Pdt Unit15 Df024(student)

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

UNIT 15: ELECTROMAGNETIC INDUCTION

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15.1 Magnetic flux, induced emf and induced current


Consider some experiments were conducted by Michael Faraday that led to the discovery of the Faraday’s law of induction as shown in figures 15a, 15b, 15c, 15d and 15e.

v =0 No movement

Fig. 15a

v

S

N I

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Move towards the coil

I Fig. 15b

2

v =0 No movement

Fig. 15c

v N

S Move away from the coil

I

I Fig. 15d

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3

v S

N

Move towards the coil

I

I Fig. 15e




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From the experiments :  When the bar magnet is in stationary, the galvanometer shows no 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.  When the bar magnet is moved relatively away from the coil, the galvanometer is seen to deflect in the opposite direction (Fig.15d).  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.

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Conclusion : • When the magnetic flux through a coil changes (magnetic field lines been cut) thus the induced e.m.f. will exist across the coil. • The magnitude of the induced e.m.f. depends on the speed of the relative motion where when

v increase v decrease

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induced e.m.f. also increase induced e.m.f. also decrease

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15.2 Electromagnetic laws 15.2.1 Faraday’s law of induction
States – the magnitude of the induced e.m.f. is proportional to the rate of change of the magnetic flux.” flux. Mathematically, or




The negative sign indicates that the direction of induced e.m.f. always oppose the change of magnetic flux producing it (Lenz’s law).

where

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(15.2a)

dΦB : change of magnetic flux dt : change of time ε : induced e.m.f. 6




For a coil of N turns, eq. (15.2a) can be written as (15.2b)




Since

dΦB = Φ f − Φi

, then eq. (15.2b) can be written as where




From the definition of magnetic flux,

ΦB = BA cos θ


Φ f : final magnetic flux Φi : initial magnetic flux

then eq. (15.2a) also can be written as

Note :  If the coil is connected in series to a resistor of resistance R and the induced e.m.f ε exist in the coil as shown in figure 15f. Therefore the induced current I is given by

Fig. 15f SF027

I

R

I

dΦB ε =− dt

and

ε = IR 7

There are several ways an e.m.f can be induced in a loop. a) Change the magnetic field, B. A and θ constant b) Change the area of the loop, A. B and θ constant. c) Change the loop orientation angle, θ with respect to the field. B and A constant.

d ( BA cos φ ) dt dB a )ε = − A cos φ dt dA b )ε = − B cos φ dt d cos φ c )ε = − BA dt

ε =−

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To calculate the magnitude of induced e.m.f., THE NEGATIVE SIGN CAN BE IGNORED. If the coil has N turns, then each of turns will have a magnetic flux, ΦB of BAcos θ through it, therefore the magnetic flux linkage (refer to the combined amount of flux through all the turns) is given by

magnetic flux linkage = NΦB


Example 1 : A rectangular coil of sides 10 cm x 5.0 cm is placed between N and S poles with the plane of the coil parallel to the magnetic field as shown in figure below. N

R

I S

Q

S

I P

If the coil is turned by 90° about its rotation axis and the magnitude of magnetic flux density is 1.0 T, find the change in the magnetic flux through the coil. SF027

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Solution: A=(10x10-2)(5.0x10-2)=50x10-4 m2,B=1.0 T r Initially, From the figure, φ =90° thus the initial r A B magnetic flux through the coil is

Φi = BA cos φ

Finally,

r B

From the figure, φ =0° thus the final magnetic flux through the coil is

r A

Φ f = BA cos φ

Therefore the change in magnetic flux through the coil is

∆ΦB = Φ f − Φi

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Example 2 : The magnetic flux passing through a coil of 1000 turns is increased quickly but steadily at rate of 2.0 x 10-2 Wb s-1. Calculate the induced e.m.f. in the coil. Solution: N=1000 turns, dΦB −2 -1

dt

= 2.0 x10 Wb s

By applying the Faraday’s law equation for a coil of N turns , thus the induced e.m.f. is




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Example 3 : A circular shaped coil 3.0 cm in radius, containing 20 turns and have a resistance of 5.0 Ω is placed perpendicular to a magnetic field of flux density of 5.0 x 10-3 T. If the magnetic flux density is reduced steadily to zero in time of 2.0 ms, calculate the induced current flows in the coil. Solution: N=20 turns, r=3.0x10-2 m, R=5.0 Ω, Bi=5.0x10-3 T , Bf=0, dt=2.0x10-3 s The area of the circular shaped coil is

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initially,

r B

From the figure, φ =0° thus the change in magnetic flux through the coil is

r A

dΦB = Φ f − Φi dΦB = B f A cos φ − Bi A cos φ dΦB = − Bi A

By applying the Faraday’s law equation for a coil of N turns , thus

ε = −N




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dΦB dt

and

ε = IR

Example 4 : (exercise) A flat coil having an area of 8.0 cm2 and 50 turns lies perpendicular to a magnetic field of 0.20 T. If the flux density is steadily reduced to zero, taking 0.50 s, find a. the initial flux through the coil. b. the initial flux linkage. c. the induced e.m.f. (Lowe&Rounce,pg.206,no.1) Ans. : 1.6 x 10-4 Wb, 80 x 10-4 Wb, 16 mV 12

15.2.2 Lenz’s law
States “ an induced electric current always flows in such a direction that it opposes the change producing it.” it.
This law is essentially a form of the law of conservation of energy. energy
An illustration of Lenz’s law can be shown by using the experiments below. First experiment : (figure 15g)

N

Direction of induced current – Right hand grip rule.

I

North pole



In figure 15g the magnitude of the magnetic field at the solenoid increases as the bar magnet is moved towards it.



An e.m.f is induced in the solenoid and galvanometer indicates that a current is flowing.



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 lenz’s law.

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Fig. 15g

Second 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 figure15h. X X X X X X

X rX Q X Xr X X X F X vX X X B X X X X XIX X X X X X X X X X X X X X X PX X Fig. 15h

X X X X X X

X X X



When the conductor moves to the left thus the induced current needs to flow in such a way to oppose the change which has induced it based on Lenz’s Law. Hence galvanometer shows a deflection.



To determine the direction of the induced current (e.m.f.) flows in the conductor PQ, the Fleming’s right hand (Dynamo) rule is used as shown in figure 15i



Therefore the induced current flows from Q to P as shown in fig. 15h.



Since the current flows in the conductor PQ and is placed in the magnetic field then this conductor will experience magnetic force.

Its direction is in opposite direction of the motion. Note: Only for the Thumb – straight conductor. First finger – 

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Second finger –

Fig. 15i

Important

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Third experiment : Consider two solenoids P and Q arranged coaxially closed to each other as shown in figure 15j.

S

N

P

I Switch, S

N

S +

I

ε ind

Q

I ind

-

I ind

Fig. 15j 

 

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At the moment when the switch S is closed, 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. time According to Faraday’s law ,an induced current due to induced e.m.f. 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 (fig. 15j) and galvanometer shows a deflection. 15





At the moment when the switch S is opened, opened the current I starts to decrease in the solenoid P and magnetic flux through the solenoid Q decreases with time. time According to Faraday’s law ,an induced current due to induced e.m.f. 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 (fig. 15k) and galvanometer seen to deflect in the opposite direction of fig.15j.

ε ind S

N

P

I Switch, S

I

S

N

-

+ Q

I ind

I ind

Fig. 15k

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Example 5 : A single turn circular shaped coil has resistance of 10 ohm and area of its plane is 5.0 cm2. It moves towards the north pole of a bar magnet as shown in figure below.

If the average rate of change of magnetic flux density through the plane of the coil is 0.50 T s-1, determine the induced current in the coil and state the direction of the induced current observed by the observer shown in figure above. Solution: N=1 turn, R=10 Ω, A=5.0x10-4 m2, dB = 0.50 T s -1 dt By applying the Faraday’s law equation for a coil of N turns , thus

dΦB o where Φ B = BA cos 180 and ε = IR dt d (− BA) IR = − N dt ε = −N

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Based on the lenz’s law, hence the direction of induced current is clockwise as shown in figure below.




Example 6 : (exercise) A bar magnet is held above a loop of wire in a horizontal plane, as shown in figure below. The south end of the magnet is toward the loop of the wire. The magnet is dropped toward the loop. Find the direction of the current through the resistor a. while the magnet falling toward the loop and b. after the magnet has passed through the loop and moves away from it. (Serway&Jewett, pg.991, no.15)

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Factors affecting the magnitude of the induced emf/current a) Change the magnetic field, B. N, A, dt and θ constant b) Change the area of the loop, A. N, B, dt and θ constant c) Change the loop orientation angle, θ with respect to the field. N, A, dt and B constant d) The number of turns, N. B, A, dt and θ constant

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d( BAcos φ ) ε =− ;Φ = NBA cos φ dt dB a )ε = − NA cos φ dt dA b )ε = − NB cos φ dt d cos φ c )ε = − BA dt dN d )ε = − BAcosφ dt

5 factors

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