8.022 (E&M) – Lecture 15
Topics:
More on Electromagnetic Inductance
Mutual and self inductance
Practical applications
Last time
Electromagnetic inductance
Faraday’s (and Lentz’s) law:
Integral form:
e .m .f . = −
1 ∂ΦB
c ∂t
Differential form: ∇ × E = −
1 ∂B
c ∂t
Let’s elaborate a bit more on this important law…
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Cu pendulum in B field (H13)
A copper pendulum is oscillating
Pendulum #1
Application of Lentz’s law
Turn on the magnetic field for the
following 3 different situations:
Pendulum #1:
B crosses area with cuts
• No effect
B crosses area above cuts
• Stops slowly: Lentz’s law
Pendulum #2
Pendulum #2:
No cuts in Cu
• Stops abruptly: Lentz’s law
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Three ways of creating e.m.f.
Faraday’s law can be used to build generators:
e .m .f . = −
1 ∂
c ∂t
∫ B ida
S
3 ways of creating e.m.f.:
Vary the area: S=S(t)
Vary the angle between B and da
Vary magnitude of B: B=B(t)
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2
e .m .f . = −
1 ∂
c ∂t
∫ B ida
S
Changing the area
Sliding rod on rails:
--
B
v
L
R
I
++
x
vBL
As derived last week: e .m .f . =
c
Because of Lentz’s law, direction of current is counterclockwise to
oppose the change of flux of B
Demo H4:
Loop + light bulb moving in B created by electromagnet
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e .m .f . = −
1 ∂
c ∂t
∫ B ida
S
Changing angle between B and S
Constant B and loop rotating around its axis with angular
velocity ω
ω
B
θ
If S is the area of the loop:
∫ B ida
= BS cos θ = BS cos ωt
S
⇒ e .m .f . =
1 ∂
ω
(BS cos ωt ) = BS sin ωt
c ∂t
c
This is an easy way to build an AC power generator
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DC vs. AC current
DC current
Electrons flow all in the same direction
at the same rate
AC current
The flow of electron varies with time in
amplitude and direction:
I(t)
DC/AC generator
Uses DC to power electromagnet and induce AC on rotating loop
Why AC? Easier to step up and down for efficient transportation
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e .m .f . = −
1 ∂
c ∂t
∫ B ida
S
Changing magnitude of B
Suppose you have a way to vary over time the magnitude
of B: B=B(t)
Flux of B:
ΦB = ∫ B (t )ida = B (t )S cos θ
S
Generated e.m.f.: e .m .f . =
1 ∂
1 ∂B (t )
ΦB = S
c ∂t
c
∂t
How to created B=B(t)?
Loop of wire: B ∝ I
If I=I(t)
B=B(t)
AC in a solenoid will do the trick!
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Induced e.m.f.
Consider a loop of wire with radius r inside a long solenoid
Solenoid:
N=# of loops, l=total length
Isol = Isol(t)
Isol
n=N/l
What is the e.m.f. generated in the loop?
Find B inside solenoid:
Bsol =
4πnI sol (t )
Q: can you derive
this in 60 sec?
c
E.m.f. generated in loop:
e .m .f . =
1 ∂
1
∂B (t ) 4π 2nr 2 ∂I sol (t )
ΦB = (π r 2 )
=
c ∂t
c
∂t
c2
∂t
The e.m.f. will depend by the geometry of the setup and
on the rate of change of the I over time
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Induced e.m.f. on solenoid itself
Isol
What if the “loop” is the solenoid itself?
Will any e.m.f. be created?
Remember Faraday’s law: e .m .f . = −
B inside solenoid: Bsol =
1 ∂
c ∂t
4πnI sol (t )
c
Flux of B through each loop: Φ1B loop = BS1 loop =
Flux of B through N loops: ΦBTot = N Φ1B loop =
Induced e.m.f. on solenoid:
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∫ B ida
S
e .m .f . =
4πnI sol (t )
c
4π 2R 2N 2
cl
πR 2
I sol (t )
4π 2R 2N 2 ∂I sol (t )
∂t
c 2l
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Back e.m.f.
Magnitude of induced e.m.f. on solenoid:
4π 2R 2N 2 ∂I sol (t )
e .m .f . =
∂t
c 2l
How about the direction? And the effect?
Isol
Use Lentz’s law to predict direction of induced current
If Isol increases
B increases
flux increases
opposite direction as Isol
Iloop will fight change
If Isol decreases
B decreases
flux decreases
same direction as Isol
Iloop will fight change
Conclusion:
The inductance always opposes the change in the current
The e.m.f. created is called back e.m.f. as it acts back on the circuit
trying to oppose changes
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Example of back e.m.f. (H17)
Fe
R
125 V
Close switch: wire jumps
I flows (30 A)
Open switch: big spark due by back emf
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Self Inductance L
Self-induced e.m.f. in the solenoid:
e .m .f . =
4 π 2R 2N
2
c l
2
∂ I sol (t )
⇒
∂t
e .m .f . = L
∂ I sol (t )
∂t
Let’s examine this in detail:
e.m.f. depends on change over time of current: dI/dt
A bunch of constants depending on geometry called self inductance L
Lsol =
For a solenoid:
Units:
cgs: [L ] =
SI:
[L ] =
4π 2R 2N 2
c 2l
[e .m .f .]
esu / cm
sec 2
=
=
[current ] /[time ] (esu / s ) / s
cm
[e .m .f .]
V
=
≡ Hen ry (H )
[current ] / [tim e ] A / s
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Energy stored in inductors
Consider an inductor L in which we start flowing a current I
As soon as the current starts flowing, a back-emf tries to fight this
current back
Power needed to fight the back-emf:
P = I × e .m .f . = IL
∂I
∂t
Calculate work to increase the current from 0
W =
t
∫t
=0
Pdt =
t
∫t
=0
LI
I
∂I
1
dt = L ∫ IdI = LI
=0
I
∂t
2
Energy stored in the inductor: W =
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I when t: 0
1
LI
2
t
2
2
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How is energy stored in inductors?
We created a magnetic field where there was none: work necessary
to create the magnetic field is the energy stored in the B itself
Same as energy stored in electric field of a capacitor
Not surprising: special relativity!
Energy density of magnetic field (solenoid example)
Energy stored in solenoid: UL=LI2/2
Self inductance of a solenoid: L=4π2R2N2 /lc2
B created by solenoid: B=4πN /lc
UL =
1
LI
2
2
=
1 4 π 2N
2 c 2l
Energy density of B:
2
I
2
=
uB =
(
)
B2
8π
Similar to energy density of the electric field:
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2
1
B2
⎛ 4π N ⎞
π R 2l ⎜
I ⎟ = Volume
8π
8π
⎝ cl
⎠
uE =
8.022 – Lecture 15
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2
8π
15
How do we calculate L in psets?
Just some examples…
Strategy 1:
L is the proportionality constant between induced emf and variation
over time of current:
e .m .f . = L
Strategy 2:
∂ I (t )
∂t
Exploit the fact that energy stored in the magnetic field is the energy
stored in the inductor:
Energy stored in B =
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B2
1
∫V 8π dV = 2 L I
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2
Mutual inductance
1
Back to the loop inside the solenoid
Label solenoid with 1 and loop with 2
e.m.f. induced on loop (ε2) depends on dI1/dt and constant M21
ε 2 = M 21
∂I 1
∂t
where M21 is the coefficient of mutual inductance
For this particular configuration we already calculated that M 21 =
4π 2r 2N
c 2l
Now do the opposite: run a current I2(t) in the loop and calculate
e.m.f. induced on solenoid (ε1):
∂I 2
ε 1 = M 12
∂t
How to calculate M12???
No need to calculate it! Reciprocity theorem: M12=M21
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Reciprocity theorem
Consider 2 loops of wire:
Loop 2
Loop 1
Current I runs through loop 1. What is ΦB through loop 2 due to 1?
Φ B21 = ∫ B 1 ida 2
S2
Now rewrite this result in terms of vector potential and use Stokes:
Φ 21 =
∫B
S2
Since A
1
=
Same fluxes
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I
c
∫C
1
1
ida 2 =
∫ ( ∇ × A )ida
1
S2
dl 1
we obtain
r
Φ 21 =
2
I
c
=
∫A
C2
∫∫
C 2 C1
1
idl 2
dl 1
idl 2 = Φ 12
r
if currents are the same: M12=M21
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Transformers
Devices to step up (or down) AC currents
N1
Practical application of mutual inductance
Simplest implementation:
Primary solenoid (black): N1 turns
Secondary solenoid (red): N2 turns
N2
I(t) in the primary will induce a varying ΦB through itself:
N d ΦB
ε1 = 1
c dt
where ΦB=magnetic flux through single turn
N 2 d ΦB
c dt
Depending on number of turns we can
• increase voltage (N2>N1)
• reduce the voltage (N2<N1)
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Flux is the same in second solenoid
Comparing: ε 2 = ε 1
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N2
N1
induced e.m.f. is: ε 2 =
Demos on mutual inductance
Single turn around primary coil (H10)
Emf: 208 V AC
Primary coil: N1= 220 turns
Secondary coil: N2= 1 turn
Effect: V goes down, but current goes up and melts the nail!
Explanation: Power = VI is conserved between the 2 coils
Variable turns around primary coil (H9)
Same primary; show how current in secondary goes as we add loops
High turn secondary (H11)
Emf: 208 V AC
Primary coil: N1= 220 turns
Secondary coil: N2= 10,000 turn
Effect: Small currents, but very large V will cause big sparks!
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Summary and outlook
Today:
Self inductance
Energy stored in inductor
Mutual inductance
And its applications: transformers
Next time:
Inductors in circuits
Quiz II-preparation supplies available here!
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