MAGNETIC FIELDS90523

NCEA Level 3

Physics

MAGNETIC FIELDS• Introduction• Electromagnetic Induction

- Magnetic flux- Faradays law- Lenz’s law

• Mutual Inductance- Introduction- The transformer

• Self Inductance- Introduction- Energy stored in inductance

• RL circuits- Time constant- Extension LC circuits

• Exercise 8 (Electromagnetic Inductance):

Page 130 - 134

Page 135 - 137

Page 145 - 149

Page 126 - 129

Page 138 - 140

Page 141 - 144

INTRODUCTION

We are all aware that there are magnetic fields all around us. We learnt last year that:

“A moving charge generates a Magnetic field”

We also know that magnetic fields generate forces on the following:

1. A north or south magnetic pole – use of a compass to navigate. Earths magnetic field interacts with tiny magnet in compass needle

2. A moving charge – used to control electron beam in a TV set.

3. A current – electric motors

Magnetic fields have the symbol, B, measured in units called, Tesla (T).

geographic north

geographic north

magnetic south

s

N

Needle of a compass is actually the north pole and the north pole of the Earth is actually magnetic south pole otherwise like poles would repel

REVISION OF LEVEL 2 ELETCROMAGNETISM EQUATIONS/RULES:

When moving charge, q, moves through a magnetic field of strength, B, with a speed, v, then the force on the charge is given by:

F = qvB

When a current, I, flows through a conductor length, L, through a mag. field strength, B, then thr force experienced by the conductor is:

F = BILsin

Note:

Often if the conductor is parallel to the field lines then the force is zero. If perpendicular then F is max. is a component between the two.

RIGHT HAND GRIP RULE:

I

current outwards current inwards

-

THE RIGHT HAND SLAP RULE:

FINGERS = MAGENTIC FIELD LINES DIRECTION (B)

THUMB = CURRENT FLOW (I)

PALM = FORCE (F)

I

F

B

+ I

FF THE FORCE IS ALWAYS

PERPENDICULAR TO THE MOTION

THE PARTICLE MUST MOVE

ALONG AN ARC OF A CIRCLE

r

mvBqv

2

qB

mvrarcofradius ,

+

PATH FOLLOWED BY A CHARGED PARTICLE IN A MAGNETIC FIELD

From F=ma

From F = Bqv

High frequency square wave to accelerate charge across the gap

CYCLOTRON

+High frequency square wave

This is a particle accelerator that accelerates a charged particle in a magnetic field in a circular path.

As the speed increases so does the circle radius. In order to keep the radius constant obviously the magnetic field strength needs to be increased.

length l

I‘n’ turns

In the middle

l

nIB 0

At the end

l

nIB

20

EXTENSION:

The magnetic field strength, B, for points in the interior of a long thin coil is given by the top equation where, n, is the number of turns per unit length, l.

= permeability for the medium; o = for a vacuum/air = 4x10-7

READ THROUGH

“THE INTRODUCTORY CONCEPTS”

PAGE 126-9 S& C

COMPLETE

QUESTIONS 1 – 3

PAGES 128 – 128

EXTENSION QUESTIONS 4 – 5

PAGE 129

S & C

ELECTROMAGNETIC INDUCTION

When moving a wire, length L, through a magnetic field, B, such as in a generator, you are in effect moving charged particles or electrons (e-) with a speed, v. A force is generated by the field, which can be determined using the “right hand slap rule”. The force produces a potential energy, V. From this the size of the energy can be calculate by:

V = BvL

V = potential energy (V)

B = magnetic field strength (T)

v = speed (ms-1)

L = length (m)

FOR DERIVATION OF THE FORMULA READ PAGE 130 S& CFOR DERIVATION OF THE FORMULA READ PAGE 130 S& C

Wire loops pushed into and pulled out of magnetic fields will generate a voltage and current.

When a loop is fully immersed there is no induced voltage or current produced.

Example 1:

A square loop of wire 0.50m by 0.50m moves into a uniform magnetic field of 0.80T, at a steady speed of 0.10ms-1. An induced voltage, V, develops between points A and B on the wire loop.

Calculate the size of V and direction of the induced current in the loop as the loop is:

a. Entering the magnetic field.

b. Completely in the field.

c. Leaving the field.

X X X X X X X X XX X X X X X X X XX X X X X X X X XX X X X X X X X XX X X X X X X X XX X X X X X X X XX X X X X X X X XX X X X X X X X X

0.50m

0.50

m

D

C A

B

0.10ms-1

Mag. field 0.80T

SOLUTION:

a. As AB enters the field:

V = BvL

= 0.80 x 0.10 x 0.50

= 0.040V

The direction of movement will be anticlockwise. (RHS rule)

b. When the coil is completely in the field, there is no induced voltage or current.

c. As CD leaves the field, the induced voltage, V, is still equal to 0.040V, but the direction of the induced current is clockwise.

MAGNETIC FLUX:

This is basically the density of the magnetic field lines within a specific area. Magnetic flux is given the symbol and is measured in units called Webers (Wb). The magnetic flux can be calculated using the following formula:

= B x A

= potential energy (V)

B = number of lines of flux per square metre (Wbm-2)

A = area (m2)

N S

Lines of magnetic flux, Φ

Note: When not perpendicular, Note: When not perpendicular, = BAsin = BAsin, is used., is used.

FARADAYS LAW:

1832 - Michael Faraday produced a magnetic field in a solenoid by passing a current through it. This solenoid was in close proximity to another solenoid but not linked. The second solenoid was linked to an ammeter. The result was the ammeter gave a reading when the current flowed through the first solenoid. Current was induced in the second.

WHY?

I

A B

FARADAY’S EXPERIMENT - 1832

Magnetic flux produced in solenoid ‘A’ that interacts with solenoid ‘B’ producing a current. The flux remains but the current reverts back to zero.

I

FARADAY’S EXPERIMENT - 1832

ANS:A current in B is only produced when the current in A is changing.

A B

When the current in ‘A’ is removed the flux collapses. Momentarily a current is induced in

circuit ‘B’. WHY?

I HAVE DISCOVERED ELECTROMAGNETIC INDUCTION

Now I understand!AN EMF IS ONLY INDUCED WHEN

THERE IS RELATIVE MOTION BETWEEN A CONDUCTOR AND A

MAGNETIC FIELD

NS

NS

NS

STRONGER FIELD (B)

NS

STRONGER FIELD (B)

NS

FASTER

NS

FASTER

THE INDUCED EMF ISDIRECTLY PROPORTIONAL TO

THE RATE OF CHANGE OF MAGNETIC FLUX CUTTING THE

CONDUCTOR

Faradays Law thus states:

“The size of the induced voltage in a conductor equals the rate of change of

magnetic flux”

From this the following equation was developed:

V = -/t

V = potential energy (V)

= change I magnetic flux (Wb)

t = change in time (s)

The negative symbol relates to Lenz’s law.

LENZ’S LAW:

HEINRICH LENZ THE INDUCED EMF GENERATES A CURRENT

THAT SETS UP A MAGNETIC FIELD WHICH ACTS TO OPPOSE THE CHANGE IN MAGNETIC

FLUX

Faradays law was further enhanced using the work of Heinrich Lenz. He used the concept of the Law of Conservation of Energy. He stated that the induced current must oppose the change in flux otherwise the energy would get bigger and bigger. Work must therefore be done in order to produce electrical energy. This doesn’t happen thus it must oppose the direction of .

Example 2:

The diagram shows a circular wire of area 0.10m2, placed in a magnetic field of strength 0.80T.

b. With the magnetic field strength back to 0.80T, the loop is rotated ¼ turn about the vertical axis. The average induced voltage during the turn is 0.40V.

i. What is the change in flux through the loop during the ¼ turn?

ii. How long does it take the loop to do the ¼ turn?

a. The magnetic field is then reduced to zero during a time of 0.40 seconds.

i. How much flux initially passes through the coil?

ii. What is the size of the induced EMF in the coil as the magnetic field strength is reduced?

iii. What is the direction of the induced current?

B = 0.80TB = 0.80T

SOLUTION:

ai. = B x A

= 0.80 x 0.10

= 0.080Wb

aii. V = / t

= 0.080 / 0.40 [the = initial when I = 0]

= 0.20V

aiii. Since the magnetic field is being reduced, the direction of the induced current will oppose this change by maintaining a magnetic field ‘out of the page’. So, using the right hand rule, the current direction is anticlockwise around the wire.

bi. When the loop has rotated through a ¼ turn, its plane is parallel to the magnetic field and so there is no flux in the loop; = 0.080Wb

bii. / t = V

t = / V

= 0.080/0.40

= 0.20s

For a coil of N turns, the induced voltage, V, is given by:

V = -N ( / t)

Thus induced voltage increase N times compared to a single turn.

Determining direction of magnetic field and hence the induced current in a loop of wire passing through a magnetic field do the following:

1.The INTERNAL magnetic field must oppose the change in field, i.e. go in the opposite direction to that of the magnetic field.

2.Reverse the right hand grip rule:

Let the thumb = direction of the internal magnetic field opposing the magnetic field of the magnet.

Let the fingers = the direction of the induced current around the loop.

READ THROUGH

“APPLICATIONS OF FARADAYS LAW”

PAGE 133-134 S& C

COMPLETE

QUESTIONS 6 – 8

PAGES 131 – 133

S & C

MUTUAL INDUCTANCE:

Faraday’s experiment showed that a current produced or present in coil A produces a magnetic flux in coil B. This changing magnetic flux induces a current & voltage in coil B. Thus the change in magnetic flux in coil B is a direct result of the current in coil A:

I A B

Thus: B IA

This requires a constant ‘M’, which is the proportionality constant. Thus:

B = M x IA

The voltage produced in coil B is:

VB = -M (IA / t)

‘M’ is also known as the mutual inductance of circuits

VB = voltage in coil B (V)

M = mutual inductance of 2 coils measured in henries (H).

IA = current in coil A (A)

t = time (s)

The above equation can be used for coil A although the constant M is the same for both. Hence the term ‘mutual’, for both coils:

VA = -M (IB / t)

The unit for mutual inductance, ‘henry’ is after the pioneer in this field Joseph Henry.

Example 3:

Two coils are arranged so that the flux in one passes into the other. When the current in coil 1 is increased from 0 to 5.0A in 2.0 seconds, an induced voltage of 100mV is induced in coil 2.

Calculate the mutual inductance between the coils.

SOLUTION:

V = -MI/t

M = (0.10 x 2.0)/5.0 [100mV = 0.10V]

M = 0.040 H

THE TRANSFORMER

A transformer is a device that uses the principles of mutual inductance to increase or decrease the induced voltage in a secondary coil. It requires two coils one, the primary coil, linked to an AC supply, which are joined by an iron core.

Why is the primary linked to an AC supply?

ANS: The AC supply produces a current which changes direction and intensity. This change in ‘I’ causes a change in flux. Changing flux generates and induced voltage in the secondary coil.

What is the purpose of the iron core?

ANS: The iron core increase the magnetic filed / flux and thus helps to enhance the induced voltage.

A.C.

SUPPLY

NP NS

VTHE

TRANSFORMER

IRON CORE

NP = Primary coil; NS = Secondary coil

Switching AC on causes mag flux change through iron core and through NS. This induces a voltage, ‘V’ in NS. Because there are less coils in NS the induced voltage is less than the voltage in NP. This transformer is said to “step down” the output voltage. To “step up” the voltage in NS there needs to be more coils on NS than NP. If the number of coils match then the voltages are the same.

From this in an ideal transformer, the voltage and number of turns are related by:

VS = NS

VP = NP

STEP-UP TRANSFORMER:

NS > NP

VS > VP

ISOLATING TRANSFORMER:

NS = NP

VS = VP

STEP-DOWN TRANSFORMER:

NS < NP

VS < VP

(Protects against electric shocks)

Transformers are used to step-up electricity leaving hydro-electric dams to 220,000V. On entering a town this is stepped- down to 11,000V to go across town and then stepped-down again on the road side to 240V.

Some appliances use transformers to step-up electricity when the mains doesn’t supply enough. The extra wire is why they are so heavy.

Example 4:

A battery charger contains a transformer to convert 240V to 12V. If the primary coil consists of 1200 turns, how many are on the secondary?

SOLUTION:

VS/VP = NS/NP

NS = NPVS/VP

= (1200 x 12) / 240

= 60 turns

THE EFFICIENCY OF THE TRANSFORMER:

This is always important whenever energy is transformed from one form to another.

Efficiency = output energy x 100Efficiency = output energy x 100

input energy 1input energy 1

Just like any other component a transformer will lose energy generally through heat. Careful design can produce efficiency ratings of 99% by:

a. Using an iron core to ensure a string magnetic field between primary and secondary coils.

b. Making the core form flat electrically insulated sheets or laminates. These reduce eddy currents which waste energy.

c. Using low-resistance copper wire for the coils.

For an ideal transformer: Power input = Power outputPower input = Power output

Thus (as P = VI): VVPPIIPP = V = VSSIISS

Example 5:

The transformer in example 4 draws 0.20A of current from the 240V supply.

SOLUTION:

a. VPIP = VSIS

IS = (240 x 0.20) / 12

= 4.0A

b. Efficiency = (power output/power/input) x (100/1)

= [(12 x 3.8) / (240 x 0.2)] x 100

= 95%

a. If the transformer were ideal, how much current could be supplied by the secondary?

b. In practice, the current that can be supplied by the secondary is 3.8A. How efficient is the transformer?

SELF-INDUCTANCE:

Just like mutual inductance a coil can produce a changing mag flux when a current is being switched on, hence it is self-inducting. How does this happen?

Inductor

12VS

B

Consider the diagram opposite. When the switch, S, is closed current flows through the inductor (L). This causes a change in mag flux. An induced voltage is produced, which according to Lenz’s law opposes the change in flux. Thus the current coming in slows down, the bulb glows dimly.

When the current eventually reaches its maximum the change in flux = 0. At this point there is no induced voltage lamp glows at its brightest.

When switch is opened then the mag field collapses, again there is a change in flux. Voltage is generated to oppose this change as the loop is open a high voltage can be produced and shocks produced.

This is often referred to as Back EMF.

SPARKTHE INDUCTOR

Has many names including choke &, torroid.

Symbol (L) Unit: Henry (H)

Inductor with an iron core running through it

As the flux is directly related to the current then I. A proportionality constant must now be used. The constant is ‘L’.

= L x I Just like in mutual inductance the voltage can be calculated by:

V = - L (I / t)

Example 6:

In the circuit shown, the coil has an inductance of 0.20H and the supply voltage is 3.0V. When switch S is opened, the current falls from 5.0A to zero in 0.010s.

a. What is the induced voltage?

b. What is the resistance of the coil?

SOLUTION:

a. V = -L (I/t)

= (0.20 x 5.0) /0/010

= 100V

b. R = V/I

= 3.0/5.0

= 0.60Ω

3.0VS

Neon lamp

R

ENERGY STORED IN AN INDUCTOR:

Obviously if the inductor has an induced voltage caused by the change in magnetic flux, it stands to reason that there must be energy present. But how much?

This can be calculated using the following formula:

E = ½ LI2

E = energy (J)

L = inductance constant (H)

I = current (A)

Example 7:

The circuit in example 6 contained a 0.20H inductor carrying a current of 5.0A. The energy stored in the inductor is:

SOLUTION:

E = ½ LI2

= ½ x 0.20 x 5.02

= 2.5J

READ THROUGH

“MUTUAL INDUCTANCE & SELF-INDUCTANCE”

FROM

PAGE 135 - 140

COMPLETE

QUESTIONS 9

PAGES 136

S & C

RL CIRCUITS

WHEN SWITCH IS CLOSED:

An RL circuit is one that contains a resistor, ‘R’ and an inductor, ‘L’.

R

L

V

CURRENT:

When current begins to flow through an inductor there is a massive surge but this slows down as a large change in magnetic flux induces a voltage that acts against the flow slowing it down. Eventually the change in flux diminishes and the maximum current can flow.

VOLTAGE:

At the beginning there is a huge change in flux producing a large induced voltage which opposes the current. This decreases as there is less of a change in flux.

When the switch is closed at t = 0 the current will also be 0. This is because the inductance, L, allows only a finite rate of increase of current.

In general:V = VR + VL

This is the voltage law for series circuits.

We know form Ohm’s law that V = IR & V = - L (I/t)

Thus:

VCIRCUIT = IR + L (I/t)

VL

time0

From this two graphs can be drawn for the flow of current and voltage through an inductor

Amps

Imax

Volts

Voltage starts at maximum VL and then decreases to zero as the change in flux diminishes. At zero there is little or no change in flux as the current is stable.

Current starts at zero and rapidly increases until it begins to reach maximum. It is slowed by the induced voltage opposing the flow until there is no opposition as it reaches its maximum.

VOLTAGE GRAPH CURRENT GRAPH

WHEN SWITCH OPENED:

The current decreases from its max. As it decreases a change in flux occurs in the opposite direction inducing a voltage.

Volts, V

Time, t0

V0

Current I

Time, t

I0

If the inductor is simply switched off the circuit is now open and its resistance enormous, due to the air gap. The inductor voltage is huge and may be big enough to cause a spark. That’s why you should never turn something off by pulling the plug.

SPARK

TIME CONSTANT:

Time constant is the measure of time it takes for an inductor to reach 63% of the total amount of voltage or current that it can induce/allow through respectively. The larger , the slower the process.

It is given the term tau, ‘’ and is measured in seconds, ‘s’. The formula for time constant is:

= L/R

Where:

= time constant (s)

R = resistance (Ω)

L = inductance (H)As the voltage never reaches max or As the voltage never reaches max or zero, then the total time taken can’t be zero, then the total time taken can’t be measured hence that is why 63% of the measured hence that is why 63% of the time is used. (time is used. (Same for currentSame for current))

Amps

63%

CURRENT GRAPH

Time, t

Imax

100%

86%

2

This works exactly the same for induced voltage as for current only the curve is reversed.

The second time constant is 63% of what is left.

Example 8:

An inductor of 0.50H is connected to a 12V supply. A steady current of 4.8A is recorded.

Find:

a. The resistance of the inductor.

b. The time constant.

c. The current one time constant after switch S is opened.

R

L = 0.50H

12V

AS

PRACTICAL:

The time constant for an inductor-resistor series circuit

Page 234-235 S & C.

SOLUTION:

a. R = V/I

= 12/4.8

= 2.5Ω

b. = L/R

= 0.50/2.5

= 0.20s

c. After 0.20s the current will fall by 63% of 4.8A.

63% of 4.8A = 3.0A

Current after 0.20s = 4.8 – 3.0

= 1.8A

READ INFORMATION PAGE 141 - 144

COMPLETE QUESTION 10

S & C

COMPLETE EXERCISE 8

PAGE 145 - 149

S & C