Download Lecture on Motional EMF, Eddy Currents, and Self Inductance and more Slides Electrical Engineering in PDF only on Docsity!
LECTURE 17
• Motional EMF
• Eddy Currents
• Self Inductance
3/5/12 2
Reminder: How to Change Magnetic Flux in a
Coil
B changes: d Φ m dt = dB dt NA cos B ,
(^ A )
- A changes: d Φ m dt = NB dA dt cos B ,
(^ A )
B ,
(^ A ) changes:^
d Φ m dt = NBA d cos B ,
⎡ ⎣ ( A )⎤ ⎦
dt
- N changes (unlikely): d Φ m dt = dN dt BA cos B ,
(^ A )
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Example Problem: Solenoid within a Coil
120 turn coil of radius 2.4 cm and resistance 5.3 " solenoid with radius 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 the current is being reduced? 3/5/12 4
Example Problem: Solenoid within a Coil
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Motional EMF
Motional EMF is any emf induced by the motion of a conductor in a magnetic field. 3/5/12 6
Calculation
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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:
- Rate of work by applied force: 3/5/12 8
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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Reduce Eddy Currents
Metal strips with insulating glue Cut slots into the metal. 3/5/12 14
Inductors & Inductance
Symbol for inductor *An inductor can be used to produce a desired B field. *Inductance: L = N Φ M i
Units of L: 1 Henry = 1 H = 1
Tm^2
A
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Self Inductance
X X X X X X X X X X X X X X a b dI/dt An induced emf, L, appears in any coil in which the current is changing. Self-Induction: Changing current through a loop induces an opposing voltag in that same loop. 3/5/12 16
Self Inductance in a Coil
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Magnetic Energy in an Inductor
power delivered by battery power delivered to inductor power dissipated by resistor Upon closing switch, S, apply Kirchoff’s loop rule:
ε − IR − L
dI
dt
Multiply through by I: ε I = I 2 R + LI dI dt 3/5/12 18
Magnetic Energy in an Inductor
2 m f I m m 0 m m m LI 2
U
U dU LI dI U LI dI dt dI LI dt dU IfU energyinthe inductor, f =
∫ ∫ energy stored in an inductor 3/5/12 19
Magnetic Energy in a Solenoid
B = μ 0 nI ⇒ I =
B
μ 0 n L = μ 0 n^2 Al Um =
LI^2 =
μ 0 n^2 Al
B
μ 0 n
2
B^2
2 μ 0 Al magnetic energy density um =
B^2
2 μ 0 electric energy density ue =
ε 0 E^2 Both of these are general results.
