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Chapter 19
Magnetic Force
on Wires
Solenoids
Mass Spectrometers
  0I
B 
2r
The magnetic field
around a very long,
current carrying wire
is given by:
Where the constant of proportionality is
known as the permeability of free space
and is found to be
 0  4  10 Tm / A
7
The magnetic field is a
vector quantity, so it has
a direction!!!!
Wrap your hand around
the wire with your thumb
pointed in the direction
of the current.
I
Your fingers curl around
the wire in the direction
the magnetic field points.
If we have two very long, current
carrying wires parallel to one
another, what force do they feel?
I1
I2
d


F  B IL sin 
 = 90o


Fmax  B IL
If the wires are parallel,
the magnetic field from
one wire will be
perpendicular
to the direction of the
current in the other wire.
I1
I2
1
d
2


F  B1@2 I 2 L2
  0I1I 2
F
L2
2d

 0 I1
B1@2 
2d

F
 0I1I 2

L2
2d
I1
I2
1
r
2
In what direction does the force act???
Using the right hand rule, we determine
that the magnetic field from wire 1 at the
location of wire 2…. Is into the page!
Now, using the OTHER right hand rule (the
one for magnetic force on a current carrying wire
in a magnetic field), we find that the direction of
the force on wire 2 ...
Is UP toward wire 1!
I1
I2
1
r
2
Parallel wires carrying current in the SAME
direction are attracted to one another.
You might be able to guess…
Parallel wires carrying current in the OPPOSITE
direction are repelled from one another.
But what if the wire isn’t a very long, straight wire?
What if it’s a loop of wire instead?
We won’t go into the details of the magnetic
field created by a single loop, but this is what
it looks like --- a lot like the field of a bar magnet!
However, we will look at the magnetic field
created by a whole bunch of such loops
combined…AKA
Solenoids produce fairly uniform magnetic
fields inside their boundaries, and generate
negligible fields outside their boundaries.

Binside   0nI
Where n is the number of turns per unit
length of the solenoid.
The magnetic field is directed parallel to the
axis of the solenoid. The right hand rule
for currents will help you determine toward
which end the field points.
A solenoid has length of 25 cm, radius
of 1 cm, 400 turns, and carries a
current of 3 A. What is the value of the
magnetic field inside the solenoid?

Binside   0nI
N 400
1
n 
 1600m
L .25

7
3
Binside  (4  10 )(1600)(3)  6  10 T
We saw that charged particles in a uniform
magnetic field will move in a circle.
x
x
x
x
x
x
x
x
x
x
x
x
x r x
x
x
x
x
x
x
We can take advantage of our understanding
of the magnetic force to create a device
which can tell us the elemental composition
of unknown particles.
x
x
x
x
x
x
x
x
x
x
x
x
x r x
x
x
x
x
x
x
mv
r
qB
We saw that the radius of the circle is
determined by the mass, speed, and charge
of the particles, and the strength of the
magnetic field.
If we know the velocity at which a particle
enters a magnetic field, the strength of that
field, and the particle’s charge, we can
measure the radius of curvature of the particle.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
d
rq B
m
v
We can measure d
using a photographic
plate (for example) to
record the location of
the end of the particle’s
path.
d=2r
If two particles with the same charge enter
the magnetic field with the same velocity,
then the difference in the radii of their paths
must be due to a difference in their masses.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
d1
d2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
mv
r
qB
d1
d2
d 2 d1
v
 
( m2  m1 )
2
2 qB
2v
d 2  d1 
(m2  m1 )
qB
Mass
Spectrometer
Two singly ionized particles enter a
mass spectrometer at a speed of
3 X 106 m/s. The strength of the
magnetic field is 0.625 T. If one of the
particles is H and the other particle hits
the photographic plate 110 cm further
away than the H atom, what chemical
element is the second particle?
Let’s first determine the distance from the
entrance point that the H atom hits the
photographic plate.
27
mv (167
.  10 kg)(3  10 m / s)
r

 5 cm
19
qB
(16
.  10 C)(0.625T)
6
Now, let’s figure out the radius of curvature
of the mystery element...
We were told that the 2nd element collides with
the photographic plate 110 cm further away from
the entrance point than the H atom.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x d2 - d1= 110 cm
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x d2 = 110 + 2r1
x
d2= 120 cm
d1
d2
r2= 60 cm
Finally, let’s determine the mass of
our mystery element...
19
rqB (0.6 m)(16
.  10 C)(0.625T)
26
m

 2  10 kg
6
v
3  10 m / s
PHYSICS
RULES!
Televisions rely upon
electromagnetic fields
to produce the images
we see.
An electron “gun” fires electrons
at the television screen. When
they collide with the material on
the back side of the screen, colored
light is emitted, producing one pixel
in the image we see on the screen.
television
screen
electromagnet
Changing the strength of the magnetic field
changes the degree to which the electron beam
is deflected.
We now make explicit a fact we’ve assumed
implicitly in dealing with magnetic fields thus far...
The superposition principle applies!
x
1
Btot = B1 + B2
x
2
Induced Voltages and Inductance
We saw in Chapter 19 that moving charges
(currents) create magnetic fields. Nature often
reveals a great deal of symmetry. So,
scientists wondered, “Can magnetic fields can
create currents?” You might guess that the
answer is, “Yes.” In this chapter, we’ll explore
this physical process known as induction.
Shortly after Oersted’s discovery that currents
create magnetic fields, Michael Faraday (England)
and Joseph Henry (US) conducted experiments
to determine if magnetic fields created electrical
currents...
Soft magnetic
material
+
A
_
primary coil
Primary Circuit
secondary coil
Secondary Circuit
A
+
_
Faraday’s Observations:
With the switch open, no reading on the ammeter.
When the switch is closed, the ammeter needle
deflects momentarily toward the right, then
returns to 0.
A
+
_
Faraday’s Observations:
With a constant current flowing through the
primary circuit, the ammeter shows no reading.
When the switch is opened, the ammeter needle
deflects to the left momentarily, then returns to 0.
A
+
_
Faraday’s Conclusions:
Steady currents in the primary produce constant
magnetic fields, which result in no current in the
secondary circuit.
Changing currents in the primary circuit result
in changing magnetic fields, which result in
induced currents in the secondary circuit.
A
If we move a bar magnet toward a loop of
wire, the ammeter deflects to the left.
When we stop moving the magnet, the
ammeter reads 0.
When we pull the magnet away from the loop,
the ammeter deflects to the right.
A
As the magnet moves toward the loop, the
magnetic field near the loop increases.
As the magnet moves away from the loop, the
magnetic field near the loop decreases.
Such results are consistent with Faraday’s
experiment: Changing magnetic fields lead
to induced currents.
Magnetic
Flux
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
side view
How much magnetic
field is passing through
the loop?
One way to think about this is
to simply count the number
of magnetic field lines
passing through a loop.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
30o
60o
top view
side view
Magnetic flux is the amount of magnetic field
perpendicular to the loop.
If you separate the magnetic field into
components parallel and perpendicular
to the loop, you see that...
30o
60o
 =| B |A =| B|A cos
top view
Btot

B
B||
This is also the direction of the normal to the loop!
 = B A = BA cos
 = B A cos
Wb 2
  T m  2 m  Wb
m
2
So, the Weber is the unit of magnetic flux.