Transcript Motion Along a Straight Line at Constant

Book Reference : Pages 108-109 & -110-111
1.
To apply are knowledge about
electromagnetic behaviour to electric motors
2.
To understand moving charges in magnetic
fields
3.
To derive an equation for the force
experienced by moving charges in magnetic
fields
The force that a current carrying conductor experiences
in a magnetic field is the basic principle behind an
electric motor
X
N
Y
S
Consider the above rectangular coil which has n turns & can
rotate about its vertical axis. The coil is arranged in a uniform
magnetic field. The coil will experience a pair of forces where the
direction is given by the left hand rule.
The vertical sides of the coil are perpendicular to the
field and will experience a force given by :
F = BIl
Each side will experience a force in the opposite
direction. Since we have a coil with n turns the force is
given by
F = BIln
We have a pair of forces in opposite direction which are
not acting through the same line (i.e. We have a couple
equal to Fd where d is the perpendicular distance
separating the forces. In this case d is the width of the
coil w)
Looking from above :
w
F
F
Y
X
Y

X
F
Coil Parallel to field.
Couple = Fw
F
Coil now at an angle  to the field.
Distance between forces now wcos 
As the coil rotates the distance between the forces reduces to wcos 
the torque is now Fwcos  which expands to BIlnwcos 
lw is the area A of the coil giving BIAncos 
When the coil is parallel ( =0, cos 0 = 1) the torque is BIAn
When the coil is perpendicular ( =90, cos 90 = 0) the torque is zero
In real motors, current must be
delivered to the rotating coil
(direct connections would
twist!). In simple motors,
sprung loaded carbon brushes
push against a rotating
commutator
Animation & clip
The second issue is that after
half a rotation the force would
change direction. We need to
change the direction of the
current so that as the coil
rotates the force is always in the
same direction. A split-ring
commutator is used
In real motors, several “armature” coils
are wound onto an iron core. Each coil has
its own section of the split commutator so
that each coil is pushed in the same
direction. The iron core makes the field
radial and each coil is parallel to the coil
most of the time (=0) making for smooth
running
A rectangular coil with 50 turns of width 60mm
& length 80mm is placed parallel to a uniform
magnetic field which has a flux density of 85mT.
The coil carries a current of 8A and the shorter
side of the coil is parallel to the field
Sketch the arrangement and determine the
forces acting on the coil
[2.72N Vertically up on one side and down on the other]
Current carrying conductors experience a force in a
magnetic field. In a similar way charge particles also
experience a force in a magnetic field and are deflected
This technology has been exploited in all sorts of
Cathode Ray Tubes (CRTs). TVs, Monitors & Oscilloscopes
Electrons are emitted by a heated cathode (-ve) & are
accelerated towards the anode (+ve). The beam is then
deflected by a magnetic field (coils “under control of the
picture”)
Notes!
Flemming’s LH rule
applies... But..
Current is in the
opposite direction to (ve) electron movement
The charge has to be
moving
Projectile like problems
Last two subtopics are a bit back to front.... The reason
why a current carrying conductor experiences a force is
because the electrons moving along the wire experience
a force and are moved to one side of the conductor
which exerts a force on it
A beam of charged particles is a flow of electric current
(Current = charge per second Q/t)
Consider a charge Q moving with a velocity v in a time t.
The distance travelled is therefore vt
If we apply this to our equation for a wire (F = BIl)
In particular we can substitute I = Q/t & l = vt
F = BIl
F = B (Q/t) vt
F = BQv
The above equation defines the force experienced by a
particle with a charge of Q as it moves with a velocity v
in a perpendicular direction to a magnetic field with flux
density B
(Note as before we can introduce a sin  term to the above
equation for when the velocity is at angle  to the field lines but it
is beyond our spec
Electrons move upwards in a vertical wire at
2.5x10-3 m/s into a uniform horizontal magnetic
field which has a flux density of 95 mT & is
oriented along a line South to North
Calculate the magnitude and direction of the
force on each electron
[3.8 x 10-23N West to East]
HW. Please read pages 111 (bottom) & 112 (top) about
the Hall effect. An application but not on the spec’