Chapter 2 - KFUPM 30.pdfTitle Chapter 2 Author ITC Created Date 1/5/2017 11:31:47 AM
Transcript of Chapter 2 - KFUPM 30.pdfTitle Chapter 2 Author ITC Created Date 1/5/2017 11:31:47 AM
Chapter 30Induction
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1. Introduction
We have seen that a current produces a magnetic field. The reverse is true: Amagnetic field can produce an electric field that can drive a current.
This link between a magnetic field and the electric field it produces (induces) iscalled Faradayβs law of induction.
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2. Two experiments
We start by discussing two simple experiments beforediscussing Faradayβs law of induction.
First Experiment
The figure shows a conducting loop connected to anammeter. A current suddenly appears in the circuit ifwe move a bar magnet toward the loop. The currentstops when the magnet stops moving. If we move themagnet away, again a sudden current appears, but inthe opposite direction.
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https://www.youtube.com/watch?v=3hj98E5K_c4
2. Two experiments
First Experiment
By experimenting this set up we can conclude thefollowing:
1. A current appears only if the loop and the magnetare in relative motion.
2. Faster motion produces greater current.
3. If moving the magnetβs north pole away produces aclockwise current, then moving the same north poletoward the loop causes a counterclockwise current.Moving the south pole toward or away also causescurrents in the reversed direction.
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2. Two experiments
First Experiment
The current produced in the loop is called an inducedcurrent. The work done per unit charge that producesthat current is called an induced emf. The process ofproducing the current and emf is called induction.
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2. Two experiments
Second Experiment
The figure shows two conducting loops close to eachother. If we close switch π, a sudden brief inducedcurrent appears in the left-hand loop. Another briefinduced current appears in the left-hand loop if weopen the switch, but in the opposite direction. Aninduced current appears only when the current in theright loop is changing.
The induced emf and current in the two experimentsappears when some physical quantity is changing. Letus find out what that quantity is.
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3. Faradayβs Law of Induction
Faraday concluded that an emf and a current can be induced in a loop by changingthe amount of magnetic field passing through the loop. He also concluded that theamount of magnetic field can be visualized in terms of the magnetic field linespassing through the loop.
Faradayβs law of induction can be stated in terms of the two experiments asfollows:
An emf is induced in the loop at the left in the two figures above when the numberof magnetic field lines that pass through the loop is changing.
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3. Faradayβs Law of Induction
Quantitative Treatment
We need a way to calculate the amount of magnetic field that passes through a
loop. In analogy with the electric flux Ξ¦πΈ = πΈ β π π΄, we define magnetic flux. Themagnetic flux through a loop of area π΄ is
Ξ¦π΅ = π΅ β π π΄.
As before, π π΄ is a vector of magnitude ππ΄ that is perpendicular to a differentialarea ππ΄.
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3. Faradayβs Law of Induction
Quantitative Treatment
If the magnetic field is perpendicular to the plane of the loop, then π΅ β π π΄= π΅ππ΄ cos 0 = π΅ππ΄. If π΅ is also uniform then we can take π΅ out of the integralsign, and ππ΄ is just the loopβs area. Thus,
Ξ¦π΅ = π΅π΄. π΅ β₯ π΄ of the loop, π΅ uniform
The SI unit for magnetic flux is the tesla-square meter, which has the special nameweber (Wb):
1 weber = 1Wb = 1 T β m2.
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3. Faradayβs Law of Induction
Quantitative Treatment
With the notion of magnetic flux, we can state Faradayβs law as:
The magnitude of the emf β° induced in a conducting loop is equal to the rate atwhich the magnetic fluxΞ¦π΅ through that loop changes with time.
You will see in the next section that the induced emf β° tends to oppose the fluxchange. Faradayβs law is therefore written as
β° = βπΞ¦π΅ππ‘
.
For an ideal π turns coil, the total emf induced in the loop is
β° = βππΞ¦π΅ππ‘
.
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3. Faradayβs Law of Induction
Quantitative Treatment
Here are the general means by which we can change the magnetic flux through acoil:
1. Changing the magnitude π΅ of the magnetic field within the coil.
2. Changing either the total area of the coil or the portion of that area that lieswithin magnetic field.
3. Changing the angle between the direction of the magnetic field π΅ and the planeof the coil.
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3. Faradayβs Law of Induction
β° = βπΞ¦π΅ππ‘
Ξ¦π΅ = π΅π΄
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π, π & π tie, π & π tie.
3. Faradayβs Law of Induction
Example 1: The long solenoid S shown (in crosssection) in the figure has 220 turns/cm andcarries a current π = 1.5 A; its diameter π· is3.2 cm . At its center we place a 130 -turnclosely packed coil C of diameter π = 2.1 cm.The current in the solenoid is reduced to zero ata steady rate in 25 ms. What is the magnitudeof the emf that is induced in coil C while thecurrent in the solenoid is changing?
The initial flux through solenoid C is
Ξ¦π΅π = π΅π΄C = π0ππSπ΄C = ππ0ππSππΆ2
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3. Faradayβs Law of Induction
Now we can write
πΞ¦π΅ππ‘
=βΞ¦π΅βπ‘
=Ξ¦π΅π βΞ¦π΅π
βπ‘
=0 β ππ0ππSππΆ
2
βπ‘= β
ππ0ππSππΆ2
βπ‘.
Substituting gives
πΞ¦π΅ππ‘
= βπ 4π Γ 10β7 T β
mA
1.5 A
25 ms
Γ 22000turn
m0.0105 m 2
= β5.76 Γ 10β4 V.14
3. Faradayβs Law of Induction
The magnitude of the induced emf is then
β° = ππΞ¦π΅ππ‘
= 130 5.76 Γ 10β4 V
= 75 mV.
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3. Faradayβs Law of Induction
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3. Faradayβs Law of Induction
Ξ¦π΅ = π΅π΄ = ππ΅π2.
πΞ¦π΅ππ‘
= 2ππ΅πππ
ππ‘= 2π 0.800 T 0.120 m β0.750
m
s= β0.452
Wb
s.
β° = βπΞ¦π΅ππ‘
= 0.452 V.
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4. Lenzβs Law
The direction of an induced current can be determined usingLenzβs law. It can be stated as follows:
An induced current has a direction such that the magnetic fielddue to the current opposes the change in the magnetic flux thatinduces the current.
To get a feel of Lenzβs law, we apply it in two different butequivalent ways to the situation shown in the figure.
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https://www.youtube.com/watch?v=3-1hiBBWgSA
4. Lenzβs Law
1. Opposition to Pole Movement:
The approach of the magnetβs north pole increases themagnetic flux through the loop and thereby induces a current inthe loop. The loop then acts as a magnetic dipole, directed fromsouth to north. The loopβs north pole must face toward theapproaching north pole of the magnet so as to repel it, tooppose the magnetic flux increase caused by the approachingmagnet. The curled-straight right-hand rule tells us that theinduced current in the loop must be counterclockwise.
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4. Lenzβs Law
2. Opposition to Flux Change:
As the north pole of the magnet nears the loop with its field π΅directed down, the flux through the loop increases. To opposethis increase in flux, the induced current π must set up its ownmagnetic field π΅ind , directed upward inside the loop. Thecurled-straight right-hand rule tells us the π must becounterclockwise.
Note that the flux of π΅ind always opposes the change in the fluxof π΅, but that does not always mean that π΅ind is opposite to π΅.
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4. Lenzβs Law
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4. Lenzβs Law
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π & π, π.
4. Lenzβs Law
Example 2: The figure shows a conducting loopconsisting of a half-circle of radius π = 0.20 mand three straight sections. The half-circle liesin a uniform magnetic field π΅ that is directedout of the page; the field magnitude is given byπ΅ = 4.0π‘2 + 2.0π‘ + 3.0, with π΅ in teslas and π‘in seconds. An ideal battery with emf β°bat= 2.0 V is connected to the loop. Theresistance of the loop is 2.0 Ξ©.
(a) What are the magnitude and direction ofthe emf β°ind induced around the loop by fieldπ΅ at π‘ = 10 s?
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4. Lenzβs Law
β°ind =πΞ¦π΅ππ‘
=π π΅π΄
ππ‘= π΄
ππ΅
ππ‘
=ππ2
28.0π‘ + 2.0 .
At π‘ = 10 s,
β°ind =π 0.20 m 2
28.0 10 s + 2.0 = 5.2 V.
By Lenzβs law, the direction of the emf isclockwise.
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4. Lenzβs Law
b) What is the current in the loop at π‘ = 10 s?
π =β°netπ =β°ind β β°bat
π =5.15 V β 2.0 V
2.0 Ξ©= 1.6 π΄.
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4. Lenzβs Law
Example 3: The figure shows a rectangular loop ofwire immersed in a nonuniform and varyingmagnetic field π΅ that is perpendicular to anddirected into the page. The fieldβs magnitude is givenby π΅ = 4π‘2π₯2, with π΅ in teslas, π‘ in seconds, and π₯in meters. (Note that the function depends on bothtime and position.) The loop has width π = 3.0 mand height π» = 2.0 m. What are the magnitude anddirection of the induced emf β° around the loop at π‘= 0.10 s?
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4. Lenzβs Law
The flux through the loop is
Ξ¦π΅ = π΅ β π π΄ = π΅ππ΄ = 0
π
4π‘2π₯2 π»ππ₯
= 4π»π‘2 0
π
π₯2ππ₯ =4
3π»π‘2π3.
Substituting for π» andπ gives,
Ξ¦π΅ =4
32.0 m 3.0 m 3π‘2 = 72π‘2.
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4. Lenzβs Law
The magnitude of the induced emf is
β° =πΞ¦π΅ππ‘
= 144 π‘.
At π‘ = 0.10 s,
β° = 144 0.10 s = 14 V.
The direction of the induced emf iscounterclockwise, by Lenzβs law.
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5. Induction and Energy Transfers
According to Lenzβs law, a magnetic force resists the movingmagnet away or toward the loop, requiring your applied forcethat moves the magnet to do positive work.
Thermal energy is then produced in the material of the loop dueto the materialβs electrical resistance to the induced current. Theenergy transferred to the closed loop + magnet system while youmove the magnet ends up in this thermal energy.
The faster the magnet is moved, the more rapidly the appliedforce does work and the greater the rate at which energy istransferred to thermal energy in the loop; the power transfer isgreater.
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5. Induction and Energy Transfers
The figure shows a situation involving an inducedcurrent. You are to pull this loop to the right at aconstant velocity π£. Let us calculate the rate at whichyou do mechanical work as you pull steadily on theloop.
The rate π at which you do work is given by
π = πΉπ£,
where πΉ is the magnitude of your force. Themagnitude of this force is equal to the magnitude ofthe magnetic force on the loop due to the emergenceof the induced current.
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5. Induction and Energy Transfers
We thus write
πΉ = πΉ1 = ππΏπ΅.
The induced current π is related to the induced emf by
π =β°
π .
The induced emf is given by
β° =πΞ¦π΅ππ‘
=π
ππ‘π΅πΏπ₯ = π΅πΏ
ππ₯
ππ‘= π΅πΏπ£.
Thus,
π =π΅πΏπ£
π .
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5. Induction and Energy Transfers
The force is then
πΉ =π΅2πΏ2π£
π .
The rate at which you do work is now
π = πΉπ£ =π΅2πΏ2π£2
π .
This rate must be equal to the rate at which thermalenergy appears in the loop:
π =π΅2πΏ2π£2
π 2π = π2π .
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5. Induction and Energy Transfers
π & π, π & π.
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β° = π΅πΏπ£