1. Electric Motors & Charged Particles
Learning Objectives
Book Reference : Pages &
To apply are knowledge about electromagnetic behaviour to electric motors
To understand moving charges in magnetic fields
To derive an equation for the force experienced by moving charges in magnetic fields

2. Electric Motors 1
The force that a current carrying conductor experiences in a magnetic field is the basic principle behind an electric motor X Y
Force
(in) N S
Force
(out)
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.

3. Electric Motors 2
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)

4. Electric Motors 3 Looking from above :w F Y X Y  X F F
Coil Parallel to field. Couple = Fw
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

5. Electric Motors 4 : Practicalities
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
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
Animation & clip

6. Electric Motors 5 : Practicalities
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

7. Problems 1
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]

8. Moving Charges in Magnetic Fields 1
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

9. Moving Charges in Magnetic Fields 2
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

10. Moving Charges in Magnetic Fields 3
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)

11. Moving Charges in Magnetic Fields 3
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

12. Problems 2
Electrons move upwards in a vertical wire at 2.5x10-3 ms-1 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’