Electromagnetic Induction. Michael Faraday Faraday was working with two coils mounted on a wooden...

Electromagnetic Induction

Michael Faraday

Faraday was working with two coils mounted on a wooden spool. Whenever he put a large amount of current into the primary coil, a small amount of current was produced in the secondary coil. This flow of current in the secondary coil was only present for a very short time.

Using a galvanometer he was also able to detect a small amount of current when the current to the primary was turned off.

After a lot of further investigations he determined that a changing magnetic field induced (created) a current flow in another completely separate solenoid.

Demo: Pass a Magnet quickly through a coil of wire attached to a galvanometer.

The galvanometer will show a current flow when the magnet moves in and out of the coil.

The amount of current that is induce depends on:

The strength of the magnetic field.

The rate of change of the magnetic field.

The number of coils in the solenoid.

This is known as Faraday's Law.

It describes the amount of emf (voltage) induced when a coil of wire is subjected to a changing magnetic field.

ξ = emf (voltage) {V}

N = number of coils in the solenoid

ΔΦ = change in magnetic flux (Webber's {Wb}

Δt = change in time (s)

ΔΦ What is “magnetic flux”?

It is a combination of the magnetic field intensity and the area in which the magnetic field lines are in.

ΔΦ = ΔBA

B = magnetic field intensity (Tesla)

A = Area of the coil or magnetic field (m2)

The combination of BA must be changed in order to generate an induced current.

Electromagnetic Inductiona)

Examine each of these two side-by-side magnetic flux situations. Do you see why the left side has the greatest magnetic flux?

The Magnetic flux is increasing as the ring is moving upwards. Why?

Electromagnetic InductionExample #1:

A 200 turn circular coil of radius 0.15 m is quickly rotated in a time of 0.12 s from being perpendicular to being parallel to a 0.56 T magnetic field as shown below.

What is the average induced emf (voltage)?

Δt = 0.12 s

Electromagnetic InductionSolution:

ξ =-200 X (0 - 0.56 x π x 0.152 )

7.9128

0.12ξ =

ξ = (The negative sign is just there to remind us in which direction the induced current is going.)

Electromagnetic InductionExample 2:

A coil having 45 loops and as area of 0.35 m2 is initially placed perpendicular to 0.75 T magnetic field. The field is reversed in direction to a magnitude of 0.62 T in a time of 0.25 s. The coil is connected to a 40 ohm resistor. What is the magnitude of the current through the resistor?

. . . . . . .

x x x x x x x

x x x x x x x R = 40 Ω

R = 40 Ω

B = -0.62 TB = 0.75 T

BeforeAfter

Solution:

ξ =-45 x {(-0.62 x 0.35) – (0.75 x 0.35)}

0.25 s

ξ =-45 x {(-0.217) – (0.2625)}

0.25 s

ξ =0.25 s

ξ =21.5775

ξ =-45 x {-0.4795}

0.25 sξ =

-45 x {-0.4795}

0.25 s

= 86.31 V

I = 86.31 V

I = 2.15 A

Remember: Ohm's Law!

But what about the direction of the current?

This will take a little explaining!

Electromagnetic InductionLenz's Law:

The induced emf always gives rise to a current whose own magnetic field opposes the original change in flux.

x x x x x x x

x x x x x x xx x x x x x xx x x x x x x

x x x x x x x

The coil is squished together reducing the area to zero. Which direction will the induced current flow in the coil during the collapse?

x x x x x x x

x x x x x x x . . . . . . .

. . . . . . .

What is the direction of the induced current in the coil as the magnetic field is reversed?

Another Example:

What direction will the current flow in the copper coil when the magnet is brought closer to the coil?

A bar magnet is quickly brought near a single coil of copper wire.

One last example:

What direction will the current flow through the meter as the bar magnet is pulled out?

Answer: Current will Downwards through the meter

Electromagnetic InductionEMF Induced in a Moving Conductor:

v x Δt

ξ =NΔBA

ΔtBut: A = L x(v x Δt) ; N = 1

ξ =B x L x(v x Δt)

B x L x v ξ =This equation will determine the voltage placed across a conductor as it moves in a magnetic field.

x x x x x x x xx x x x x x x xx x x x x x x xx x x x x x x xx x x x x x x x B = 0.75 T

v = 5.6 m/s

0.75 x 0.45 x 5.6 ξ =

ξ = 1.9 V

Example:

(The top of the conductor will become positively charged! {right hand rule})

Electromagnetic InductionThe back (or Counter) Emf in a Motor:

Motor Demo:

A motor simply a set of coils with current running through them placed in a magnetic field.

The split-ring commutator allows the current to change direction every ½ turn keeping the motor coil turning.

But the turning coil crossing the field lines generate a induced current that opposes the original applied current that makes the motor turn.

Electromagnetic InductionThis induced current produces an opposing voltage called back emf. It acts against the voltage causing the motor to spin in the first place.

As the motor spins up to speed the back emf increases in value until most of the applied voltage is cancelled out.

At start up the motor only sees the applied voltage. Since the resistance of the motor is relatively constant the motor draws a lot of current when starting up.

Most heavy duty motors found in refrigerators or table saws have a start capacitor which supplies some of this current at start up so that it doesn't trip the circuit breaker when the motor is first turned on. (In older homes you may have noticed that some lights in a kitchen dim when the refrigerator motor (compressor) turns on.)

When the motor is running at full speed there is lots of back emf. The effective voltage across the motor is much less than at start up.

Since the effective voltage is reduced when the motor is running at full speed, so then is the current required. A motor running at full speed requires much less current than at start up. (Once the refrigerator motor is running at full speed the lights go back to their usual brightness).

A very similar effect can cause an electric drill to “burn out its motor”. If the drill is being used to drill a hole and the drill bit is grabbed and held stationary, the back emf through the drill's motor is reduced to zero. The current through the motor goes very high. High enough in some cases to melt the wires in the motor windings and “burn out”the drill's motor.

Vb = Emf - IR

Vb = back emf

Emf = voltage applied to the motor

I = current applied to the motor

R = resistance of the motor (constant)

The armature windings of a motor have a resistance of 8.0 ohms. The motor is connected to 120 V. The motor draws 2.0 A while running at full speed.

Example:

a) What current is drawn by the motor at start up?

b) What is the back Emf when the motor is running at full speed?

8.0 Ω= 15 A

Vb = 120 V - 2.0 x 8.0 = 104 V

Transformers:

A transformer is used to either increase (Step-UP) or decrease (Step-Down) the voltage for various uses.

Transformers consist of two separate coils of wire and usually a conducting iron metal support.

It is the arrangement of the two coils that creates either a step-up or step-down transformer.

Transformers only work with AC (alternating Current)!

The transformer is very efficient device. It has no moving parts. The two coils that make up the transformer are not in contact with each other so how is it able to change the voltage?

The answer lies in using AC (alternating current) to create a changing current flow which produces a changing magnetic field on one side of the transformer. This induces a changing magnetic field on the other coil which induces a new current flow in this coil.

Since the only a changing magnetic field will induce a new current, the transformer only works if the current is AC (which is changed 60 times every second).

Let us examine how to work with the transformer equations.

For all ideal Transformers:

Power x time In = Power x time Out

And Energy = Power x Time

Energy In = Energy Out

Power In = Power Out

And Power = V x I

Vinput

x Iinput

= Voutput

x Ioutput

Transformer Equation:

Vi x I

ξ =NΔBA

Let us have another look at Faraday's law and apply it to transformers:

Remember: ξ = voltage

oΔBA x I

Vi x I

Replace the V on each side with Faraday's Law:

NiΔBA x I

The B and A and t are the same on both sides and cancel each other out.

This results in a new equation:

NiIi = N

Putting this equation with the Power one results in:

Electromagnetic InductionExample:

A small step-down transformer is plugged into a wall outlet supplying 120 V. The transformer has 1800 windings on the input side and 40 windings on the output side. The input current is 0.0050 A.

a) What is the output voltage of this transformer?

Vo = (V

Vo = (120 x 40) / 1800 = 2.7 V

Electromagnetic Inductionb) What is the output current of this transformer?

c) What is the power of this transformer?

Io = (V

i) / V

Io = (120 V x 0.0050 A) / 2.7 V

Io = 0.22 A

P = V x I = 2.7 V x 0.22 A = 0.59 W

Large Transformers used in Power and Sub-stations.

These transformers convert high voltage to lower household voltage.

Small Transformers used in converting 120 V to 6.0 v for small devices.

Transformers are used for stepping up voltage to very high values for efficient transmission over long distances from power stations and also used to reduce voltage to more household friendly voltage of 240 V.

But why is electrical energy transported at such high voltage? Why do we have high voltage power lines that cross our province?

Second of all no electrical transportation system is 100% efficient. All electrical wires have some resistance and hence generate heat when current is flowing in the wires.

Third the amount of energy lost per second in a power line called Power loss can be calculated from this equation:

PL = I2R

I = current in the wires

R = resistance of the wires

First of all large scale power production is usually done far away from populated centers. Dams are usually created across large rivers at some distance from a large urban center. Like wise for large thermal generating plants or Nuclear power plants.

Typically the resistance in power line wires is quite small at around 5 – 10 ohms.

The current flow on the other hand can be quite significant as power (energy) has to be supplied to all those who need it.

Lets us do a sample calculation as to why high voltage is used to transport electrical energy

Typically power stations generate power at 5 - 15000 MW (5 million watts to 15 billion watts) at around 500 V and around 10000 or more Amps

Assuming a resistance of 10 Ω in the wires what would be the power loss if the electrical energy was transported at this voltage and current.

PL = I2R

PL = 100002 x 10

PL = 1000000000 W

PL = 1000 MW

This is 200 x the available energy!!!!

The wires would get very hot and maybe melt. This would be one huge toaster and no power would reach those who needed it!

Assuming Current is 10000 A and resistance is 10 Ω then the power loss would be:

Now let us redo the calculation but use a transformer to increase the voltage up to high values.

Remember: P = VI Since the same power has to be delivered by increasing the voltage the current has to be decrease!

5 x 106 W = 250000 x I

Let's increase Voltage to 250000 V

I = 5 x 106 / 250000

I = 20 A At a high voltage of 250 KV only 20 amps of current is needed

Electromagnetic InductionRe-doing the Power Loss calculation at the new current:

PL = I2R

PL = 202 x 10

PL = 4000 W (A good size electric Heater)

This an insignificant amount of power loss and would allow a delivery of 99.92% of the energy produced!

Remember transformers themselves are very efficient as well so very little power is lost in using them.

The high voltage that power is transmitted at is too dangerous to use the consumer level so step-down transformers change it back to a more manageable level.

Have another look at the transmission diagram. You should now have a good understanding of why all the changes in voltage have to happen.

I hope you found it both interesting and challenging.

This the end of electromagnetism and of Physics 12!

Good luck on your final exams and next year!

Electromagnetic Induction. Michael Faraday Faraday was working with two coils mounted on a wooden...

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