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PPT – Chapter 7:Electromagnetic induction PowerPoint presentation | free to view - id: 573656-NzA5M

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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)

Learning Outcome

20.1 Magnetic flux (1/2 hour)

- At the end of this chapter, students should be

able to - Define and use magnetic flux,

- 20.1.1 Magnetic flux of a uniform magnetic field
- is defined as the scalar product between the

magnetic flux density, B with the vector of the

area, A. - Mathematically,

where

- It is a scalar quantity and its unit is weber

(Wb) OR tesla meter squared ( T m2). - Consider a uniform magnetic field B passing

through a surface area A of a single turn coil as

shown in Figures 7.2a and 7.2b. - From the Figure 7.2a, the angle ? is 0? thus the

magnetic flux is given by

Figure 7.2a

maximum

- From the Figure 7.2a, the angle ? is 90? thus

the magnetic flux is given by

Figure 7.2b

- 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. 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.

Figure 7.3

Solution The area of the coil is Initially,

Finally, Therefore the change in

magnetic flux through the coil is

From the figure, ? thus the initial

magnetic flux through the coil is

From the figure, ? thus the final magnetic

flux through the coil is

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

Learning Outcome

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 Faradays law and Lenzs law to determine

the direction of induced current. - Apply formulae,
- Derive and use induced emf
- I) in straight conductor,
- ii) in coil, OR
- iii) in rotating coil,

20.2.1 Magnetic flux

- 20.1.1(a) Phenomenon of electromagnetic induction
- Consider some experiments were conducted by

Michael Faraday that led to the discovery of the

Faradays law of induction as shown in Figures

7.1a, 7.1b, 7.1c, 7.1d and 7.1e.

Figure 7.1a

S

N

Figure 7.1b

Figure 7.1c

N

S

Figure 7.1d

S

N

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 - Therefore v is proportional to the induced emf.

v increases

induced emf increases

v decreases

induced emf decreases

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

20.2.2 Induced emf

- 20.2.2(a) Faradays law of electromagnetic

induction - states that the magnitude of the induced emf is

proportional to the rate of change of the

magnetic flux. - Mathematically,
- The negative sign indicates that the direction of

induced emf always oppose the change of magnetic

flux producing it (Lenzs law).

OR

where

, then eq. (7.3) can be written as

where

and

- For a coil of N turns is placed in a uniform

magnetic field B but changing in the coils area

A, the induced emf ? is given by

and

- For a coil is connected in series to a resistor

of resistance R and the induced emf ? exist in

the coil as shown in Figure 7.5,

the induced current I is given by

and

Figure 7.5

- To calculate the magnitude of induced emf, the

negative sign can be ignored. - For a coil of N turns, each turn will has a

magnetic flux ? of BAcos? through it, therefore

the magnetic flux linkage (refer to the combined

amount of flux through all the turns) is given by

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

Faradays law equation for a coil of N turns ,

thus

Example 5

A coil having an area of 8.0 cm2 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 a. 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

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 (b) Lenzs law

- states that an induced electric current always

flows in such a direction that it opposes the

change producing it. - This law is essentially a form of the law of

conservation of energy.

- An illustration of lenzs law can be explained by

the following experiments.

Direction of induced current Right hand grip

rule.

- 1st experiment
- In Figure 7.6 the magnitude of the magnetic field

at the solenoid increases as the bar magnet is

moved towards it. - An emf is induced in the solenoid and the

galvanometer indicates that a current is flowing.

N

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 and obey the lenzs law.

- 2nd 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 based on

lenzs law. Hence galvanometer shows a deflection.

Figure 7.7

- To determine the direction of the induced current

(induced emf) flows in the conductor PQ, the

Flemings right hand (Dynamo) rule is used as

shown in Figure 7.8. - 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.

Thumb direction of Motion First finger

direction of Field Second finger direction of

induced

current OR induced emf

Figure 7.8

- 3rd experiment
- Consider two solenoids P and Q arranged coaxially

closed to each other as shown in 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.

Figure 7.9a

- The magnetic flux through the solenoid Q

increases with time. According to Faradays 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

Lenzs law, the induced current flows in circuit

consists of solenoid Q is anticlockwise (Figure

7.9a) and the galvanometer shows a deflection.

Figure 7.9b

- 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 Faradays 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

Lenzs 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 cm2. It

moves toward the north pole of a bar magnet as

shown in 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.

Figure 7.10

Solution a. By applying the Faradays law of

induction, thus Therefore the

induced current in the coil is given by

Solution b. Based on the lenzs law, hence the

direction of induced current is clockwise as

shown in figure below.

20.2.3 Induced emf in a straight conductor

- Consider a straight conductor PQ of length l is

moved perpendicular with velocity v across a

uniform magnetic field B as shown in Figure 7.11. - When the conductor moves through a distance x in

time t, the area swept out by the conductor is

given by

Area, A

Figure 7.11

- Since the motion of the conductor is

perpendicular to the magnetic field B hence the

magnetic flux cutting by the conductor is given

by - According to Faradays law, the emf is induced in

the conductor and its magnitude is given by

and

and

- In general, the magnitude of the induced emf in

the straight conductor is given by - This type of induced emf is known as motional

induced emf. - The direction of the induced current or induced

emf in the straight conductor can be determined

by using the Flemings right hand rule (based on

Lenzs 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.

where

- 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 N turns,

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. 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.

Figure 7.12

Solution a. By applying the equation for

motional induced emf, thus b. Given i. By

applying the Ohms law, thus By using

the Flemings right hand rule, ii. Given

The total charge passing through the resistor is

given by

20.2.4 Induced 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 about an axis as shown in

Figure 7.13. - When the vector of area, A is at an angle ? to

the magnetic field B, the magnetic flux ? through

each turn of the coil is given by

Figure 7.13 side view

and

- By applying the equation of Faradays law for a

coil of N turns, thus the induced emf is given by - The induced emf is maximum when

hence

where

where

- Eq. (7.11) also can be written as
- 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.

where

This phenomenon was the important part in the

development of the electric generator or dynamo.

Figure 7.14

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

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,6th

edition, SerwayJewett, Q15, p.991) ANS. U think

Figure 7.15

2. 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?10?2 V 3. A coil of area 0.100 m2

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,6th edition,SerwayJewett, Q35,

p.996) ANS. 7.54?103 V 4. 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. ANS. 1.77?10?5 V

Learning Outcome

20.3 Self-inductance (1 hour)

- At the end of this chapter, students should be

able to - Define self-inductance.
- Apply formulae
- for a loop and solenoid.

20.3 Self-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. - The current produces a magnetic field whose field

lines through the solenoid and generate the

magnetic flux linkage. - If the resistance of the variable resistor

changes, thus the current flows in the solenoid

also changed, then so too does magnetic flux

linkage.

Figure 7.16a initial

- According to the Faradays law, an emf has to be

induced in the solenoid itself since the flux

linkage changes. - In accordance with Lenzs law, the induced emf

opposes the changes that has induced it and it is

known as a back emf. - For the current I increases

Figure 7.16b I increases

- For the current I decreases
- 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.

Figure 7.16c I decreases

20.3 SELF-INDUCTANCE

Iinduced

Iinduced

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

(b) If the current increases, the increasing

magnetic flux creates an induced emf having

the polarity shown by the dashed battery.

(b) If the current increases, the increasing

magnetic flux creates an induced emf having

the polarity shown by the dashed battery.

(c) The polarity of the induced emf reverses if

the current decreases.

(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. - 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 A2 with

variable resistor R is seen to become bright

almost immediately but the lamp A1 with iron-core

coil L increases slowly to full brightness.

Figure 7.17a

- 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 A1 increases slowly. - The resistor R, however has no back emf, hence

the lamp A2 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.

Figure 7.17b

Example 10

A circuit contains an iron-cored coil L, a

switch S, a resistor R and a dc source ? arranged

in series as shown in Figure 7.18.

The switch S is closed for a long time and is

suddenly opened. Explain why a spark jump across

the switch contacts S .

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 Self-inductance, L

- From the self-induction phenomenon, we get
- From the Faradays law, thus

where

- Self-inductance is defined as the ratio of the

self induced (back) emf to the rate of change of

current in the coil. - OR
- For the coil of N turns, thus

and

- It is a scalar quantity and its unit is henry

(H). - Unit conversion
- The value of the self-inductance depends on
- the size and shape of the coil,
- the number of turn (N),
- 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 Self-inductance of a solenoid

- The magnetic flux density at the centre of the

air-core solenoid is given by - The magnetic flux passing through each turn of

the solenoid always maximum and is given by - Therefore the self-inductance of the solenoid is

given by

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,6th edition,SerwayJewett

, Q6, p.1025) ANS. 1.92?10?5 Wb 2. 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. the magnetic flux

through each turn, c. the inductance of the

solenoid. ANS. 1.88?10?4 T 3.33?10?8 Wb

3.75?10?4 H

3. 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?10?2 J m?1 4. 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

V. 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?10?4 Wb 6.48?10?2

J 5. 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

Learning Outcome

20.4 Mutual inductance (2 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, - Explain the working principle of transformer and

the effect of eddy current in transformer.

20.4 Mutual 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 I1 flows in coil 1, produced by the

battery in the external circuit. - The current I1 produces a magnetic field lines

inside it and this field lines also pass through

coil 2 as shown in Figure 7.20.

Figure 7.20

- If the current I1 changes with time, the magnetic

flux through coils 1 and 2 will change with time

simultaneously. - Due to the change of magnetic flux through coil

2, an emf is induced in coil 2. This is in

accordance to the Faradays 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 in

coil 1 since the magnetic flux through it changes.

20.4.2 Mutual inductance, M

- From the Figure 7.20, consider the coils 1 and 2

have N1 and N2 turns respectively. - If the current I1 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 - If vice versa, the induced emf in coil 1, ?1 is

given by - It is a scalar quantity and its unit is henry (H).

where

Mutual inductance

- 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 Faradays law for the coil 2, thus

magnetic flux linkage through coil 1

magnetic flux linkage through coil 2

and

20.4.3 Mutual inductance for two solenoids

- Consider a long solenoid with length l and cross

sectional area A is closely wound with N1 turns

of wire. A coil with N2 turns surrounds it at its

centre as shown in Figure 7.21. - When a current I1 flows in the primary coil (N1),

it produces a magnetic field B1,

N1 primary coil N2 secondary coil

Figure 7.21

- and then the magnetic flux ?1,
- If no magnetic flux leakage, thus
- If the current I1 changes, an emf is induced in

the secondary coils, therefore the mutual

inductance occurs and is given by

Mutual inductance, M

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 Transformer

- 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. - If NP gt NS the transformer is a step-down

transformer. - If NP lt NS the transformer is a step-up

transformer.

Figure 7.22

- The symbol of transformer in the electrical

circuit is shown in 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.

Figure 7.23

- 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 (I2R) heat

losses occur. - Overcome The transformer coils are made of

thick copper wire. - Hysteresis
- The magnetization of the core is repeatedly

reversed by the alternating magnetic field. The

resulting expenditure of energy in the core

appears as heat. - Overcome By using a magnetic material (such as

Mumetal) which has a low hysteresis loss. - Flux leakage
- The flux due to the primary may not all link the

secondary. Some of the flux loss in the air. - Overcome By designing one of the insulated

coils is wound directly on top of the other

rather than having two separate coils.

Learning Outcome

20.5 Energy stored in an inductor (½ hour)

- At the end of this chapter, students should be

able to - Derive and use formulae for energy stored in an

inductor,

23.4 Energy 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 in overcoming the back emf

in the circuit is given by

and

and analogous to

in capacitor

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 By applying the equation

of energy stored in the inductor, thus

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