Download Motors and Generators

Motors and Generators
Check Your Learning
FOR THESE QUESTIONS ASSUME DIRECTIONS ARE IN
A FLAT PLANE.
1. A proton is travelling South in a magnetic field that is
directed upward. What is the direction of the force on
the proton?
a. down
b. East
c. West
d. South
The answer is c.
Check Your Learning
2. An electron is travelling upward through a magnetic field
that is directed west to east. The force on the electron
will be
a. East
b. North
c. South
d. Zero
The answer is c – remember that it is an electron so if
using your right hand your thumb must point downward.
Check Your Learning
3. A proton is travelling west in a magnetic field that is
directed east. The force on the proton is
a. East
b. North
c. South
d. Zero
The answer is d – there is no force since the proton is
travelling parallel to the magnetic field.
Check Your Learning
4. A wire is carrying a current across this page in a
magnetic field that is directed down the page. The force
on the wire is into the page. The current must be
a. Left to right
b. Right to left
c. Zero
The answer is a.
Motors
• A motor is a device that will convert
electrical potential energy into kinetic energy
• Motors work on the principle of torque on a
coil (remember the RHT rule 3 example?)
– When a coil of wire carrying a current is
placed in a magnetic field it will
experience force.
– If the coil is free to rotate, it will, as the
force exerts a torque.
Motors
• All electric motors contain 2 key
components
– A coil of wire
– Permanent magnets
• Kinetic energy is possible because when an
electric current occurs in a magnetic field,
the induced magnetic field in the wire will
interact with the magnetic field
How does a motor work?
• http://www.explainthatstuff.com/electricmot
ors.html
Calculating Motor Force (FM)
FM  B IL
• Where FM = motor force (N)
• B┴ = magnetic field intensity in Tesla (T) (must
be perpendicular component)
• I = current in amperes (amps, A)
• L = length of conductor in the magnetic field (m)
Motor Force
• Note 1: For a looped conductor (ie, coil):
L = nC
• Where n = number of loops (an integer)
• C = circumference of one loop
• Note 2: If the magnetic field is not
perpendicular to the current, you need to
modify the equation:
FM  B IL sin 
Example 1
• A conductor carries a current of 4.8A to the
right through a magnetic field pointing into
the board. The length of wire inside the
field is 25cm and it experiences a force of
0.60N. What is the intensity of the field?
What direction is the force in?
Answer
• Force points up.
FM  B IL
FM
B 
IL
0 .6
B 
(4.8)(0.25)
 0.50T
Example 2
• A 3.2 cm wire is carrying a 34 mA current as shown in
the magnetic field below that is directed to the right. The
strength of the magnetic field is 0.23 T. Find the force
(magnitude and direction) on the wire.
Solution
l  3.2cm  0.032m
I  34mA  0.034 A
B  0.23T
  40.0
F ?
F  IlB sin 
 (0.034)(0.032)(0.23)sin 40.0
 1.6 104 N
The magnetic field is to the right, and
a component of the current is up the
page. Using the third right hand rule,
• Fingers point to the right (with the
magnetic field)
• Thumb points up the page
• Palm is left facing into the page
The force is therefore into the page.
F  1.6 104 N , into the page
Example 5 (from other sheet)
• A strong current is suddenly switched on in
a wire, but no force acts on the wire. Can
you conclude that there is no magnetic
field at the location of the wire?
• No. It is possible that there is a magnetic
field but that it is parallel to the wire.
There is no force when a magnetic field
and a wire carrying a current are parallel.
Practice Problems
• Page 778, questions 1 to 4
• Worksheet (given out on Friday):
• Questions 7, 8, 9 (current is 1.9A – we no
longer cover this section of the textbook)
Calculating Motor Force on Moving
Charged Particles
• This is how devices like particle
accelerators work.
• Any moving charged particle generates a
magnetic field (like electrons moving in a
wire – current).
• https://www.youtube.com/watch?v=G6mmI
zRz_f8
Motor Force on Moving Charged
Particles
FM  qBv
•
•
•
•
Where FM = magnetic force (N)
q = charge of particle (C)
B┴ = magnetic field intensity (T)
v = speed of particle (m/s)
Example 3
• An electron moving at 1.0 x 105 m/s
(initially moving to the left) enters a 0.25T
field pointed out of the board. Calculate
the force on the electron and draw its path
through the field.
• ** NOTE: If the moving particle is negative, you must
point your thumb in the direction opposite the motion of
the particle.
Answer
FM  qB v
 (1.60 x10
 4.0 x10
UP
19
15
N
5
)(0.25)(1.0 x10 )
Example 4
• A particle, moving with a velocity of
8.00×104 m/s at an angle of 30’ with
respect to a magnetic field of 5.60 × 10−5
T, experiences a force of 2.00×10−4 N.
• Calculate the magnitude of the particle’s
charge. (NOTE: use sinθ to get
perpendicular angle)
• 8.93 x 10-5 C
Practice Problems
• Worksheet: Questions 11, 12, 13