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is defined as

the production of an induced e.m.f. in a conductor/coil whenever the magnetic flux through the conductor/coil changes.

CHAPTER 20: Electromagnetic induction (6 Hours)

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20.1 Magnetic flux (

1/2

hour)

At the end of this chapter, students should be able to:



Define and use magnetic flux,

  

 



cos



20.1.1 Magnetic flux of a uniform magnetic field

 is defined as

the scalar product between the magnetic flux density,

with the vector of the area,

Mathematically,

Φ

 

 



cos



(7.1)

where  

B A

: magnetic flux : angle between th e direction of



and



: magnitude of : area of the coil the magnetic flux density

  It is a

scalar quantity

and its unit is

weber (Wb) meter squared

( T m 2 ).

tesla

Consider a uniform magnetic field

B

passing through a surface area

A

area 



Figure 7.2a

From the Figure 7.2a, the angle  given by Φ    

cos

BA BA

cos  0  is 0  thus the magnetic flux is

maximum 4



  90 area  



Note: Figure 7.2b

From the Figure 7.2a, the angle  is given by Φ    

BA BA

0 cos cos  90 is 90   thus the magnetic flux   Direction of vector

A

always

perpendicular (normal)

to the surface area,

A

The

magnetic flux is proportional to the number of field lines passing through the area.

Example 1 :

A single turn of rectangular coil of sides 10 cm  5.0 cm is placed between north and south poles of a permanent magnet. Initially, the plane of the coil is parallel to the magnetic field as shown in Figure 7.3. R

I I

S P

Figure 7.3

If the coil is turned by 90  about its rotation axis and the magnitude of magnetic flux density is 1.5 T, Calculate the change in the magnetic flux through the coil.

Solution :

B

 1

.

5 T The area of the coil is Initially, 



From the figure,  = thus the initial magnetic flux through the coil is Finally, 



From the figure,  = thus the final magnetic flux through the coil is Therefore the change in magnetic flux through the coil is  Φ  Φ f  Φ i

Example 2 :

A single turn of circular coil with a diameter of 3.0 cm is placed in the uniform magnetic field. The plane of the coil makes an angle 30  to the direction of the magnetic field. If the magnetic flux through the area of the coil is 1.20 mWb, calculate the magnitude of the magnetic field.

Solution :

The area of the coil is

Solution :

The angle between the direction of magnetic field,

B

and vector of area,

A

is given by Therefore the magnitude of the magnetic field is

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20.2 Induced emf (2 hours)

At the end of this chapter, students should be able to:

  

Use Faraday's experiment to explain induced emf.

State Faraday’s law and Lenz’s law to determine the direction of induced current.

Apply formulae,

  





Derive and use induced emf: I) in straight conductor,

 

lvB

sin    

A dt

iii) in rotating coil,

 

NAB

 sin 

  

B dA dt

20.2.1 M

AGNETIC FLUX

20.1.1(a) Phenomenon of electromagnetic induction

 Consider some experiments were conducted by Michael Faraday that led to the discovery of the Faraday’s law of induction as shown in Figures 7.1a, 7.1b, 7.1c, 7.1d and 7.1e.

 0

No movement Figure 7.1a

I I v

N Move towards the coil Figure 7.1b

 0

No movement Figure 7.1c

N N

I I v

Move away from the coil Figure 7.1d

Move towards the coil

13 Figure 7.1e

  From the experiments:   When the

bar magnet is stationary

, the galvanometer not show any deflection (

no current flows in the coil

When the bar magnet is moved relatively towards the coil, the galvanometer shows a momentary deflection to the right (Figure 7.1b). When the bar magnet is moved relatively away from the coil, the galvanometer is seen to deflect in the opposite direction (Figure 7.1d).

 Therefore when there is

any relative motion between the coil and the bar magnet

, the current known as

induced current will flow momentarily

through the galvanometer. This

current due to an induced e.m.f

across the coil.

Conclusion :  When the

magnetic field lines through a coil changes

thus the

induced emf will exist

across the coil.

 The magnitude of the

induced e.m.f. depends on the speed of the relative motion

where if the

v

increases induced emf increases

v

decreases induced emf decreases Therefore

is proportional to the induced emf

Example 3 :

Figure 7.4

The three loops of wire as shown in Figure 7.4 are all in a region of space with a uniform magnetic field. Loop 1 swings back and forth as the bob on a simple pendulum. Loop 2 rotates about a vertical axis and loop 3 oscillates vertically on the end of a spring. Which loop or loops have a magnetic flux that changes with time? Explain your answer.

Solution : 17

20.2.2 I

NDUCED EMF

20.2.2(a) Faraday’s law of electromagnetic induction

 states that

the magnitude of the induced emf is proportional to the rate of change of the magnetic flux

Mathematically,   

OR   

(7.2)

 where

Φ



: change of the magnetic flux : change of time : induced emf

The

negative sign

indicates that the

direction of induced emf

always

oppose the change of magnetic flux producing it (Lenz’s law)

(7.3)

d

Φ  Φ f  Φ i , then eq. (7.3) can be written as where

Φ

: final magnetic flux : initial magnetic flux

  

N d

and Φ 

cos 

(7.4) 19

   

N d



cos  

 

 cos  

dB dt

(7.5)

 For a

coil of

turns is placed in a uniform magnetic field

but

changing in the coil’s area

, the induced emf  is given by   

N d

and Φ 

cos    

N d



cos

    

 cos  

dA dt

(7.6) 20

 For

a coil is connected in series to a resistor of resistance

and the induced emf  exist in the coil as shown in Figure 7.5,

I R I

 the induced current

I

 

N d

and  is given by 

IR IR

 

N d

(7.7) Note:

 

Figure 7.5

To calculate the

magnitude of induced emf

, the

negative sign can be ignored

For a coil of

N BAcos

 turns, each turn will has a magnetic flux  through it, therefore the

magnetic flux linkage

of (refer to the combined amount of flux through all the turns) is given by

magnetic flux linkage



Φ

Example 4 :

The magnetic flux passing through a single turn of a coil is increased quickly but steadily at a rate of 5.0

 10  2 Wb s  1 . If the coil have 500 turns, calculate the magnitude of the induced emf in the coil.

Solution :

By applying the Faraday’s law equation for a coil of

N

turns , thus

Example 5 :

A coil having an area of 8.0 cm 2 and 50 turns lies perpendicular to a magnetic field of 0.20 T. If the magnetic flux density is steadily reduced to zero, taking 0.50 s, determine a. the initial magnetic flux linkage.

b. the induced emf.

Solution :



a. The initial magnetic flux linkage is given by

Solution :

b. The induced emf is given by

Example 6 :

A narrow coil of 10 turns and diameter of 4.0 cm is placed perpendicular to a uniform magnetic field of 1.20 T. After 0.25 s, the diameter of the coil is increased to 5.3 cm.

a. Calculate the change in the area of the coil.

b. If the coil has a resistance of 2.4  , determine the induced current in the coil.

Solution :



Initial   0  Final

Solution :

a. The change in the area of the coil is given by

Solution :

b. Given The induced emf in the coil is Therefore the induced current in the coil is given by

20.2.2 (

) L

ENZ

’

S LAW  states that

an induced electric current always flows in such a direction that it opposes the change producing it

 

1 st

 This law is essentially a form of the law of

conservation of energy

An illustration of lenz’s law can be explained by the following experiments.

experiment:

In Figure 7.6 the magnitude of the magnetic field at the

Direction of induced current – Right hand grip rule.

solenoid increases as the bar magnet is moved towards it.

North pole

 An emf is induced in the solenoid and the galvanometer indicates that a current is flowing.

Figure 7.6

 To determine the direction of the current through the galvanometer which corresponds to a deflection in a particular sense, then the

current through the solenoid seen is in the direction that make the solenoid upper end becomes a north pole

. This

opposes the motion of the bar magnet obey the lenz’s law

and

2 nd experiment:

 Consider a straight conductor PQ is placed perpendicular to the magnetic field and move the conductor to the left with constant velocity

v

as shown in Figure 7.7.

 When the conductor move to the left thus the

induced current needs to flow in such a way to oppose the change which has induced it

deflection.

based on lenz’s law. Hence galvanometer shows a

X X X X X X X X X X X X X

X X



X X X

X X X X X



X X X X X X X X X

X X Figure 7.7

X X X X X X X X X X X X X 29

 To determine the

direction of the induced current (induced emf)

flows in the conductor PQ, the

Fleming’s right hand (Dynamo) rule





(

is used as shown in Figure 7.8.

motion )

Note: Thumb – direction of Motion First finger – direction of Field

induced induced

OR emf

Figure 7.8



Second finger – direction of induced current OR induced emf

Therefore the induced current flows from Q to P as shown in Figure 7.7.

 Since the induced current flows in the conductor PQ and is placed in the magnetic field then this

conductor will experience magnetic force

 Its direction is in the

opposite direction of the motion

3 rd experiment:

 Consider two solenoids P and Q arranged coaxially closed to each other as shown in Figure 7.9a.

 ind

P

Switch , S

N N +

ind

Q

S -

ind 

Figure 7.9a

At the moment when the

switch S is closed

, current

I

begins to flow in the solenoid P and producing a magnetic field inside the solenoid P. Suppose that the field points towards the solenoid Q.

  The magnetic flux through the solenoid Q

increases with time

. According to Faraday’s law ,an induced current due to induced emf will exist in solenoid Q.

The induced current flows in solenoid Q must produce a magnetic field that oppose the change producing it (increase in flux). Hence based on Lenz’s law, the induced current flows in circuit consists of solenoid Q is

anticlockwise

(Figure 7.9a) and the galvanometer shows a deflection.

 ind

P

Switch, S

N S Figure 7.9b

ind

Q

ind

+ 32

  At the moment when the

switch S is opened

, the current

I

starts to decrease in the solenoid P and magnetic flux through the solenoid Q

decreases with time

. According to Faraday’s law ,an induced current due to induced emf will exist in solenoid Q.

The induced current flows in solenoid Q must produce a magnetic field that oppose the change producing it (decrease in flux). Hence based on Lenz’s law, the induced current flows in circuit consists of solenoid Q is

clockwise

(Figure 7.9b) and the galvanometer seen to deflect in the opposite direction of Figure 7.9a.

Example 7 :

A single turn of circular shaped coil has a resistance of 20  and an area of 7.0 cm 2 . It moves toward the north pole of a bar magnet as shown in Figure 7.10.

Figure 7.10

If the average rate of change of magnetic flux density through the coil is 0.55 T s  1 , a. determine the induced current in the coil b. state the direction of the induced current observed by the observer shown in Figure 7.10.

Solution :

a. By applying the Faraday’s law of induction, thus Therefore the induced current in the coil is given by

Solution :

b. Based on the lenz’s law, hence the direction of induced current is

clockwise

as shown in figure below.

20.2.3 I

NDUCED EMF IN A STRAIGHT CONDUCTOR  Consider a straight conductor PQ of length

l

is moved perpendicular with velocity

v

as shown in Figure 7.11.

X X X X X

across a uniform magnetic field

B

X X





X X X X X X X X

X X X X X X

Area,

A

X X X



X X X X X X X X X X

ind

X X X X

 ind

X X X X X

X Figure 7.11

X X

When the conductor moves through a distance

x

area swept out by the conductor is given by



in time

t

, the

  Since the motion of the conductor is perpendicular to the magnetic field

B

hence the magnetic flux cutting by the conductor is given by Φ 

cos Φ 

Blx

cos  0  and  Φ   0 

Blx

According to Faraday’s law, the emf is induced in the conductor and its magnitude is given by  



 

d dt

 

Bl dx dt

and

dx dt



 

Blv

(7.8) 38

 In general, the

magnitude

of the induced emf in the straight conductor is given by

Note:

       where

lvB θ

sin 

: angle between



(7.9)

and



This type of induced emf is known as

motional induced emf.

The

direction

of the

induced current

induced emf

in the straight conductor can be determined by using the

Fleming’s right hand

rule (based on Lenz’s law). In the case of Figure 7.11, the direction of the induced current or induced emf is from Q to P. Therefore P is higher potential than Q.

Eq. (7.9)

also can be used for a

single turn of rectangular coil moves across the uniform magnetic field

For a

rectangular coil of

turns

,  

NlvB

sin



(7.10) 39

Example 8 :

A 20 cm long metal rod CD is moved at speed of 25 m s  1 across a uniform magnetic field of flux density 250 mT. The motion of the rod is perpendicular to the magnetic field as shown in Figure 7.12.

C 

2 5 m s  1

Figure 7.12

D a. Calculate the motional induced emf in the rod.

b. If the rod is connected in series to the resistor of resistance 15  , determine i. the induced current and its direction.

ii. the total charge passing through the resistor in two minute.

Solution :

a. By applying the equation for motional induced emf, thus b. Given



15

 i . By applying the Ohm’s law, thus By using the Fleming’s right hand rule, ii. Given The total charge passing through the resistor is given by

20.2.4 I

NDUCED EMF IN A ROTATING COIL  Consider a rectangular coil of

N

turns, each of area

A

, being rotated mechanically with a constant angular velocity  in a uniform magnetic field of flux density

B

in Figure 7.13.

B ω

 

 coil When the vector of area,

A

field

B

the magnetic flux  by

Figure 7.13: side view

is at an angle     

BA BA

cos

 

and  to the magnetic through each turn of the coil is given  

  By applying the equation of Faraday’s law for a coil of

N

thus the induced emf is given by   

N d



 

N d dt



 

NBA d dt

cos  cos 



  turns,  

NBA

 sin 

(7.11)

where

: time

 max 

NBA

 where   2 

 2 



(7.12) 43

  Eq. (7.11) also can be written as  

NBA

 sin 

(7.13)

where 

: angle between



and



Conclusion

: A coil rotating with constant angular velocity in a uniform magnetic field produces a

sinusoidally alternating emf

as shown by the induced emf  against time

t

graph in Figure 7.14.

 



max

sin

ωt

 max

Note:

This phenomenon was the important  part in the  max

0

development of the

electric generator or dynamo

0 .

T T

1 .

T

Figure 7.14

T



Example 9 :

A rectangular coil of 100 turns has a dimension of 10 cm  15 cm. It rotates at a constant angular velocity of 200 rpm in a uniform magnetic field of flux density 5.0 T. Calculate a. the maximum emf produced by the coil, b. the induced emf at the instant when the plane of the coil makes an angle of 38  to the magnetic field.

Solution :

The area of the coil is and the constant angular velocity in rad s  1 is

Solution :

a. The maximum emf produced by the coil is given by b.



From the figure, the angle  is Therefore the induced emf is given by

Exercise 20.1 :

A bar magnet is held above a loop of wire in a horizontal plane, as shown in Figure 7.15.

The south end of the magnet is toward the loop of the wire. The magnet is dropped toward the loop. Determine the direction of the current through the resistor a. while the magnet falling toward the loop, b. after the magnet has passed through the loop and moves away from it.

(Physics for scientists and engineers,6 th edition, Serway&Jewett, Q15, p.991) ANS. : U think Figure 7.15

A straight conductor of length 20 cm moves in a uniform magnetic field of flux density 20 mT at a constant speed of 10 m s -1 . The velocity makes an angle 30  to the field but the conductor is perpendicular to the field. Determine the induced emf.

ANS. : 2.0





2 V

A coil of area 0.100 m 2 is rotating at 60.0 rev s -1 with the axis of rotation perpendicular to a 0.200 T magnetic field.

a. If the coil has 1000 turns, determine the maximum emf generated in it.

b. What is the orientation of the coil with respect to the magnetic field when the maximum induced emf occurs?

(Physics for scientists and engineers,6 th edition,Serway&Jewett, Q35, p.996) ANS. : 7.54



10 3 V

A circular coil has 50 turns and diameter 1.0 cm. It rotates at a constant angular velocity of 25 rev s  1 in a uniform magnetic field of flux density 50  T. Determine the induced emf when the plane of the coil makes an angle 55  to the magnetic field.

48 ANS. : 1.77





5 V

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20.3 Self-inductance (1 hour)

At the end of this chapter, students should be able to:



Define self-inductance.



Apply formulae

 



  0

A l

for a loop and solenoid.

20.3 S

ELF

-

INDUCTANCE

20.3.1 Self-induction

 Consider a solenoid which is connected to a battery , a switch S and variable resistor

R

, forming an open circuit as shown in Figure 7.16a.

 When the switch S is closed, a current

I

begins to flow in the solenoid.

R

 The

current produces a magnetic field whose field lines through the solenoid

and generate the

magnetic flux linkage

Figure 7.16a: initial

 If the

resistance

of the variable resistor

changes

, thus the

current

flows in the solenoid also

changed

, then

so too does magnetic flux linkage

   According to the Faraday’s law, an

emf

has to be

induced in the solenoid itself since the flux linkage changes

In accordance with Lenz’s law, the

induced emf opposes

the

changes that has induced it

and it is known as a

back emf

For the current

I

increases :

ind

N N +

I I

ind

S Figure 7.16b:

increases

Direction of the induced emf is in the

opposite direction

of the current

I

 For the current

I

decreases :

ind

S S N N

ind

Figure 7.16c:

I I

ind

decreases

Direction of the induced emf is in the

same direction

of the current

I

  This process is known as self-induction.

Self-induction

is defined as

the process of producing an induced emf in the coil due to a change of current flowing through the same coil

20.3 SELF-INDUCTANCE

I induced I induced

(a) A current in the coil produces a magnetic field directed to the left.

( (b) If the current increases, the increasing magnetic (c) The polarity of the induced emf reverses if the current decreases.

Self-induction experiment

 The effect of the self-induction can be demonstrated by the circuit shown in Figure 7.17a.

 switch, S iron-core lamp A 1 coil, L

lamp A 2 

Figure 7.17a

Initially variable resistor

R

is adjusted so that the two lamps have the same brightness in their respective circuits with steady current flowing.

 When the switch S is closed, the lamp A 2 with variable resistor

R

is seen to become bright almost immediately but the lamp A 1 iron-core coil L increases slowly to full brightness.

with

 Reason:    The coil L undergoes the

self-induction and induced emf

in it. The induced or back emf

opposes the growth of current so the glow in the lamp A 1 increases slowly

The resistor

, however has no back emf, hence the lamp A 2 glow fully bright as soon as switch S is closed

This effect can be shown by the graph of current

I

against time

t

through both lamps in Figure 7.17b.

I I

0 lamp A 2 with resistor

R

0

lamp A 1 with coil L

Figure 7.17b

Example 10 :

A circuit contains an iron-cored coil L, a switch S, a resistor

R

a dc source  arranged in series as shown in Figure 7.18.

and The switch S is closed for a long time and is suddenly opened.

Explain why a spark jump across the switch contacts S .

 switch, S coil, L

Figure 7.18

Solution :

 When the switch S is suddenly opened, the ……………………… ……………………………………… and ….................................. .....………………………..…. which tends to maintain the current.

 This back emf is high enough to …………………………………… …………………………….……………….and a ………………… ………………………………………………………………………..

20.3.2 S

ELF

-

INDUCTANCE

,

 From the self-induction phenomenon, we get Φ L 

Φ L 

(7.14)

 where

: self inductance of the coil



: current : magnetic flux linkage

L From the Faraday’s law, thus   



L  

d dt

  

L dI dt

(7.15) 57



Self-inductance

is defined as

the ratio of the self induced (back) emf to the rate of change of current in the coil

 

 /

 For the coil of

N



L L



dI dt



  turns, thus 

N d

 

N d dt

 

N d dt



 



  

I I

and L   

L dI dt

magnetic flux linkage

(7.16) 58

     It is a

scalar quantity

and its unit is

henry (H)

Unit conversion :

1 H



1 Wb A

 1 

1 T m

A

 1 The value of the

self-inductance depends

on  the

size and shape of the coil

,   the

number of turn (

)

, the

permeability of the medium in the coil (



)

A circuit element which possesses mainly self-inductance is known as an

inductor

. It is used to

store energy in the form of magnetic field

The symbol of inductor in the electrical circuit is shown in Figure 7.19.

Figure 7.19

20.3.3 S

ELF

-

INDUCTANCE OF A SOLENOID    The magnetic flux density at the

centre of the air-core solenoid

is given by

  0

l NI

The

magnetic flux

passing through each turn of the solenoid

always maximum

and is given by      

 0 cos

l NI

0   

Therefore the

self-inductance of the solenoid

   0

NIA l

is given by





I L



N I

   0

NIA l

 

  0

(7.17)

Example 11 :

A 500 turns of solenoid is 8.0 cm long. When the current in the solenoid is increased from 0 to 2.5 A in 0.35 s, the magnitude of the induced emf is 0.012 V. Calculate a. the inductance of the solenoid, b. the cross-sectional area of the solenoid, c. the final magnetic flux linkage through the solenoid.

(Given  0 = 4   10  7 H m  1 )

Solution :

a. The change in the current is Therefore the inductance of the solenoid is given by

Solution :

b. By using the equation of self-inductance for the solenoid, thus c. The final magnetic flux linkage is given by

Exercise 20.2 :

Given  0 = 4   10  7 H m  1 1.

An emf of 24.0 mV is induced in a 500 turns coil at an instant when the current is 4.00 A and is changing at the rate of 10.0 A s -1 . Determine the magnetic flux through each turn of the coil.

(Physics for scientists and engineers,6 th Q6, p.1025) ANS. : 1.92





5 Wb edition,Serway&Jewett,

A 40.0 mA current is carried by a uniformly wound air-core solenoid with 450 turns, a 15.0 mm diameter and 12.0 cm length. Calculate a.

the magnetic field inside the solenoid, b. c.

ANS. :

the magnetic flux through each turn, the inductance of the solenoid.

1.88





4 T; 3.33





8 Wb; 3.75





4 H 63

A current of 1.5 A flows in an air-core solenoid of 1 cm radius and 100 turns per cm. Calculate a. the self-inductance per unit length of the solenoid.

b. the energy stored per unit length of the solenoid.

ANS. : 0.039 H m



1 ; 4.4





2 J m



At the instant when the current in an inductor is increasing at a rate of 0.0640 A s  1 , the magnitude of the back emf is 0.016

a. Calculate the inductance of the inductor.

b. If the inductor is a solenoid with 400 turns and the current flows in it is 0.720 A, determine i. the magnetic flux through each turn, ii. the energy stored in the solenoid.

ANS. : 0.250 H; 4.5





4 Wb; 6.48





2 J

At a particular instant the electrical power supplied to a 300 mH inductor is 20 W and the current is 3.5 A. Determine the rate at which the current is changing at that instant.

ANS. : 19 A s



1 64

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20.4 Mutual inductance (

hours)

At the end of this chapter, students should be able to:



Define mutual inductance.



Derive and use formulae for mutual inductance of two coaxial coils,

12 

2  12

1   0

A l



Explain the working principle of transformer and the effect of eddy current in transformer.

20.4 M

UTUAL INDUCTANCE

20.4.1 Mutual induction

 Consider two circular close packed coils near each other and sharing a common central axis as shown in Figure 7.20.

 A current

I

1 flows in coil 1, produced by the battery in the external circuit.

 The current

1 produces a magnetic field lines

inside it and this field lines

also pass through coil 2

as shown in Figure 7.20.



1 

1 Coil 1 Coil 2

Figure 7.20

 If the current

1 changes with time

, the

magnetic flux

coils 1 and 2 will

change with time

simultaneously.

through  Due to the change of magnetic flux through coil 2, an

emf is induced in coil 2

. This is in accordance to the

Faraday’s law of induction

    In other words, a

change of current in one coil leads to the production of an induced emf in a second coil

which is

magnetically linked to the first coil

This process is known as mutual induction.

Mutual induction

is defined as

the process of producing an induced emf in one coil due to the change of current in another coil.

At the same time, the

self-induction occurs magnetic flux through it changes

in coil 1 since the

20.4.2 M

UTUAL INDUCTANCE

,

 From the Figure 7.20, consider the coils 1 and 2 have

N

2 turns respectively.

and  If the current

I

1 in coil 1 changes, the magnetic flux through coil 2 will change with time and an induced emf will occur in coil 2,  2 where  2  

dI dt

1   2  

If vice versa, the induced emf in coil 1,  1 is given by

(7.21)

 1  

(7.22)

 It is a

scalar quantity

and its unit is

henry (H)

where

12 

21 

: Mutual inductance 68

 

Mutual inductance

is defined as

the ratio of induced emf in a coil to the rate of change of current in another coil

From the Faraday’s law for the coil 2, thus  2 

M dt

 12

12     

 

d dt

 

dt N N

2 2  

 2 2 2 2

2  2

1 and

21 

1  1

2 magnetic flux linkage through coil 2



2  2

1  magnetic flux linkage through coil 1

1  1

(7.23) 69

20.4.3 M

UTUAL INDUCTANCE FOR TWO SOLENOIDS  Consider a long solenoid with length

l

and cross sectional area

A

is closely wound with

N

1 turns of wire. A coil with surrounds it at its centre as shown in Figure 7.21.

N

2 turns

N

1 : primary coil

N

2 : secondary coil 

Figure 7.21

When a current

I

1 magnetic field

B

1 , flows in the primary coil (

N

1 ), it produces a

1   0

  and then the magnetic flux Ф 1 ,  1 

cos 0   1   0

A l

no magnetic flux leakage

, thus  1   2 If the current

I

1 changes, an emf is induced in the secondary coils, therefore the mutual inductance occurs and is given by

12 

1  2

12   

1    0

A l M

12 

  0

A l

(7.24) 71

Mutual inductance, M



N 2 Φ 2 I 1



N 1 Φ 1 I 2 M

 

o N

A l

 1  

Example 13 :

A current of 3.0 A flows in coil C and is produced a magnetic flux of 0.75 Wb in it. When a coil D is moved near to coil C coaxially, a flux of 0.25 Wb is produced in coil D. If coil C has 1000 turns and coil D has 5000 turns.

a. Calculate self-inductance of coil C and the energy stored in C before D is moved near to it. b. Calculate the mutual inductance of the coils.

c. If the current in C decreasing uniformly from 3.0 A to zero in 0.25 s, calculate the induced emf in coil D.

Solution :

a. The self-inductance of coil C is given by

Solution :

a. and the energy stored in C is b. The mutual inductance of the coils is given by

Solution :

c. Given The induced emf in coil D is given by

20.4.4 T

RANSFORMER   is an electrical instrument to

increase or decrease the emf (voltage) of an alternating current.

Consider a structure of the transformer as shown in Figure 7.22.

laminated iron core alternating voltage source primary coil

N

P turns

N

S turns secondary coil

Figure 7.22

  If

N

S the transformer is a

step-down transformer

N

S the transformer is a

step-up transformer

 The symbol of transformer in the electrical circuit is shown in Figure 7.23.

Figure 7.23

Working principle of transformer

   When an alternating voltage source is applied to the primary coil, the alternating current produces an alternating magnetic flux concentrated in the iron core.

Without no magnetic flux leakage from the iron core, the same changing magnetic flux passes through the secondary coil and inducing an alternating emf.

After that the induced current is produced in the secondary coil.

 The characteristics of an ideal transformer are: 

Zero resistance of primary coil

 

No magnetic flux leakage from the iron core

No dissipation of energy and power

Energy losses in transformer

 Although transformers are very efficient devices, small energy losses do occur in them owing to four main causes: 

Resistance of coils

The wire used for the primary and secondary coils has

resistance and so ordinary (

) heat losses

occur.

 

Overcome : copper wire

The transformer coils are made of

thick Hysteresis

The

magnetization of the core

repeatedly reversed by the alternating magnetic field

. The

resulting expenditure of energy in the core appears as heat

Overcome :

By using a Mumetal) which has a

magnetic material low hysteresis loss

(such as

Flux leakage

The flux due to the primary may not all link the secondary.

Some of the

flux loss in the air

Overcome :

wound directly on top of the other rather than having two separate coils.

By designing one of the insulated coils is

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20.5 Energy stored in an inductor (

hour)

At the end of this chapter, students should be able to:



Derive and use formulae for energy stored in an inductor,



1 2

23.4 E

NERGY STORED IN AN INDUCTOR

   Consider an inductor of inductance

L

. Suppose that at time

t

, the current in the inductor is in the process of building up to its steady value

I

at a rate

dI/dt

. The magnitude of the back emf  is given by The electrical power

P

 

L dI dt

in overcoming the back emf in the circuit is given by





dI P



LI dt Pdt



LIdI

and

Pdt



dU dU



LIdI

(7.18) 81

 0

U dU



 0

I IdI

and analogous to

(7.19)



1 2

in capacitor

  0

A l U

 1 2

(7.20) 82

Example 12 :

A solenoid of length 25 cm with an air-core consists of 100 turns and diameter of 2.7 cm. Calculate a. the self-inductance of the solenoid, and b. the energy stored in the solenoid, if the current flows in it is 1.6 A.

(Given  0 = 4   10  7 H m  1 )

Solution :

a. The cross-sectional area of the solenoid is given by Hence the self-inductance of the solenoid is

Solution :

b. Given



1 .

6 A

By applying the equation of energy stored in the inductor, thus