Electro-Magnetic Induction © David Hoult 2009

Magnetic flux © David Hoult 2009

If the magnitude of the flux density is B then the magnitude of the magnetic flux (  ) linked with the area A is defined to be © David Hoult 2009

If the magnitude of the flux density is B then the magnitude of the magnetic flux (  ) linked with the area A is defined to be   = A B © David Hoult 2009

units of flux   = A B © David Hoult 2009

  = A B units of flux T m 2 © David Hoult 2009

  = A B units of flux T m 2 1 T m 2 = 1 Weber (Wb) © David Hoult 2009

Now, the magnitude of the component of the flux density perpendicular to the area is B cos  © David Hoult 2009

Now, the magnitude of the component of the flux density perpendicular to the area is B cos    = A B cos  so the magnetic flux (  ) linked with the area is now alternatively © David Hoult 2009

In practice, the area in question is usually surrounded by a conductor, often a coil of wire. © David Hoult 2009

In practice, the area in question is usually surrounded by a conductor, often a coil of wire. If the coil of wire has N turns, we define the flux linkage as follows © David Hoult 2009

Flux linkage = N  In practice, the area in question is usually surrounded by a conductor, often a coil of wire. If the coil of wire has N turns, we define the flux linkage as follows © David Hoult 2009

EMF induced in a conductor moving through a uniform magnetic field © David Hoult 2009

The wire moves distance  s in time  t. In this time, a charge  q moves past any point in the wire. © David Hoult 2009

work done = F  s The wire moves distance  s in time  t. In this time, a charge  q moves past any point in the wire. © David Hoult 2009

work done = F  s The wire moves distance  s in time  t. In this time, a charge  q moves past any point in the wire. work done per unit charge = F sF s qq © David Hoult 2009

work done = F  s The wire moves distance  s in time  t. In this time, a charge  q moves past any point in the wire. work done per unit charge = F sF s qq work done per unit charge is the induced emf © David Hoult 2009

If the wire moves at constant speed, the force F must be © David Hoult 2009

If the wire moves at constant speed, the force F must be equal but opposite to the force acting on it due to the current I, induced in it © David Hoult 2009

If the wire moves at constant speed, the force F must be equal but opposite to the force acting on it due to the current I, induced in it F = - I L B © David Hoult 2009

If the wire moves at constant speed, the force F must be equal but opposite to the force acting on it due to the current I, induced in it F = - I L B F sF s qq E = © David Hoult 2009

If the wire moves at constant speed, the force F must be equal but opposite to the force acting on it due to the current I, induced in it F = - I L B F sF s qq E = - I LB  s qq E = © David Hoult 2009

I =I =

I =I = qq tt

I =I = qq tt - (  q/  t) L B  s qq E = and © David Hoult 2009

I =I = qq tt - (  q/  t) L B  s qq E = and ss tt = © David Hoult 2009

I =I = qq tt - (  q/  t) L B  s qq E = and ss tt = v © David Hoult 2009

I =I = qq tt - (  q/  t) L B  s qq E = and ss tt = v E = - B L v © David Hoult 2009

The Laws of Electro-magnetic Induction © David Hoult 2009

When e.m.i. occurs, any induced current will flow in such a direction as to Lenz’s Law © David Hoult 2009

When e.m.i. occurs, any induced current will flow in such a direction as to oppose the change producing it Lenz’s Law © David Hoult 2009

When e.m.i. occurs, any induced current will flow in such a direction as to oppose the change producing it Lenz’s Law It should be clear that Lenz’s law is an “electro- magnetic version” of © David Hoult 2009

When e.m.i. occurs, any induced current will flow in such a direction as to oppose the change producing it Lenz’s Law It should be clear that Lenz’s law is an “electro- magnetic version” of the law of conservation of energy © David Hoult 2009

When e.m.i. occurs, any induced current will flow in such a direction as to oppose the change producing it Lenz’s Law Faraday’s Law The induced emf is directly proportional to the © David Hoult 2009

When e.m.i. occurs, any induced current will flow in such a direction as to oppose the change producing it Lenz’s Law Faraday’s Law The induced emf is directly proportional to the rate of change of flux linking the conductor © David Hoult 2009

The sense of the induced current can be predicted using Fleming’s RIGHT hand rule © David Hoult 2009

The sense of the induced current can be predicted using Fleming’s RIGHT hand rule which is pretty much like Fleming’s left hand rule © David Hoult 2009

The sense of the induced current can be predicted using Fleming’s RIGHT hand rule which is pretty much like Fleming’s left hand rule except, guess what... © David Hoult 2009

The sense of the induced current can be predicted using Fleming’s RIGHT hand rule which is pretty much like Fleming’s left hand rule except, guess what... using the right hand instead of the left hand ! © David Hoult 2009

Motion ThuMb Field Current First finger SeCond finger © David Hoult 2009

Both the laws of e.m.i. can be combined in a single mathematical statement © David Hoult 2009

Faraday Both the laws of e.m.i. can be combined in a single mathematical statement © David Hoult 2009

E   tt Faraday Both the laws of e.m.i. can be combined in a single mathematical statement © David Hoult 2009

E   tt Faraday Faraday + Lenz Both the laws of e.m.i. can be combined in a single mathematical statement © David Hoult 2009

E   tt Faraday E  -  tt Faraday + Lenz Both the laws of e.m.i. can be combined in a single mathematical statement © David Hoult 2009

Using the S.I. system of units the constant of proportionality is 1 so, if (as is often the case) the conductor is a coil of wire having N turns, we have © David Hoult 2009

E = -N  tt Using the S.I. system of units the constant of proportionality is 1 so, if (as is often the case) the conductor is a coil of wire having N turns, we have © David Hoult 2009

E = -N  tt Using the S.I. system of units the constant of proportionality is 1 so, if (as is often the case) the conductor is a coil of wire having N turns, we have This equation is often referred to as Neumann’s Law © David Hoult 2009

For a wire moving at 90° to a field with speed v, the induced emf is given by © David Hoult 2009

For a wire moving at 90° to a field with speed v, the induced emf is given by E = - B L v © David Hoult 2009

E = - B L v © David Hoult 2009

E = - B L v = - B L ss tt © David Hoult 2009

E = - B L v = - B L ss tt = - B AA tt © David Hoult 2009

E = - B L v = - B L  tt ss tt = - B AA tt = - © David Hoult 2009

E = - B L v = - B L  tt ss tt = - B AA tt = - This suggests that change of flux linking a conductor and flux cutting by a conductor are equivalent actions © David Hoult 2009

Simple a.c. generator © David Hoult 2009

Simple a.c. generator © David Hoult 2009

Simple a.c. generator © David Hoult 2009

Simple a.c. generator © David Hoult 2009

Simple a.c. generator © David Hoult 2009

Simple a.c. generator © David Hoult 2009

Simple a.c. generator Fleming’s right hand rule gives the sense of the current to be... © David Hoult 2009

Simple a.c. generator © David Hoult 2009

Simple a.c. generator Fleming’s right hand rule gives the sense of the current to be... which means that, at the instant shown in the diagram, terminal Q is the positive terminal of the generator © David Hoult 2009

Simple a.c. generator Fleming’s right hand rule gives the sense of the current to be which means that, at the instant shown in the diagram, terminal Q is the positive terminal of the generator When the coil has turned through 180°, P will be the positive terminal so the generator gives us alternating current © David Hoult 2009

The Transformer © David Hoult 2009

weak induced alternating emf © David Hoult 2009

iron core © David Hoult 2009

The Ideal Transformer If a transformer is described as ideal we mean it is 100% efficient © David Hoult 2009

The Ideal Transformer If a transformer is described as ideal we mean it is 100% efficient In other words, power output (at secondary coil) equals power input (to primary coil) © David Hoult 2009

The Ideal Transformer Therefore, in an ideal transformer: © David Hoult 2009

The Ideal Transformer Therefore, in an ideal transformer: i) the coils have zero resistance © David Hoult 2009

The Ideal Transformer Therefore, in an ideal transformer: i) the coils have zero resistance ii) all the magnetic flux  produced by the primary current I p is linked with the secondary coil © David Hoult 2009

The Ideal Transformer Therefore, in an ideal transformer: i) the coils have zero resistance ii) all the magnetic flux  produced by the primary current I p is linked with the secondary coil iii) no current flows in the iron core © David Hoult 2009

The Ideal Transformer © David Hoult 2009

The Ideal Transformer When I p changes,  changes. © David Hoult 2009

The Ideal Transformer When I p changes,  changes. When  changes, an emf is induced in both coils. © David Hoult 2009

At the primary coil, the magnitude of the induced emf is given by © David Hoult 2009

At the primary coil, the magnitude of the induced emf is given by E p = - N p  tt © David Hoult 2009

At the primary coil, the magnitude of the induced emf is given by E p = - N p  tt as we are assuming zero resistance coils, the supply voltage must also have this magnitude in order to maintain the flow of current © David Hoult 2009

At the primary coil, the induced emf is given by as we are assuming zero resistance coils, the supply voltage must also have this magnitude in order to maintain the flow of current At the secondary coil, the magnitude of the induced emf is given by E s = - N s  tt E p = - N p  tt © David Hoult 2009

EsEs EpEp = NpNp NsNs

EsEs EpEp = NpNp NsNs here we are assuming that  is the same for both coils © David Hoult 2009

The power input to the primary coil is given by power in = © David Hoult 2009

The power input to the primary coil is given by power in = E p I p © David Hoult 2009

The power input to the primary coil is given by power in = E p I p The power output from the secondary coil is given by power out = © David Hoult 2009

The power input to the primary coil is given by power in = E p I p The power output from the secondary coil is given by power out = E s I s © David Hoult 2009

The power input to the primary coil is given by power in = E p I p The power output from the secondary coil is given by power out = E s I s as we are considering an ideal transformer © David Hoult 2009

The power input to the primary coil is given by power in = E p I p The power output from the secondary coil is given by power out = E s I s as we are considering an ideal transformer E p I p = E s I s © David Hoult 2009

E p I p = E s I s Therefore, with an ideal transformer, when the secondary coil is open circuit (not connected to anything), there is no net energy taken from the supply. © David Hoult 2009

E p I p = E s I s Therefore, with an ideal transformer, when the secondary coil is open circuit (not connected to anything), there is no net energy taken from the supply. Energy is stored in the magnetic field during the time the current is increasing but is recovered from the field when it “collapses”. © David Hoult 2009

Transmission of Electrical Energy © David Hoult 2009

If the voltage across the bulb is 6V the current will be © David Hoult 2009

If the voltage across the bulb is 6V the current will be 1A © David Hoult 2009

If the voltage across the bulb is 6V the current will be 1A Power wasted in R = R I 2 = R © David Hoult 2009

If the voltage across the bulb is still 6V the current flowing through the bulb is still 1A © David Hoult 2009

If the step down transformer is 100% efficient the current flowing through R is © David Hoult 2009

If the step down transformer is 100% efficient the current flowing through R is 1/8 A © David Hoult 2009

If the step down transformer is 100% efficient the current flowing through R is 1/8 A Power wasted in R = R I 2 = © David Hoult 2009

If the step down transformer is 100% efficient the current flowing through R is 1/8 A Power wasted in R = R I 2 = R 64 © David Hoult 2009

64 times less energy wasted ! © David Hoult 2009