Book Reference : Pages 108-109 & -110-111Book Reference : Pages 108-109 & -110-111

1.1. To apply are knowledge about To apply are knowledge about electromagnetic behaviour to electric motorselectromagnetic behaviour to electric motors

2.2. To understand To understand movingmoving charges in magnetic charges in magnetic fieldsfields

3.3. To derive an equation for the force To derive an equation for the force experienced by moving charges in magnetic experienced by moving charges in magnetic fieldsfields

The force that a current carrying conductor experiences The force that a current carrying conductor experiences in a magnetic field is the basic principle behind an in a magnetic field is the basic principle behind an electric motorelectric motor

N S

Consider the above rectangular coil which has n turns & can Consider the above rectangular coil which has n turns & can rotate about its vertical axis. The coil is arranged in a uniform rotate about its vertical axis. The coil is arranged in a uniform magnetic field. The coil will experience a pair of forces where the magnetic field. The coil will experience a pair of forces where the direction is given by the left hand rule.direction is given by the left hand rule.

Force(out)

Force(in)

X Y

The vertical sides of the coil are perpendicular to the The vertical sides of the coil are perpendicular to the field and will experience a force given by :field and will experience a force given by :

F = BIF = BIll

Each side will experience a force in the opposite Each side will experience a force in the opposite direction. Since we have a coil with n turns the force is direction. Since we have a coil with n turns the force is given bygiven by

F = BIF = BIllnn

We have a pair of forces in opposite direction which are We have a pair of forces in opposite direction which are not acting through the same line (i.e. We have a couple not acting through the same line (i.e. We have a couple equal to Fd where d is the perpendicular distance equal to Fd where d is the perpendicular distance separating the forces. In this case d is the width of the separating the forces. In this case d is the width of the coil w)coil w)

Looking from above :Looking from above :F

F

w

Coil Parallel to field. Coil Parallel to field. Couple = FwCouple = Fw

F

F

Coil now at an angle Coil now at an angle to the field. to the field. Distance between forces now wcos Distance between forces now wcos

As the coil rotates the distance between the forces reduces to wcos As the coil rotates the distance between the forces reduces to wcos

the torque is now Fwcos the torque is now Fwcos which expands to BI which expands to BIllnnwcos wcos

llw w is the area A of the coil giving BIAncos is the area A of the coil giving BIAncos

When the coil is parallel (When the coil is parallel ( =0 =0, cos 0, cos 0 = 1) the torque is BIAn = 1) the torque is BIAn

When the coil is perpendicular (When the coil is perpendicular ( =90 =90, cos 90, cos 90 = 0) the torque is zero = 0) the torque is zero

X

Y

X

Y

In real motors, current must be In real motors, current must be delivered to the rotating coil delivered to the rotating coil (direct connections would (direct connections would twist!). In simple motors, twist!). In simple motors, sprung loaded carbon sprung loaded carbon brushesbrushes push against a rotating push against a rotating commutatorcommutator

The second issue is that after The second issue is that after half a rotation the force would half a rotation the force would change direction. We need to change direction. We need to change the direction of the change the direction of the current so that as the coil current so that as the coil rotates the force is always in rotates the force is always in the same direction. A the same direction. A split-ring split-ring commutatorcommutator is used is used

Animation & clip

In real motors, several “In real motors, several “armaturearmature” coils ” coils are wound onto an iron core. Each coil has are wound onto an iron core. Each coil has its own section of the split commutator so its own section of the split commutator so that each coil is pushed in the same that each coil is pushed in the same direction. The iron core makes the field direction. The iron core makes the field radial and each coil is parallel to the coil radial and each coil is parallel to the coil most of the time (most of the time (=0) making for smooth =0) making for smooth runningrunning

A rectangular coil with 50 turns of width 60mm A rectangular coil with 50 turns of width 60mm & length 80mm is placed parallel to a uniform & length 80mm is placed parallel to a uniform magnetic field which has a flux density of 85mT.magnetic field which has a flux density of 85mT.

The coil carries a current of 8A and the shorter The coil carries a current of 8A and the shorter side of the coil is parallel to the fieldside of the coil is parallel to the field

Sketch the arrangement and determine the Sketch the arrangement and determine the forces acting on the coilforces acting on the coil

[2.72N Vertically up on one side and down on the other][2.72N Vertically up on one side and down on the other]

Current carrying conductors experience a force in a Current carrying conductors experience a force in a magnetic field. In a similar way charge particles also magnetic field. In a similar way charge particles also experience a force in a magnetic field and are deflectedexperience a force in a magnetic field and are deflected

This technology has been exploited in all sorts of This technology has been exploited in all sorts of Cathode Ray Tubes (CRTs). TVs, Monitors & Cathode Ray Tubes (CRTs). TVs, Monitors & OscilloscopesOscilloscopes

Electrons are emitted by a heated cathode (-ve) & are Electrons are emitted by a heated cathode (-ve) & are accelerated towards the anode (+ve). The beam is then accelerated towards the anode (+ve). The beam is then deflected by a magnetic field (coils “under control of the deflected by a magnetic field (coils “under control of the picture”)picture”) Notes!Notes!

Flemming’s LH rule Flemming’s LH rule applies... But..applies... But..

Current is in the Current is in the opposite direction to (-opposite direction to (-ve) electron movementve) electron movement

The charge The charge hashas to be to be movingmoving

Projectile like problemsProjectile like problems

Last two subtopics are a bit back to front.... The reason Last two subtopics are a bit back to front.... The reason why a current carrying conductor experiences a force is why a current carrying conductor experiences a force is because the electrons because the electrons movingmoving along the wire experience along the wire experience a force and are moved to one side of the conductor a force and are moved to one side of the conductor which exerts a force on itwhich exerts a force on it

A beam of charged particles is a flow of electric current A beam of charged particles is a flow of electric current (Current = charge per second Q/t)(Current = charge per second Q/t)

Consider a charge Q moving with a velocity v in a time t. Consider a charge Q moving with a velocity v in a time t. The distance travelled is therefore vtThe distance travelled is therefore vt

If we apply this to our equation for a wire (F = BIIf we apply this to our equation for a wire (F = BIll) )

In particular we can substitute I = Q/t & In particular we can substitute I = Q/t & l l = vt = vt

F = BIF = BIll

F = B (Q/t) vtF = B (Q/t) vt

F = BQvF = BQv

The above equation defines the force experienced by a The above equation defines the force experienced by a particle with a charge of Q as it moves with a velocity v particle with a charge of Q as it moves with a velocity v in a perpendicular direction to a magnetic field with flux in a perpendicular direction to a magnetic field with flux density Bdensity B

(Note as before we can introduce a sin (Note as before we can introduce a sin term to the above term to the above equation for when the velocity is at angle equation for when the velocity is at angle to the field lines but it to the field lines but it is beyond our specis beyond our spec

Electrons move upwards in a vertical wire at Electrons move upwards in a vertical wire at 2.5x102.5x10-3-3 m/s into a uniform horizontal magnetic m/s into a uniform horizontal magnetic field which has a flux density of 95 mT & is field which has a flux density of 95 mT & is oriented along a line South to Northoriented along a line South to North

Calculate the magnitude and direction of the Calculate the magnitude and direction of the force on each electronforce on each electron

[3.8 x 10[3.8 x 10-23-23N West to East]N West to East]

HW. Please read pages 111 (bottom) & 112 (top) about HW. Please read pages 111 (bottom) & 112 (top) about the Hall effect. An application but not on the spec’the Hall effect. An application but not on the spec’