field is very large (about the largest possible today in a lab).
The surface of a neutron star is believed to be ~108T
Example
A uniform magnetic field points
into the screen. The direction is
indicated by the crosses, dots
would be coming out of the screen
(imagine arrows, you see the tail
feathers, not the points).
A positive charge moves from
point A to point C, the direction of
the magnetic force is:
a) up and right,
d) down and left
b) up and left,
c) down and right,
Example
A uniform magnetic field points
into the screen.
A positive charge moves from
point A to point C, the direction of
the magnetic force is:
a) up and right,
d) down and left
b) up and left,
c) down and right,
r
r r
The magnetic force is given by F = qv ! B
The cross produce of the velocity and the magnetic field vector
is up and to the left. As the charge is positive, the force is in
the same direction
Motion of charged particles in EM fields
r r r
v
F = q( E + v ! B )
EM force – “Lorentz” Force
+
v
r
d2 r
F = ma = m 2 r
dt
Newton’s 2nd Law
r r
d r r r
d2 r
q[ E ( r ) + r ! B( r )] = m 2 r
dt
dt
r
r (t )
Differential equation
Solution = “Equation of Motion”
For constant force there are two important simple cases:
v
r
Uniform linear acceleration
F is parallel to v
v
r
F is perpendicular to v
Uniform circular motion
r r r
v = v0 + at
Uniform circular motion (see Y&F chapter 3)
r
!v !s
=
v1
R
r v1
!v = !s
R
r
!v v1 !s
Average acceleration
=
!t
R !t
r
!v v1
r
!s
a = lim !t "0
= lim !t "0
!t
R
!t
r v2
a =
R
Similar triangles (i.e.
same angle) gives
Uniform Circular Motion implies a acceleration that
is always directed towards the center of the circle.
r
r mv 2
Amplitude of acceleration is constant: F = m a =
R
Direction of acceleration changes with time
Angular Velocity (see Y& F chapter 9)
Uniform Circular Motion is described by an “Angular Velocity”
Since
s = r!
v = r!
and
ds
v=
dt
v2
a=
= r! 2
r
v2
F = m = mr! 2
r
"
Frequency f =
2!
1
Period
T=
f
"!
Angular Velocity (radians/s)
cycles/s=Hertz
seconds
d!
"=
dt
Angular velocity as a vector
Motion due to a Magnetic Force
• What is the motion like if the velocity is not
perpendicular to B?
• Break up the velocity into components along the
magnetic field and perpendicular to it
• The component perpendicular will still produce
circular motion
• The component parallel will
produce no force, and this
motion will be unaffected
• The combination of these
two types of motion result in
a helical motion
Velocity Selector
A neat device for selecting ions by their velocities (actually
used in research!) has crossed electric and magnetic
fields.
A charged particle (ion) experiences both the E and the B
r
r
field.
The forces acting on the ion are:
FE = qE
r
r r
FB = qv " B
For a positive charge, force due the electric field is to the
left and force due to the magnetic field is to the right
If the velocity of the ion is precisely right then the forces
cancel out
!
r
r
FE = FB
E
v=
B
Thomson’s e/m Experiment (1897)
J.J. Thomson used the idea of a velocity selector to measure the
ratio of charge to mass for the electron. The hot cathode releases
electrons which are accelerated towards the two anodes.
1 2
mv = eV
2
E
v=
B
e
E2
=
2
m 2VB
Measure E,V and B, find e/m = 1.7588 x 1011C/kg
regardless of material on cathode. Discovery of
the electron!
!
!
Magnetic force on a current-carrying conductor
• We’ve seen that there is a force on a
charge moving in a magnetic field
• Now we’re going to consider multiple
charges moving together, such as a
current in a conductor
• We start with a wire of length l and cross
section area A in a magnetic field of
strength B with the charges having a
drift velocity of vd. The total number of
charges in this section is then nAl where
n is the charge density. The force on a
single charge is given by F=qvdB. So,
the total force on this segment is:
F = nqvdAlB
Magnetic force on a current-carrying conductor
• We’ve found: F = nqvdAlB; however we already
know J=nqvd.
• So F=JAIB = IlB
• The force is proportional to the current through the
wire, the length of the wire in the field and the
magnetic field strength
Magnetic force on a current-carrying conductor
• But what if the magnetic field and the wire are not
perpendicular? Only the component of B ( B" = Bsin #)
perpendicular to the wire exerts a force.
!
F = IlB"
F = IlBsin #
r r r
F = Il $ B