motional emf • eddy currents • self inductance
TRANSCRIPT
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3/25/201
ANNOUNCEMENT*Exam 2: Tuesday March 31, 2020, 1 PM - Midnight*Location: Online*Covers all readings, lectures, homework from Chapters
23 through 28.1.*The exam will be multiple choice (15 questions).
The equation sheet is posted on the course homepage. Click on the link on the left labeled “Equationsheet”
Exam 2 will now be given online through CHIP. The exam will be available on March 31st from 1 PM until Midnight. Once you click on the link, the exam must be completed in 2 hours. During this 2-hour period time, you cannot pause or stop the exam.
The exam will be open book but you cannot discuss the exam with anyone. The Purdue Honor Code applies.
Those with adaptive learner status as certified by the Office of the Dean of Students will be given appropriate time accommodations.
LECTURE 17
• Motional EMF• Eddy Currents• Self Inductance
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Reminder: How to Change Magnetic Flux in a Coil
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Example Problem: Solenoid within a Coil120 turn coil of radius 2.4 cm andresistance 5.3 Ω
solenoid withradius 1.6 cm and n = 220 turns/cm
Initial current in the solenoid is 1.5 A.Current is reduced to zero in 25 ms.What is the current in the coil while thecurrent is being reduced?
B = µonIdBdt
= µondIdt
2
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Example Problem: Ideal Solenoid within a Coil
dφdt
= µondIdt
n = 120 turnscm
= 1.2x104 turnsm
ε = −Ncoildφdt
= 109Volts
I = VR= 49mA
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Example Problem: Solenoid within a Coil
What is the direction of the current in the coil?
(A) Same direction as the current in the solenoid
(B) Opposite the direction of the current in the solenoid
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Motional EMF
Motional EMF is any emf induced by the motion of a conductor in a magnetic field.
v0
X X X X X X X X X
B
X X X X X X X X X
+ −F
F
Fb = qv0B
Charge Separation
Current flows counterclockwise….Resulting in a magnetic force F pointing to the left
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CalculationA thin conducting rod is pulled with velocity v along conducting rails that are connected by a resistor, R. A uniform magnetic field B is directed into the page. Surface S is increasing therefore flux is increasing. What will be the magnitude & direction of the induced current? dA = vldt
dAdt
= vl
dφdt
= Bvl
ε = −Bvl
I = εR= Bvl
R
3
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Energy Conservation
• Energy is dissipated in circuit at rate P’:
• The induced current gives rise to a net magnetic force F in the loop which opposes the motion:
• External Agent must exert equal but opposite force FR to move the loop with velocity v; therefore, agent does work at rate P:
• Rate of work by applied force:
P = F i v
F = ILB
P = F i v = ILBv = B2l2v2
R
P = I 2R = B2l2v2
R 3/25/20 10
Eddy Currents*Relative motion between a B field and a conductor induces a current in the conductor.
*The induced current give rises to a net magnetic force, FM, which opposes the motion.
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Eddy Currents
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Demonstration
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Demonstration: Eddy Currents
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Eddy Currents
B field into the page
v
v
Induced current anti-clockwise
Induced current clockwise
Flux increasing Flux
decreasing
FF
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 x
•During the last lecture we considered currents confined to a loop or a coil.•Electrical equipment contains other metal part which are often located in changing B fields. Currents in such parts are called Eddy currents
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Reduce Eddy Currents
Metal strips with insulating glue
Cut slots intothe metal.
DEMO: Lenz’s Law
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Inductors & Inductance
Symbol for inductor
*An inductor can be used to produce a desired B field.
*Inductance:
Units of L:
5
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Self Inductance
X XX X X
X X
X XX X X
X X
a b
dI/dt
An induced emf, εL, appears in any coil in which the current is changing.
Self-Induction: Changing current through a loop induces an opposing voltage in that same loop.
ε = − Δ(NφB )Δt
= −L ΔIΔt
U = 12LI 2
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Self Inductance in a Coil
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Magnetic Energy in an Inductor
powerdeliveredby battery
powerdelivered
to inductorpower
dissipatedby resistor
Upon closing switch, S, apply Kirchoff’s loop rule:
Multiply through by I:
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Magnetic Energy in an Inductor
energy stored inan inductor
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Magnetic Energy in a Solenoid
Both of these aregeneral results.